Tính a) $\sqrt{\left(4-2\sqrt{3}\right).\left(13 4\sqrt{3}\right)}$ b) $\left(\sqrt{3}-2\right).\left(\sqrt{6} \sqrt{2}\right).\sqrt{\sqrt{3} 2}$ c)\(\left(3 \sqrt{5}\right).\left(\sqrt{10}-\sq...

B

bbbbbb

3 tháng 8 2020

Tính a) $\sqrt{\left(4-2\sqrt{3}\right).\left(13+4\sqrt{3}\right)}$ b) $\left(\sqrt{3}-2\right).\left(\sqrt{6}+\sqrt{2}\right).\sqrt{\sqrt{3}+2}$ c)$\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}$ d) $\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}$ Giúp mk với, mk hứa sẽ tick cho.Cảm ơn...

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#Toán lớp 9

1

NV

Nguyễn Việt Lâm

Giáo viên

4 tháng 8 2020

a/ $=\sqrt{\left(\sqrt{3}-1\right)^2\left(2\sqrt{3}+1\right)^2}$

$=\left(\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)=5-\sqrt{3}$

b/ $=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}$

$=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}+1\right)^2}$

$=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}+1\right)=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)^2$

$=\left(\sqrt{3}-2\right)\left(4+2\sqrt{3}\right)=2\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)$

$=2\left(3-4\right)=-2$

c/ $=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}$

$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}$

$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)$

$=2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=2.\left(9-5\right)=8$

d/ $=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}$

$=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}$

$=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)$

$=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2$

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QE

Quynh Existn't

26 tháng 6 2021

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#Toán lớp 9

2

NN

Nguyễn Ngọc Lộc

26 tháng 6 2021

$A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|$

$=54+8-32=30$

$B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}$

$=5-2\sqrt{2}$

$C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}$

$=2-2\sqrt{3}$

$D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}$

$=2\sqrt{6}$

$E=\sqrt{15^2}-\sqrt{5^2}=15-5=10$

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Y

Yeutoanhoc

26 tháng 6 2021

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

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NT

Nguyễn Trà Giang

24 tháng 8 2020

Rút gọn : a) $\left(\sqrt{6}+\sqrt{2}\right).\left(\sqrt{3}-2\right)\left(\sqrt{2+\sqrt{3}}\right)$ b) $\sqrt{2}.\left(\sqrt{2-\sqrt{3}}\right).\left(\sqrt{3}+1\right)$ c) $\left(\sqrt{10}-\sqrt{6}\right).\left(\sqrt{4-\sqrt{15}}\right)$ d)$\left(\sqrt{3}-\sqrt{12}\right).\left(\sqrt{5+2\sqrt{6}}\right)$ e) $\sqrt{2-\sqrt{3}}.\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)$ f)...

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#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

24 tháng 8 2020

a) Ta có: $\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)$

$=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}$

$=\sqrt{4+2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)$

$=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)$

$=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)$

$=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)$

$=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)$(Vì $\sqrt{3}>1>0$)

$=\left(4+2\sqrt{3}\right)\cdot\left(\sqrt{3}-2\right)$

$=2\cdot\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)$

$=2\cdot\left(3-4\right)$

$=-2$

b) Ta có: $\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)$

$=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)$

$=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)$

$=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)$

$=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}+1\right)$

$=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)$(Vì $\sqrt{3}>1$)

$=3-1=2$

c) Ta có: $\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)$

$=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)$

$=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)$

$=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}-\sqrt{3}\right)$

$=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)$

$=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)$

$=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)$(Vì $\sqrt{5}>\sqrt{3}$)

$=8-2\sqrt{15}$

d) Ta có: $\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)$

$=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}$

$=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}$

$=-\sqrt{3}\cdot\left|\sqrt{3}+\sqrt{2}\right|$

$=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)$(Vì $\sqrt{3}>\sqrt{2}>0$)

$=-3-\sqrt{6}$

e) Ta có: $\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)$

$=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)$

$=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)$

$=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)$

$=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)$

$=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)$

$=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)$(Vì $\sqrt{3}>1$)

$=\frac{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{2}$

$=\frac{16-12}{2}=\frac{4}{2}=2$

f) Ta có: $\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}$(Vì $2>\sqrt{3}>0$)

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}$(Vì $5>\sqrt{3}$)

$=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}$

$=\sqrt{4+\sqrt{25}}$

$=\sqrt{4+5}=\sqrt{9}=3$

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2

ＴＲ ᗩ ＮＧ ²ᵏ⁶

17 tháng 6 2021

Dạng 3.Chứng minh đẳng thức Bài 1: CMa)$\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2$b)$\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8$Bài 2...

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#Toán lớp 9

1

D

๖ۣۜDũ๖ۣۜN๖ۣۜG

17 tháng 6 2021

Bài 1

a) Đặt VT = A

<=> $2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}$

<=> $2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}$

<=> $2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2$

<=> 2A = $\left(5-3\right)^2=4$

<=> A = 2

b) Đặt VT = B

<=> $2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}$

<=> $2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}$

<=> $2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16$

<=> B = 8

Bài 2

Đặt VT = A

<=> A² = $\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}$

<=> A² = $\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1$

<=> $A=\sqrt{\sqrt{5}+1}$

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LT

Lê Thu Hiền

22 tháng 7 2021

giải pt :

a,$\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3$

b, $\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4$

c, $\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}$

#Toán lớp 11

0

MT

Mai Thị Thúy

22 tháng 7 2021

giải pt:

a,$\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}$

b,$\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4$

c, $\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3$

#Toán lớp 11

0

AQ

Anh Quynh

4 tháng 8 2021

Tính:$A=\sqrt{27}-2\sqrt{48}+3\sqrt{75}$$B=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}-3\right)^2}$$C=\sqrt{\left(2\sqrt{3}+1\right)^2}+\sqrt{\left(2\sqrt{3}-5\right)^2}$$D=\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}$$E=\dfrac{4}{\sqrt{5}-2}-\dfrac{32}{\sqrt{5}+1}$$M=\dfrac{10}{3\sqrt{2}-4}+\dfrac{28}{3\sqrt{2}+2}$ please help...

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#Toán lớp 9

0

TN

Tam Nguyen

12 tháng 8 2017

Tính (rút gọn) a) $\left(\sqrt{3+\sqrt{5}}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)$ b) $\left(4+\sqrt{15}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-}\sqrt{15}$ c) $\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3+2}}$ d)...

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#Toán lớp 9

4

BL

Bình Lê

12 tháng 11 2017

$f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}$

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BL

Bình Lê

12 tháng 11 2017

$a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8$

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AQ

Anh Quynh

30 tháng 7 2021

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