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mình mới học lớp 7 thui à
Nếu lớp 8 thì sẽ giúp bạn liền
\(\frac{x^2-3x+2}{x^3-1}=\frac{x^2-2x-x+2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\frac{x.\left(x-2\right)-\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{\left(x-1\right).\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\frac{x-2}{x^2+x+1}\)
a, Xét tử thức \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left[\left(x-z\right)-\left(y-z\right)\right]\)
\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-z\right)-z^2\left(y-z\right)\)
\(=\left(x^2-z^2\right)\left(y-z\right)-\left(y^2-z^2\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(x+z\right)\left(y-z\right)-\left(y-z\right)\left(y+z\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(xy-xz+yz-z^2-y^2-yz+yz+z^2\right)\)
\(=\left(x-z\right)\left(xy-xz+yz-y^2\right)=\left(x-z\right)\left[x\left(y-z\right)-y\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)
Mẫu thức \(x^2y-x^2z+y^2z-y^3=x^2\left(y-z\right)-y^2\left(y-z\right)=\left(x-y\right)\left(x+y\right)\left(y-z\right)\)
Vậy \(\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}=\frac{x-z}{x+y}\)
b, \(\frac{x^5+x+1}{x^3+x^2+x}=\frac{x^5-x^2+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}=\frac{x^3-x^2+1}{x}\)
\(a,\frac{3.\left(x-y\right)}{y-x}=\frac{-3.\left(y-x\right)}{y-z}=-3\)
\(b,\frac{x^2-x}{1-x}=\frac{x.\left(x-1\right)}{1-x}=\frac{-x.\left(1-x\right)}{1-x}=-x\)
\(\frac{3\left(x-y\right)}{y-x}=\frac{3\left(x-y\right)}{-1\left(x-y\right)}=-3\)
\(\frac{x^2-x}{1-x}=\frac{x\left(x-1\right)}{-1\left(x-1\right)}=-x\)
1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)