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\(\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}\)
\(ĐK:x\ne\pm y\)
\(=\dfrac{2\left|x+y\right|}{2\left(x+y\right)\left(x-y\right)}=\dfrac{\sqrt{3}\left|x+y\right|}{\left(x+y\right)\left(x-y\right)}\)
Nếu x > -y thì x + y > 0 , ta có :\(\dfrac{\sqrt{3}}{x-y}\)
Nếu x < -y thì x + y < 0 , ta có :\(\dfrac{-\sqrt{3}}{x-y}\)
\(a.A=\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x+y\right)\sqrt{3}}{2}=\dfrac{\sqrt{3}}{x-y}\) ( x # y )
\(b.\dfrac{1}{2x-1}.\sqrt{5a^4\left(1-4x+4a^2\right)}=\dfrac{1}{2a-1}.\left(2a-1\right)a^2\sqrt{5}=a^2\sqrt{5}\) ( a # \(\dfrac{1}{2}\) )
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 \(A=\frac{\sqrt{3}}{y-x}\)
1)
\(\dfrac{1}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{1+\sqrt{5}}{4}-\dfrac{4\sqrt{5}-4}{4}\) \(=\dfrac{5-3\sqrt{5}}{4}\)
2)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\9x+3y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7\\y=4-3x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(1;1\right)\)
1: Rút gọn
Ta có: \(\dfrac{1}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}\)
\(=\dfrac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-4\sqrt{5}+4}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\dfrac{5-3\sqrt{5}}{4}\)
2)
Ta có: \(\left\{{}\begin{matrix}2x+3y=5\\3x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+9y=15\\6x+2y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7y=7\\3x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\3x=4-y=4-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
ĐKXĐ : \(x^4+\left(\sqrt{3}-\sqrt{2}\right).x^2-\sqrt{6}\ne0\)
\(\Leftrightarrow x\ne\sqrt[4]{2}\)
\(P=\dfrac{x^2-\sqrt{2}}{x^4+\left(\sqrt{3}-\sqrt{2}\right).x^2-\sqrt{6}}\)
\(=\dfrac{x^2-\sqrt{2}}{\left(x^4-\sqrt{2}x^2\right)+\sqrt{3}\left(x^2-\sqrt{2}\right)}\)
\(=\dfrac{x^2-\sqrt{2}}{\left(x^2+\sqrt{3}\right)\left(x^2-\sqrt{2}\right)}=\dfrac{1}{x^2+\sqrt{3}}\)
a: \(M=1:\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{3x}{2\left(x-4\right)}+\dfrac{1}{2\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{4-2\sqrt{x}}{1}\)
\(=1:\left(\dfrac{2\sqrt{x}-4-3x+\sqrt{x}+2}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{-2\left(\sqrt{x}-2\right)}{1}\)
\(=\dfrac{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\cdot\left(-2\right)\cdot\left(\sqrt{x}-2\right)}{-3x+3\sqrt{x}-2}\)
\(=\dfrac{-4\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}{-3x+3\sqrt{x}-2}\)
b: M=20
=>\(-4\left(x-4\right)\left(\sqrt{x}-2\right)=-60x+60\sqrt{x}-40\)
=>\(x\sqrt{x}-2x-4\sqrt{x}+8=-15x+15\sqrt{x}-10\)
=>\(x\sqrt{x}+13x-19\sqrt{x}+18=0\)
=>\(x\in\varnothing\)
1) Ta có: \(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=-7\\2x+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x=5-y=5-\left(-1\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
2) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+3\sqrt{x}+2+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x-2\sqrt{x}+2x-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{1}\)
\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}-2}\)
\(=3\sqrt{x}\)
Hỏi nhiều thế.
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\sqrt{3}}{2}\cdot\left|x+y\right|\)
\(=\pm\dfrac{\sqrt{3}}{\left(x-y\right)}\)