\(x=\frac{4}{5};x=-\frac{5}{6};x=\frac{7}{15}\)
\(M+N=9\left(xz\right)^2+14\left(yz\right)^2=z^2\left(9x^2+14y^2\right)\)
Câu 1 : \(x-\frac{1}{5}=\frac{3}{5}\)
\(x=\frac{3}{5}+\frac{1}{5}\)
\(x=\frac{4}{5}\)
\(\frac{1}{2}+x=-\frac{1}{3}\)
\(x=-\frac{1}{3}-\frac{1}{2}\)
\(x=-\frac{5}{6}\)
\(-x+\frac{1}{3}=-\frac{2}{15}\)
\(x=\frac{1}{3}+\frac{2}{15}\)
\(x=\frac{7}{15}\)
Câu 2 : Ta có : \(M+N=5x^2z^2+5y^2z+8x+4x^2z^2+9y^2z-8x\)
\(=9x^2x^2+14y^2z\)