Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{6}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(A=\frac{\left(-\sqrt{a}+1\right)^2}{\left(-a+1\right)^2}.\left(\sqrt{a}+\frac{-a\sqrt{a}+1}{-\sqrt{a}+1}\right)\)
\(A=\frac{\left(1-\sqrt{a}\right)^2\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(A=\frac{\frac{-a\sqrt{a}+\sqrt{a}.\left(-\sqrt{a}+1\right)+1}{-\sqrt{a}+1}.\left(-\sqrt{a}+1\right)^2}{\left(1-a\right)^2}\)
\(A=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(A=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)
\(A=1\)
\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}-2\right)=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
Tương tự \(y^2-1=\frac{1}{4}\left(b-\frac{1}{b}\right)^2\)
\(P=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}\)
\(=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab+\frac{a}{b}+\frac{b}{a}-\frac{1}{ab}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab-\frac{a}{b}-\frac{b}{a}+\frac{1}{ab}}=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)
1: \(=\dfrac{1}{a+b}\cdot a^2\cdot\left|a-b\right|=\dfrac{a^2\left|a-b\right|}{a+b}\)
2: \(=\sqrt{9}\cdot\sqrt{a^2}\cdot\sqrt{\left(b-2\right)^2}=9\cdot\left|a\right|\cdot\left|b-2\right|\)
3: \(=\sqrt{13a\cdot\dfrac{52}{a}}=\sqrt{52\cdot13}=2\sqrt{13}\cdot\sqrt{13}=26\)
4: \(=4x-2\sqrt{2}-a\)(vì a>1>0)
1) \(\left(a-b\right)\cdot\sqrt{\frac{ab}{\left(a-b\right)^2}}=\left(a-b\right)\cdot\frac{\sqrt{ab}}{a-b}=\sqrt{ab}\)
2) \(\frac{x-y}{y}\cdot\sqrt{\frac{y^4}{x^2-2xy+y^2}}=\frac{x-y}{y}\cdot\frac{\sqrt{y^4}}{\sqrt{\left(x-y\right)^2}}=\frac{x-y}{y}\cdot\frac{y^2}{x-y}=y\)
điều kiện : \(x>0;x\ne1\)
ta có : \(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)
\(\Leftrightarrow A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)=\left(\dfrac{a-1}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)\)\(\Leftrightarrow A=\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}\)
ĐK: \(\hept{\begin{cases}1-a\ge0\\a\left(a-1\right)\ge0\\\frac{a-1}{a}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}a\le1\\a\le0\vee a\ge1\\a< 0\vee a\ge1\end{cases}}\Leftrightarrow a< 0\)
Khi đó \(A=\sqrt{1-a}+\sqrt{a\left(a-1\right)}-\sqrt{\frac{a^2\left(a-1\right)}{a}}\)
\(=\sqrt{1-a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}\)
\(=\sqrt{1-a}\)