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\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}-3+\sqrt{x}\)
\(=\sqrt{x}+2-3+\sqrt{x}=2\sqrt{x}-1\)
Ủa đáp số là\(\sqrt{2x-2}\) với \(\sqrt{2}\) mà bạn thử A2 đi
a) Ta có: \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}-1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}+1}-1\right):\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}+1\right)+\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}-1\right)-2x+1}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}:\left(\dfrac{2x-1+\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)-\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}\right)\)
\(=\dfrac{x\sqrt{2}+\sqrt{x}+\sqrt{2x}+1+2x-\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{2x-1}:\dfrac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-\left(2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}\right)}{2x-1}\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{x}+2}{-2-2\sqrt{x}}\)
Với \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
\(P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\\ =\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1-\sqrt{a}}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}\right)^2\\ =\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(\dfrac{1}{1+\sqrt{a}}\right)^2\\ =\left[\left(1+\sqrt{a}\right)+\sqrt{a}\left(\sqrt{a}+1\right)\right]\left(\dfrac{1}{1+\sqrt{a}}\right)^2\\ =\dfrac{\left(1+\sqrt{a}\right)\left(1+\sqrt{a}\right).1^2}{\left(1+\sqrt{a}\right)^2}=1\)
a) Ta có: \(M=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\dfrac{\sqrt{x}}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)+\sqrt{x}\left(\sqrt{2x}+1\right)^2-2x+1}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}\right):\left(\dfrac{2x-1+\sqrt{x}\left(\sqrt{2x}-1\right)-\sqrt{x}\left(\sqrt{2x}+1\right)^2}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}\right)\)
\(=\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+\sqrt{x}\left(2x+2\sqrt{2x}+1\right)-2x+1}{2x-1+x\sqrt{2}-\sqrt{x}-\sqrt{x}\left(2x+2\sqrt{2x}+1\right)}\)
\(=\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-2x+2x\sqrt{x}+2\sqrt{2x}+\sqrt{x}}{2x-1+x\sqrt{2}-\sqrt{x}-2x\sqrt{x}-2\sqrt{2x}-\sqrt{x}}\)
\(=\dfrac{x\sqrt{2}+3\sqrt{2x}-2x+2x\sqrt{x}}{x\sqrt{2}-2\sqrt{2x}+2x-2\sqrt{x}-2x\sqrt{x}}\)
Lời giải:
ĐKXĐ: $x>0; x\neq 1$
\(A=\frac{\sqrt{x}-(1-\sqrt{x})}{\sqrt{x}(1-\sqrt{x})}\left[\frac{(2\sqrt{x}-1)(\sqrt{x}+1)}{(1-\sqrt{x})(1+\sqrt{x})}+\frac{\sqrt{x}(2\sqrt{x}-1)(\sqrt{x}+1)}{(1+\sqrt{x})(x-\sqrt{x}+1)}\right]\)
\(=\frac{2\sqrt{x}-1}{\sqrt{x}(1-\sqrt{x})}\left[\frac{2\sqrt{x}-1}{1-\sqrt{x}}+\frac{\sqrt{x}(2\sqrt{x}-1)}{x-\sqrt{x}+1}\right]\)
Nghe biểu thức cứ sai sai ấy bạn. Có phải giữa 2 ngoặc lớn là dấu chia không?