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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(A=\left(\frac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)
\(=\left(\frac{4\sqrt{xy}+\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\) (Quy đồng biểu thức đầu và đổi dấu số hạng cuối)
\(=\left(\frac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1.\)
Vậy giá trị biểu thức \(A=1.\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐK : tự ghi nha
b)
\(Q=\left(\frac{2\sqrt{xy}}{x-y}-\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}-2\sqrt{y}}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{4\sqrt{xy}}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{4\sqrt{xy}-\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{4\sqrt{xy}-\left(x+y+2\sqrt{xy}\right)}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{4\sqrt{xy}-x-y-2\sqrt{xy}}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=\left(\frac{2\sqrt{xy}-x-y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=-\left(\frac{x-2\sqrt{xy}+y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=-\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=-\left(\frac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(Q=-\frac{1}{2\left(\sqrt{x}+\sqrt{y}\right)}.2\sqrt{x}\)
\(Q=-\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
P /s : Các bạn tham khảo nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
\(=\sqrt{\left(\frac{x}{y}\right)^2+\left(\frac{y}{x}\right)^2-2.\frac{x}{y}.\frac{y}{x}}-\sqrt{\left(\frac{x}{y}\right)^2+\left(\frac{y}{x}\right)^2+2.\frac{x}{y}.\frac{y}{x}}\)
\(=\sqrt{\left(\frac{x}{y}-\frac{y}{x}\right)^2}-\sqrt{\left(\frac{x}{y}+\frac{y}{x}\right)^2}\)
\(=\left|\frac{x}{y}-\frac{y}{x}\right|-\left|\frac{x}{y}+\frac{y}{x}\right|\)
Yêu tao ư kiếp sau nhé ọe