Rút gọn biểu thức :
A = \(\frac{\sqrt{x-2.\sqrt{2.x-4}}}{\sqrt{2}}\) với x \(\ge\)2
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AH
Akai Haruma
Giáo viên
11 tháng 4 2023
Lời giải:
$A=\frac{10\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+4)}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}-1)(\sqrt{x}+4)}$
$=\frac{10\sqrt{x}-(2\sqrt{x}-3)(\sqrt{x}-1)-(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}+4)(\sqrt{x}-1)}$
$=\frac{-3x+10\sqrt{x}-7}{(\sqrt{x}+4)(\sqrt{x}-1)}$
$=\frac{-(\sqrt{x}-1)(3\sqrt{x}-7)}{(\sqrt{x}+4)(\sqrt{x}-1)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}$
8 tháng 4 2021
a, Với \(x\ge0,x\ne4\)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
b, Ta có \(x=6+4\sqrt{2}=2^2+4\sqrt{2}+\left(\sqrt{2}\right)^2=\left(2+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2+\sqrt{2}\right|=2+\sqrt{2}\)do \(2+\sqrt{2}>0\)
\(\Rightarrow A=\frac{2+\sqrt{2}-4}{2+\sqrt{2}-2}=\frac{-2+\sqrt{2}}{\sqrt{2}}=\frac{-2\sqrt{2}+2}{2}=\frac{-2\left(\sqrt{2}-1\right)}{2}=1-\sqrt{2}\)
30 tháng 6 2021
1, A = \(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
2 , A = \(1-\sqrt{2}\)
14 tháng 10 2021
\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)
20 tháng 10 2023
a: \(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\)
\(=\dfrac{2\sqrt{x}-4}{x-4}=\dfrac{2}{\sqrt{x}+2}\)
b: A=1/2
=>\(\sqrt{x}+2=4\)
=>\(\sqrt{x}=2\)
=>x=4(loại)
MN
10 tháng 7 2020
Sửa đề :
a) \(A=\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-5}{x-\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{x-\sqrt{x}+4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-4-x+\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}\)
b) \(A=4\)
\(\Leftrightarrow\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}=4\)
\(\Leftrightarrow x+3\sqrt{x}+4=4\sqrt{x}+4\)
\(\Leftrightarrow x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(A=4\Leftrightarrow x\in\left\{0;1\right\}\)
Y
7 tháng 8 2023
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
Ta có : \(A=\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}=\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(2x-4\right)-4\sqrt{2x-4}+4}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}=\frac{\left|\sqrt{2x-4}-2\right|}{2}\)