Rút gọn biểu thức \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)^2\times\dfrac{x^2-1}{2}-\sqrt{1-x^2}\)
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Những câu hỏi liên quan
5 tháng 5 2021
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
\(A=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\) (ĐK: \(x>0;x\ne2;x\ne1\))
\(A=\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(A=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{-3}\)
\(A=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
T
5 tháng 9 2023
\(A=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\left(ĐKXĐ:x>0;x\ne1;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)\(=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\dfrac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{-3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
#Urushi☕
10 tháng 11 2021
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
14 tháng 9 2021
\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)
14 tháng 9 2021
a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) \(\dfrac{P}{A}\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)
Vậy \(S=\varnothing\)
TA
21 tháng 7 2021
ĐK: `x>0`
`A=((\sqrtx)/(\sqrtx+2) - 4/(x+2\sqrtx)):(1+1/(\sqrtx))`
`=((\sqrtx)/(\sqrtx+2)-4/(\sqrtx(\sqrtx+2))):((\sqrtx+1)/(\sqrtx))`
`=(x -4)/(\sqrtx(\sqrtx+2)) . (\sqrtx)/(\sqrtx+1)`
`=((\sqrtx+2)(\sqrtx-2))/(\sqrtx+2) . 1/(\sqrtx+1)`
`=(\sqrt-2)/(\sqrtx+1)`
21 tháng 7 2021
Ta có:\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{x-4}{x\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2}{x}.\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
20 tháng 7 2021
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
ĐKXĐ : \(0\le x\le1\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)^2.\dfrac{x^2-1}{2}-\sqrt{1-x^2}\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)^2.\dfrac{\left(x-1\right)\left(x+1\right)}{2}-\sqrt{\left(1-x\right)\left(1+x\right)}\)
\(=\dfrac{4x}{\left(x-1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2}-\sqrt{\left(1-x\right)\left(1+x\right)}\)
\(=\dfrac{2x}{x-1}-\sqrt{1-x}.\sqrt{1+x}\)
\(=\dfrac{2x-\sqrt{1-x^2}.\left(x-1\right)}{x-1}\)