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a) điều kiện xác định : \(x>0;x\ne1\)
ta có : \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(\Leftrightarrow A=\left(\dfrac{x}{2\sqrt{x}}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\)
\(\Leftrightarrow A=\left(\dfrac{x-1}{2\sqrt{x}}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow A=\left(\dfrac{x-1}{2\sqrt{x}}\right)\left(\dfrac{-4x}{x-1}\right)=-2\sqrt{x}\)
b) để \(A>-6\Leftrightarrow-2\sqrt{x}>-6\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0< x< 9\) và \(x\ne1\)
vậy ....
Đk: x >0 ; x khác 1
sau khi rút gọn ra -2\(\sqrt{x}\)
b, 9>x>0
a: ĐKXĐ: x>0; x<>1
\(A=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)-\sqrt{x}\left(x+2\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1-x-2\sqrt{x}-1\right)}{2\sqrt{x}}=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
b: Để A>-6 thì -2 căn x>-6
=>2 căn x<6
=>0<x<9
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)
b: Vì x+căn x+1>0
nên A>0
Nguyễn Huy Tú và phương An chắc h o onl đâu .
h bn nên tag DƯƠNG PHAN KHÁNH DƯƠNG ; Nhã Doanh ; Nguyễn Thanh Hằng ...
a. H=\(\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+1}\dfrac{\sqrt{x^3}-x}{\sqrt{x}-1}\left(đkxđ:x\ge1\right)\)
H=\(-2\sqrt{x-1}+x\)
b. Với x=\(\dfrac{53}{9-2\sqrt{7}}:\)
H=\(-2\sqrt{\dfrac{53}{9-2\sqrt{7}}-1}+\dfrac{53}{9-2\sqrt{7}}\)
H\(=7\)
c. \(-2\sqrt{x-1}+x=16\)
\(\sqrt{x-1}=\dfrac{x-16}{2}\)
\(4x-4=x^2-32x+256\)
\(x^2-36x+260=0\)
x=26
d. Để H>1 thì x>3
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\cdot\left(\dfrac{1}{1+\sqrt{x}}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
b) Để P>0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}>0\)
mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}\left(\sqrt{x}-1\right)>0\)
mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
hay x>1
Kết hợp ĐKXĐ,ta được: x>1
Vậy: Để P>0 thì x>1