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S
6 tháng 9 2018
\(a.\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=2x^2+2y^2=2\left(x^2+y^2\right)\)
\(b.2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2=4x^2\)
\(c.\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z+y-z\right)^2=x^2\)
KB
6 tháng 9 2018
a ) \(\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=2x^2+2y^2=2\left(x^2+y^2\right)\)
b ) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2\)
\(=\left(2x\right)^2=4x^2\)
c ) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z\right)^2+\left(y-z\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z+y-z\right)^2\)
\(=x^2\)
TN
6 tháng 6 2017
\(a,\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)\(b,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3x^2\)\(c,\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2=\left(x-2y\right)^2\)
17 tháng 6 2017
a) \(\left(x+y\right)^2+\left(x-y\right)^2\)
=\(\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
=\(x^2+2xy+y^2+x^2-2xy+y^2\)
\(2x^2+2y^2=2\left(x^2+y^2\right)\)
b) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
=\(\left[\left(x-y\right)+\left(x+y\right)\right]^2\)
= \(\left(x-y+x+y\right)^2\)
\(=2x^2\)
c) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(z-y\right)+\left(z-y\right)^2\)
\(=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2\)
= \(\left(x-y+z-z+y\right)^2=x^2\)
K
15 tháng 11 2018
\(\frac{x^2-3x+2}{x^3-1}=\frac{x^2-2x-x+2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\frac{x.\left(x-2\right)-\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{\left(x-1\right).\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\frac{x-2}{x^2+x+1}\)
2 tháng 1 2019
\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)
\(x^2+y^2+z^2=-2\left(xy+yz+zx\right)\)
\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\frac{-2\left(xy+yz+zx\right)}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{2\left[-2\left(xy+yz+zx\right)\right]-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{-4\left(xy+yz+zx\right)-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{-6\left(xy+yz+zx\right)}\)
\(=\frac{1}{3}\)
2 tháng 1 2019
Ta có: \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^2=\left(-z\right)^2\)
\(x^2+2xy+y^2=z^2\)
\(x^2+y^2-z^2=-2xy\)
\(\frac{2x^2y+2xy^2}{x^2+y^2-z^2}\)
\(=\frac{2xy\left(x+y\right)}{-2xy}\)
\(=\frac{-2xyz}{-2xy}\)
\(=z\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2+x^2-y^2=3x^2+y^2\\ B=\left(x-y-x+y-z\right)^2=\left(-z\right)^2=z^2\)
câu b mình ko hiểu lắm bạn ơi. Bạn có thể làm cụ thể hơn giúp mình được không? cảm ơn bạn