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\(B=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(xy+\frac{1}{xy}\right)^2\)
\(-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)
\(\Rightarrow B=x^2+2+\frac{1}{x^2}+y^2+2+\frac{1}{y^2}+x^2y^2+2+\frac{1}{x^2y^2}-x^2y^2\)
\(-2-x^2-y^2-\frac{1}{y^2}-\frac{1}{x^2}-\frac{1}{x^2y^2}\)
\(\Rightarrow B=x^2y^2-x^2y^2+x^2-x^2+1.\frac{1}{x^2}+1.\frac{1}{x^2y^2}-1.\frac{1}{x^2}-1\)
\(.\frac{1}{x^2y^2}+1.\frac{1}{y^2}-1.\frac{1}{y^2}+y^2-y^2+2+2+2-2\)
\(\Rightarrow B=4\)
a: \(A=\dfrac{2}{xy}:\left(\dfrac{y-x}{xy}\right)^2-\left(\dfrac{x^2+y^2}{\left(x-y\right)^2}\right)\)
\(=\dfrac{2}{xy}\cdot\dfrac{\left(xy\right)^2}{\left(x-y\right)^2}-\dfrac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\dfrac{2xy-x^2-y^2}{\left(x-y\right)^2}=-1\)
2:
\(P=\dfrac{\left(5x+3\right)^2}{3x-2}\cdot\dfrac{\left(3x-2\right)\left(3x+2\right)}{5x+3}=\left(5x+3\right)\left(3x+2\right)\)
\(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)=ab\)
<=> \(ab\left(x^2+y^2+2xy\right)-2xy.ab+xy\left(a^2+b^2\right)=ab\)
<=> \(ab\left(x+y\right)^2+xy\left(a^2+b^2-2ab\right)=ab\)
<=> \(ab+xy\left(a-b\right)^2=ab\)
<=> \(xy\left(a-b\right)^2=0\)
<=> a - b = 0
<=> a = b
Bài này có cho x, y >0 hay không? Nếu không có thì sai đề nhé.
\(\left\{{}\begin{matrix}x^2-yz=a\\y^2-xz=b\\z^2-xy=c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^3-xyz=ax\\y^3-xyz=by\\z^3-xyz=cz\end{matrix}\right.\) \(\Rightarrow ax+by+cz=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)⋮\left(x+y+z\right)\)
\(x-y=1\Rightarrow x^2-2xy+y^2=1\Rightarrow x^2+xy+y^2=19\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=1.19=19\)
\(2,a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0ma:\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow a=b=c\)
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=0\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2+4abc\left(a+b+c\right)=4a^2b^2+4c^2a^2+4b^2c^2\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=\left(a^2+b^2+c^2\right)^2\left(dpcm\right)\)
Đề sai rồi em, đề đúng phải là:
\(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)=ab\)
Vế phải em thiếu a