Bài 1 rút gọn
a)\(3\sqrt{5a}-\sqrt{20a}+\sqrt{45a}\)với a
≥0
b)\(\sqrt{160a^2}+\dfrac{1}{2}\sqrt{40a^2}-3\sqrt{90a^2}\)
c)\(\sqrt{x^2-2x+1}-\sqrt{x^2-4x+4}\)
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12 tháng 6 2017
a) \(2x-\sqrt{4x^2+4x+1}=2x-\sqrt{\left(2x+1\right)^2}=2x-\left|2x+1\right|\)
Vì \(x< -\frac{1}{2}\)nên \(\left|2x+1\right|=-\left(2x+1\right)\)
\(\Rightarrow2x+2x+1=4x+1\)
b) \(3x+2-\sqrt{9x^2-12x+4}=3x+2-\sqrt{\left(3x-2\right)^2}=3x+2-\left|3x-2\right|\)
Khi \(x\ge\frac{2}{3}\)thì \(\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-3x+2=4\)
Khi \(x< \frac{2}{3}\) thì \(\left|3x-2\right|=2-3x\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-\left(2-3x\right)=6x\)
c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
Đặt \(\sqrt{a}=x\) ta được : \(3x-4x+7x=6x\)\(=6\sqrt{a}\)( Do \(a\ge0\))
d) \(\sqrt{160a}+2\sqrt{40a}-3\sqrt{90a}=4\sqrt{10a}+4\sqrt{10a}-9\sqrt{10a}\)\(=-\sqrt{10}\)
TK NKA !!!
11 tháng 7 2021
Bài 1:
a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)
b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)
Bài 2:
a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)
b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
19 tháng 7 2021
Bài 1:
a.
\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)
b.
\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)
AH
Akai Haruma
Giáo viên
19 tháng 7 2021
Bài 2.
a.
\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)
b.
\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)
22 tháng 5 2022
a: \(=4a-4\sqrt{10a}-9\sqrt{10a}=4a-13\sqrt{10a}\)
b: \(=\sqrt{x}\left(4-\sqrt{2}\right)\cdot\sqrt{x}\left(1-\sqrt{2}\right)\)
\(=x\cdot\left(4-4\sqrt{2}-\sqrt{2}+2\right)\)
\(=\left(6-5\sqrt{2}\right)x\)
c: \(=\dfrac{2}{2x-1}\cdot x\sqrt{5}\cdot\left(2x-1\right)=2x\sqrt{5}\)
AT
30 tháng 6 2021
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(x>0,x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2x}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}:\dfrac{7-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{7-\sqrt{x}}=\dfrac{x}{\sqrt{x}-7}\)
\(B=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\left(x>0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}+1\)
\(=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}+1=-\dfrac{\sqrt{x}+1}{\sqrt{x}}+1\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}}=-\dfrac{1}{\sqrt{x}}\)
AH
Akai Haruma
Giáo viên
30 tháng 6 2021
A. ĐKXĐ: $x>0; x\neq 1; x\neq 4$
\(A=\left[\frac{x-\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}-2)}-\frac{x}{\sqrt{x}(\sqrt{x}-2)}\right].\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\left[\frac{x-\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}-2)}-\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-2)}\right].\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{-2(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}-1}=\frac{-2}{\sqrt{x}+1}\)
AH
Akai Haruma
Giáo viên
30 tháng 6 2021
B.
ĐKXĐ: $x\geq 0, x\neq \frac{1}{4}$
\(B=\frac{2\sqrt{x}-1+2\sqrt{x}+1}{(2\sqrt{x}+1)(2\sqrt{x}-1)}.(1-4x)=\frac{4\sqrt{x}}{4x-1}(1-4x)=-4\sqrt{x}\)
a: Ta có: \(3\sqrt{5a}-\sqrt{20a}+\sqrt{45a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+3\sqrt{5a}\)
\(=4\sqrt{5a}\)
b: Ta có: \(\sqrt{160a^2}+\dfrac{1}{2}\sqrt{40a^2}-3\sqrt{90a^2}\)
\(=4a\sqrt{10}+\dfrac{1}{2}\cdot2a\sqrt{10}-3\cdot3a\sqrt{10}\)
\(=-4a\sqrt{10}\)
c: Ta có: \(\sqrt{x^2-2x+1}-\sqrt{x^2-4x+4}\)
\(=\left|x-1\right|-\left|x-2\right|\)