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a) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b) \(\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}+1}\cdot\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}+1}\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
a)\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x+1}\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b)\(\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}\cdot\frac{\sqrt{\left(\sqrt{y}-1\right)^{2^2}}}{\sqrt{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
1) Ta có: \(\left\{{}\begin{matrix}2x-y=3\\x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=3\\x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-x=-1\end{matrix}\right.\)
Vậy: (x,y)=(1;-1)
2) Ta có: \(A=\dfrac{x+20}{x-4}+\dfrac{2}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}-2}\)
\(=\dfrac{x+20+2\left(\sqrt{x}-2\right)-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+20+2\sqrt{x}-4-6\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
a) \(\frac{2}{x^2-y^2}\cdot\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{\sqrt{3}\left(x+y\right)}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)
b) \(\frac{2}{2a-1}\cdot\sqrt{5a^2\left(1-4a+4a^2\right)}=\frac{2}{2a-1}\cdot\sqrt{5a^2\left(1-2a\right)^2}\)
\(=\frac{2}{2a-1}\cdot\sqrt{5}a\left(1-2a\right)=-2\sqrt{5}a\)
\(a,\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}=\left|\sqrt{x}-\sqrt{y}\right|\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=y-x\)
\(b,\frac{3-\sqrt{x}}{x-9}=\frac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{1}{\sqrt{x}+3}\)
\(c,\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(d,6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-3+x=3-x\)
\(a,\)\(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(=|\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)|\)
\(=|\sqrt{x}^2-\sqrt{y}^2|\)
\(=|x-y|\)
Vì \(x\le y\)\(\Rightarrow x-y\ge0\)
\(\Rightarrow|x-y|=x-y\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
tớ ra kết quả là 2+\(\frac{5\sqrt{xy}}{x-\sqrt{xy}+y}\) mà thấy số xấu quá :(
a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)
\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)
\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)
\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)
a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)
do \(x\ge0;y\ge0\)
\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)
Ta có : \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2\sqrt{3}.\left|x+y\right|}{\sqrt{2}.\left(x-y\right)\left(x+y\right)}\)
Vì \(x\ge y\ge0\) nên ta có : \(\left|x+y\right|=x+y\)
\(\Rightarrow\frac{2\sqrt{3}\left|x+y\right|}{\sqrt{2}\left(x-y\right)\left(x+y\right)}=\frac{\sqrt{2}.\sqrt{6}\left(x+y\right)}{\sqrt{2}\left(x-y\right)\left(x+y\right)}=\frac{\sqrt{6}}{x-y}\)