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1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
a, \(VT=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b=VP\) đpcm
b,\(VT=1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}-\dfrac{a^2-a}{a-1}=1-\sqrt{a}+\sqrt{a}-a=1-a=VP\) đpcm
\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}\cdot\sqrt{a}\right)\cdot\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{1^3-\left(\sqrt{a}\right)^3}{1-\sqrt{a}}\cdot\sqrt{a}\right)\cdot\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}\cdot\sqrt{a}\cdot\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(1+\sqrt{a}+a\right)\cdot\sqrt{a}\cdot\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\sqrt{a}+a+a\sqrt{a}\cdot\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2\left(1+\sqrt{a}\right)^2}\)
\(=\sqrt{a}+a+a\sqrt{a}\cdot\dfrac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=\dfrac{\sqrt{a}+a+a\sqrt{a}}{1+2\sqrt{a}+a}\)
Với \(a\ge0,a\ne1\) ta có:
\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\\ =\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(đk:a>0,x\ne1\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}.\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{a-1}\)
\(=\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{2\sqrt{a}}\)
\(=\dfrac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)
b) \(A=-2\sqrt{a}>-6\)
\(\Leftrightarrow\sqrt{a}< 3\Leftrightarrow0\le a< 9\) và \(a\ne1\)
c) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)
\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)
\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
=1-a
a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)
\(=1-x\sqrt{x}-x\sqrt{x}\)
\(=1-2x\sqrt{x}\)
b: \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\cdot\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\left(\dfrac{\left(1-\sqrt{a}\right)\cdot\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\)
\(=\left(\dfrac{1}{\sqrt{a}+1}\right)^2\cdot\left(a+\sqrt{a}+1+\sqrt{a}\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
a) \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{\left(2+\sqrt{a}-\sqrt{a}-1\right)\left(2+\sqrt{a}+\sqrt{a}+1\right)}{2\sqrt{a}+3}\)
\(=\dfrac{1.\left(2\sqrt{a}+3\right)}{2\sqrt{a}+3}=1\)
b) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}=\left(a+2\sqrt{a}+1\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)
a, \(VT=\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)
\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1=VP\)
Vậy ta có đpcm
b, \(VT=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=\dfrac{\left(1+\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2}=1=VP\)
Vậy ta có đpcm
Ta có:
\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{a+\sqrt{a}+\sqrt{a}+1}{\sqrt{a}+1}\right)\left(\dfrac{\sqrt{a}-1-a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{a+2\sqrt{a}+1}{\sqrt{a}+1}.\dfrac{-\left(a-2\sqrt{a}+1\right)}{\sqrt{a}-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\dfrac{-\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\)
\(=-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\)
\(=-\left(a-1\right)\)
\(=1-a\)
\(\rightarrowđpcm\)
\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) \(=\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right].\left[1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)