A=\(\left(\dfrac{\sqrt{21}-\sqrt{7}}{\sqrt{3}-1}\right)+\left(\dfrac{\sqrt{18}-\sqrt{3}}{\sqrt{6}-1}\right):\dfrac{2}{\sqrt{7}-\sqrt{3}}\)
B=\(\left(\sqrt{9}+4\sqrt{5}\right)+\sqrt{14-6\sqrt{5}}\)
C=\(3\sqrt{12}-\sqrt{27}-\sqrt{4-2\sqrt{3}}\)
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19 tháng 5 2022
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
2 tháng 6 2021
`a)sqrt{28a^4}`
`=sqrt{7.4.a^4}`
`=2sqrt7a^2`
`b)A=((sqrt{21}-sqrt7)/(sqrt3-1)+(sqrt{10}-sqrt5)/(sqrt2-1)):1/(sqrt7-sqrt5)`
`=((sqrt7(sqrt3-1))/(sqrt3-1)+(sqrt5(sqrt2-1))/(sqrt2-1)).(sqrt7-sqrt5)`
`=(sqrt7+sqrt5)(sqrt7-sqrt5)`
`=7-5=2`
`c)` $\begin{cases}\dfrac{3}{2x}-y=6\\\dfrac{1}{x}+2y=-4\end{cases}$
`<=>` $\begin{cases}\dfrac{3}{x}-2y=12\\\dfrac{1}{x}+2y=-4\end{cases}$
`<=>` $\begin{cases}\dfrac{4}{x}=8\\2y+\dfrac{1}{x}=-4\end{cases}$
`<=>` $\begin{cases}x=\dfrac12\\2y=-4-2=-6\end{cases}$
`<=>` $\begin{cases}x=\dfrac12\\y=-3\end{cases}$
Vậy HPT có nghiệm `(x,y)=(1/2,-3)`.
26 tháng 9 2021
bài 1:
a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)
\(=-33\sqrt{2}\)
b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
26 tháng 9 2021
Bài 2:
a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)
\(\Leftrightarrow\left|2x+3\right|=8\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}=8\)
hay x=4
c: Ta có: \(\sqrt{9x-9}+1=13\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow x-1=16\)
hay x=17
31 tháng 10 2021
\(a,=\sqrt{17}-4-\sqrt{17}-2=-6\\ b,=7\left(\sqrt{3}+\sqrt{2}\right)-7\sqrt{3}-6\sqrt{2}\\ =7\sqrt{3}+7\sqrt{2}-7\sqrt{3}-6\sqrt{2}=\sqrt{2}\\ c,=\dfrac{6\sqrt{5}+12-6\sqrt{5}+12}{3}+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}\\ =8+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}=\dfrac{56+14\sqrt{2}-4\sqrt{7}}{7}\\ d,=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{-5\sqrt{2}+32\sqrt{2}}{4}\cdot8=54\sqrt{2}\)
1 tháng 6 2023
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
24 tháng 8 2021
a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)
\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)
b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)
c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=2-\sqrt{2}-\left(2+\sqrt{2}\right)=-2\sqrt{2}\)
a, \(A=\left(\dfrac{\sqrt{7}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)+\left(\dfrac{\sqrt{3}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}\right):\dfrac{2\left(\sqrt{7}+\sqrt{3}\right)}{4}=\sqrt{7}+\sqrt{3}:\dfrac{\sqrt{7}+\sqrt{3}}{4}=\sqrt{7}+\dfrac{4\sqrt{3}\left(\sqrt{7}-\sqrt{3}\right)}{4}=\sqrt{7}+\sqrt{21}-3\)
b, \(B=3+4\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}=3+4\sqrt{5}+3-\sqrt{5}=6+3\sqrt{5}\)
c, \(D=6\sqrt{3}-3\sqrt{3}-\left(\sqrt{3}-1\right)=3\sqrt{3}-\sqrt{3}+1=2\sqrt{3}+1\)
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