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\(\frac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)
\(=\frac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}\)
\(=\frac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}\)
\(=\frac{x-y+z}{x-y-z}\)
A=\(\frac{2xy-x^2+z^2-y^2}{x^2+z^2-y^2+2xz}\)=\(\frac{z^2-\left(x^2-2xy+y^2\right)}{\left(x^2+2xz+z^2\right)-y^2}\)=\(\frac{z^2-\left(x-y\right)^2}{\left(x+z\right)^2-y^2}\)=\(\frac{\left(z+x-y\right)\left(z-x+y\right)}{\left(x+z-y\right)\left(x+z+y\right)}\)=\(\frac{\left(z-x+y\right)}{\left(x+z+y\right)}\)
c) hang dang thuc ( x -y+z)^2
o duoi phan h hang dang thuc luon
a) phan h nhan tu ra sao cho co tử la (x-1)(3x^2 -4x +1)
mau la (x-1)(2x^2 -x-3)
b ) k nhin dc de
a)= \(\frac{\left(2x+3\right)^2}{2x^2+3x-4x-6}\)
=\(\frac{\left(2x+3\right)^2}{x\left(2x+3\right)-2\left(2x+3\right)}\)
= \(\frac{\left(2x+3\right)^2}{\left(x-2\right)\left(2x+3\right)}\)
=\(\frac{2x+3}{x-2}\)
b) = \(\frac{3\left|x-4\right|}{3x^2-3x-1296}\)
= \(\frac{3\left|x-4\right|}{3\left(x^2-x-432\right)}\)
=\(\frac{\left|x-4\right|}{x^2-x-432}\)
Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
Tương tự thay vào mà quy đồng
Ta có: \(A=\frac{2a^3b^5}{3a^3b^2}=\frac{2b^3}{3}\)
Ta có:
\(B=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x-y+z\right)\left(x+y+z\right)}\)
\(=\frac{x+y-z}{x-y+z}\)
A= \(\frac{2b^3}{3}\)
B= \(\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+z+y\right)\left(x+z-y\right)}=\frac{x+y-z}{x+z-y}\)
\(\frac{2xy-x^2+z^2-y^2}{-x^2+y-z^2+2xz}\)
\(=\frac{-\left[\left(x^2-2xy+y^2\right)-z^2\right]}{-\left[\left(x^2-2xz+z^2\right)-y\right]}\)
\(=\frac{-\left[\left(x-y\right)^2-z^2\right]}{-\left[\left(x-z\right)^2-y\right]}\)
\(=\frac{-\left(x-y-z\right)\left(x-y+z\right)}{-\left(x-z\right)^2+y}\)