Question

4/(y-x)(x-z)+3/(y-x)(y-z)+3/(y-z)(x-z)

Answer 1

Ta có: \(\frac{4}{\left(y-x\right)\left(x-z\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)

\(=\frac{4\left(y-z\right)}{\left(y-x\right)\left(x-z\right)\left(y-z\right)}+\frac{3\left(x-z\right)}{\left(y-x\right)\left(y-z\right)\left(x-z\right)}+\frac{3\left(y-x\right)}{\left(y-z\right)\left(x-z\right)\left(y-x\right)}\)

\(=\frac{4y-4z+3x-3z+3y-3x}{\left(y-x\right)\left(y-z\right)\left(x-z\right)}=\frac{7y-7z}{\left(y-x\right)\left(y-z\right)\left(x-z\right)}\)

\(=\frac{7\left(y-z\right)}{\left(y-x\right)\left(y-z\right)\left(x-z\right)}=\frac{7}{\left(y-x\right)\left(x-z\right)}\)

\(=\frac{7}{yx-yz-x^2+xz}\)

Answer 2

\(=\frac{4\left(y-z\right)+3\left(x-z\right)+3\left(y-x\right)}{\left(y-x\right)\left(y-z\right)\left(x-z\right)}=\frac{7\left(y-z\right)}{\left(y-x\right)\left(y-z\right)\left(x-z\right)}=\frac{7}{\left(y-x\right)\left(x-z\right)}\)