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ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
Ta có \(P=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{\left(1+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(a+2\sqrt{a}+1\right).\left(a-2\sqrt{a}+1\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}.\frac{1}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2}=\frac{\sqrt{a}}{1+a}\)
tham khao nha
\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{ab}-a}\right):\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\)
\(A=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right):\left(\frac{\sqrt{b}+\sqrt{a}}{\sqrt{ab}}\right)\)
\(A=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}\right).\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{a-2\sqrt{ab}+b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
vay \(A=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
ĐK : tự ghi nha
\(\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{ab}-a}\right):\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\)
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
Lời giải:
a)
\(=\frac{(\sqrt{x}+1)\sqrt{x}(\sqrt{x}-\sqrt{y}))\sqrt{x}+\sqrt{y})}{(x-y)x(\sqrt{x}+1)}=\frac{(\sqrt{x}+1)\sqrt{x}(x-y)}{(x-y)x\sqrt{x}+1)}=\frac{1}{\sqrt{x}}\)
b)
\(=\frac{(2-\sqrt{x}-\sqrt{x}-3)(2-\sqrt{x}+\sqrt{x}+3)}{1+2\sqrt{x}}=\frac{(-1-2\sqrt{x}).5}{2\sqrt{x}+1}=\frac{-5(2\sqrt{x}+1)}{2\sqrt{x}+1}=-5\)
\(a,\frac{\left(\sqrt{x}+1\right)\cdot\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\sqrt{x}\left(x+1\right)}\)\(=\frac{\left(\sqrt{x}+1\right)\sqrt{x}\left(x-y\right)}{\left(x-y\right)\sqrt{x} \left(x+1\right)}\)\(=\frac{\sqrt{x}+1}{x+1}\)
\(b,\frac{\left(2-\sqrt{x}\right)^2-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{4+x-4\sqrt{x}-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{1+x-5\sqrt{x}}{1+2\sqrt{x}}\)
a) \(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{6}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(A=\frac{\left(-\sqrt{a}+1\right)^2}{\left(-a+1\right)^2}.\left(\sqrt{a}+\frac{-a\sqrt{a}+1}{-\sqrt{a}+1}\right)\)
\(A=\frac{\left(1-\sqrt{a}\right)^2\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(A=\frac{\frac{-a\sqrt{a}+\sqrt{a}.\left(-\sqrt{a}+1\right)+1}{-\sqrt{a}+1}.\left(-\sqrt{a}+1\right)^2}{\left(1-a\right)^2}\)
\(A=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(A=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)
\(A=1\)
