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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.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
Ta có VT =\(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
=\(\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\) =\(\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\dfrac{a-b}{a-b}=1=VP\)
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
ta có : \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(=\dfrac{-4\sqrt{a}}{a-4}\dfrac{a-4}{\sqrt{a}}=-4\)
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
\(6\sqrt{a}+\dfrac{2}{3}\sqrt{\dfrac{a}{4}}-a\sqrt[]{\dfrac{9}{a}}+\sqrt[]{7}\)
\(=6\sqrt[]{a}+\dfrac{\sqrt[]{a}}{3}-3\sqrt[]{a}+\sqrt[]{7}=\dfrac{19}{3}\sqrt[]{a}+\sqrt[]{7}\) ( a > 0 )
a. \(A=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
\(A=\dfrac{\left(\sqrt{15}-\sqrt{12}\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(A=\dfrac{5\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{3}}{5-4}-\dfrac{2+\sqrt{3}}{4-3}\)
\(A=\sqrt{3}-2-\sqrt{3}=-2\)
b.
\(B=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(B=\left[\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right].\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(B=\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right).\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(B=\dfrac{-8\sqrt{a}}{a-4}.\dfrac{a-4}{\sqrt{a}}\)
\(B=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
\(a.A=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}=\sqrt{3}-\dfrac{1}{2-\sqrt{3}}=\dfrac{2\sqrt{3}-4}{2-\sqrt{3}}\)
\(b.\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}.\dfrac{a-4}{\sqrt{a}}=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\left(a>0;a\ne4\right)\)
a) P = \(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
P = \(\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}+\dfrac{4\sqrt{a}-4}{4-a}\)
⇒ P = \(\sqrt{a^2}+2\sqrt{a}+3\sqrt{a}+6-\sqrt{a^2}-2\sqrt{a}-\sqrt{a}+2+4\sqrt{a}-4\)
P = \(4+6\sqrt{a}\)
b) Ta thay a = 9 vào P = \(4+6\sqrt{a}\)
P = \(4+6\sqrt{9}\) = \(4+6\sqrt{9}\) = \(4+6.3\) = \(22\)
điều kiện xác định : \(x\ge0;x\ne4\)
a) ta có : \(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
\(\Leftrightarrow P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}-\dfrac{4\sqrt{a}-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(\Leftrightarrow P=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(\Leftrightarrow P=\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(\Leftrightarrow P=\dfrac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{4}{\sqrt{a}-2}\)
b) thế \(a=9\) vào \(P\) ta có : \(P=\dfrac{4}{\sqrt{9}-2}=\dfrac{4}{3-2}=4\)
a) \(A=\left(\dfrac{\sqrt{a}+2}{2-\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}+2}-\dfrac{4a+2\sqrt{a}-4}{4-a}\right):\left(\dfrac{-2}{2-\sqrt{a}}+\dfrac{2+\sqrt{a}}{2\sqrt{a}-a}\right)=\left[\dfrac{\left(\sqrt{a}+2\right)^2}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}\right]:\left[\dfrac{-2\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}+\dfrac{2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right]=\left[\dfrac{a+4\sqrt{a}+4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{2\sqrt{a}-a}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}\right]:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}=\dfrac{a+4\sqrt{a}+4+2\sqrt{a}-a-4a-2\sqrt{a}+4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\dfrac{2-\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}=\dfrac{-4a+4\sqrt{a}+8}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\dfrac{1}{\sqrt{a}}=\dfrac{4\left(2-\sqrt{a}\right)\left(\sqrt{a}+1\right).\sqrt{a}}{\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)}=\dfrac{4a+4\sqrt{a}}{\sqrt{a}+2}\)
Ta có A=\(\sqrt{a}+2\Leftrightarrow\dfrac{4a+4\sqrt{a}}{\sqrt{a}+2}=\sqrt{a}+2\Leftrightarrow4a+4\sqrt{a}=\left(\sqrt{a}+2\right)^2\Leftrightarrow4a+4\sqrt{a}=a+4\sqrt{a}+4\Leftrightarrow3a=4\Leftrightarrow a=\dfrac{4}{3}\left(tm\right)\)Vậy a=\(\dfrac{4}{3}\) thì A=\(\sqrt{a}+2\)
\(A=\left(\dfrac{\sqrt{a}+2}{2-\sqrt{a}}+\dfrac{\sqrt{a}}{2+\sqrt{a}}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\right):\left(\dfrac{-2}{2-\sqrt{a}}+\dfrac{2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)\(=\left(\dfrac{\left(2+\sqrt{a}\right)^2+\sqrt{a}\left(2-\sqrt{a}\right)-4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\right)\)\(:\left(\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)
\(=\dfrac{4+4\sqrt{a}+a+2\sqrt{a}-a-4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\) . \(\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
\(=\dfrac{-4a+8\sqrt{a}}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\) .\(\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
=\(\dfrac{4\sqrt{a}\left(2-\sqrt{a}\right)}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}.\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
=\(\dfrac{4a}{2+\sqrt{a}}\)
b, Để A=\(\sqrt{a}+2\)
<=> \(\dfrac{4a}{2+\sqrt{a}}\) =\(\sqrt{a}+2\)
<=> 4a=\(\left(\sqrt{a}+2\right)^2\)
<=> \(a+4\sqrt{a}+4-4a=0\)
<=> \(-3a+4\sqrt{a}+4=0\)
<=>\(-3a+6\sqrt{a}-2\sqrt{a}+4=0\)
<=> \(-3\sqrt{a}\left(\sqrt{a}-2\right)-2\left(\sqrt{a}-2\right)=0\)
<=> \(\left(\sqrt{a}-2\right)\left(-3\sqrt{a}-2\right)=0\)
<=>\(\left[{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=\dfrac{-2}{3}\left(vl\right)\end{matrix}\right.\)
<=> a=4
\(a.P=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{2-\sqrt{a}}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}=\sqrt{a}+2+\sqrt{a}+2=2\sqrt{a}+4\) \(b.P=a+1\)
⇔ \(2\sqrt{a}+4=a+1\)
⇔ \(a-2\sqrt{a}-3=0\)
⇔ \(a+\sqrt{a}-3\sqrt{a}-3=0\)
⇔ \(\sqrt{a}\left(\sqrt{a}+1\right)-3\left(\sqrt{a}+1\right)=0\)
⇔ \(a=9\left(TM\right)\)
KL.............