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Điều kiện \(x\ne y\)
\(A=\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}.\left|x+y\right|}{2}=\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right)\left(x+y\right)}\)
Nếu \(x+y>0\) thì \(A=\frac{\sqrt{3}}{x-y}\)
Nếu \(x+y< 0\) thì \(A=\frac{\sqrt{3}}{y-x}\)
\(\frac{\sqrt{3x^2+6xy+3y^2}}{x^2-y^2}\)
<=>\(\frac{\sqrt{3.\left(x+y\right)^2}}{\left(x-y\right).\left(x+y\right)}\)
<=>\(\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right).\left(x+y\right)}.\)
<=>\(\frac{\sqrt{3}}{x-y}\)
\(\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}\)
\(ĐK:x\ne\pm y\)
\(=\dfrac{2\left|x+y\right|}{2\left(x+y\right)\left(x-y\right)}=\dfrac{\sqrt{3}\left|x+y\right|}{\left(x+y\right)\left(x-y\right)}\)
Nếu x > -y thì x + y > 0 , ta có :\(\dfrac{\sqrt{3}}{x-y}\)
Nếu x < -y thì x + y < 0 , ta có :\(\dfrac{-\sqrt{3}}{x-y}\)
mk nghĩ bạn chép sai đề hình như đề bài phải là \(A=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\)
ta xét \(A^3=\left(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\right)^3\)
<=> \(A^3=x^3-3x+3A\cdot\sqrt[3]{\frac{4}{4}}\)
<=> \(A^3=x^3-3x+3A\)
<=> \(A^3-3A-x^3+3x=0\)
<=>\(\left(A^3-x^3\right)-3A+3x=0\)
<=> \(\left(A-x\right)\left(A^2+Ax+x^2\right)-3\left(A-x\right)=0\)
<=> \(\left(A-x\right)\left(A^2+Ax+x^2-3\right)=0\)
<=> \(\orbr{\begin{cases}A=x\\A^2+Ax+x^2-3=0\end{cases}}\)(vô lí )
vậy \(A=x\)
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
