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điều kiện : \(x>0;x\ne1\)
ta có : \(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)
\(\Leftrightarrow A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)=\left(\dfrac{a-1}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)\)\(\Leftrightarrow A=\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}\)
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a: \(A=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(x+1\right)\left(y+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
b: \(x=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
\(A=\dfrac{2\sqrt{\sqrt{2}-1}}{\sqrt{2}-1+1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
Câu 1:
a: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để Q>0 thì \(\sqrt{a}-2>0\)
=>a>4
a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{1-\sqrt{x}}=\dfrac{3\sqrt{x}}{1-\sqrt{x}}\)
b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)
\(\Leftrightarrow x+2\sqrt{x}=0\)
hay x=0
1)
DKCĐ: a>0,\(a\ne1\)
\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)
Lời giải:
ĐK: \(a\geq 0; a\neq 1\)
Ta có:
\(P=\frac{1}{2(1+\sqrt{a})}+\frac{1}{2(1-\sqrt{a})}-\frac{a^2+2}{1-a^2}\)
\(=\frac{(1-\sqrt{a})+(1+\sqrt{a})}{2(1+\sqrt{a})(1-\sqrt{a})}-\frac{a^2+2}{1-a^2}\)
\(=\frac{1}{1-a}-\frac{a^2+2}{1-a^2}\)
\(=\frac{1+a}{(1-a)(1+a)}-\frac{a^2+2}{(1-a)(1+a)}=\frac{1+a-(a^2+2)}{(1-a)(1+a)}\)
\(=\frac{a^2-a+1}{a^2-1}\)
b)
\(P=\frac{7}{8}\Leftrightarrow \frac{a^2-a+1}{a^2-1}=\frac{7}{8}\)
\(\Rightarrow 8(a^2-a+1)=7(a^2-1)\)
\(\Leftrightarrow a^2-8a+15=0\Rightarrow \left[\begin{matrix} a=5\\ a=3\end{matrix}\right.\) (đều thỏa mãn)
A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
Ta có:
A = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
= \(\dfrac{-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-1\)
Kết luận: ...
ĐK của nó còn là: x ≥ 0 nữa dung doan nhé, mình viết thiếu...
\(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)=\left[\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\dfrac{\left(1+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{1+\sqrt{a}}-\sqrt{a}\right]=\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2=\left[\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\right]^2=\left(a-1\right)^2\)
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