a) \(\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}\)

\(=\frac{2x-7}{10x-4}-\frac{-\left(3x+5\right)}{-\left(4-10x\right)}\)

\(=\frac{2x-7}{10x-4}-\frac{5-3x}{10x-4}\)

\(=\frac{2x-7-\left(5-3x\right)}{10x-4}\)

\(=\frac{2x-7-5+3x}{10x-4}\)

\(=\frac{5x-12}{10x-4}\)

Tự trình bày nha

a) MC : x^3 +1 

KQ: (x+1)^3 / (x^3 +1)

b) MC: 2x-4x^2 

KQ: -1/(2x)

\(\frac{2x^2-11x}{2xy}+\frac{5y-x}{y}+\frac{x+2y}{x}\)

\(=\frac{2x^2-11x+\left(5y-x\right)2x+\left(x+2y\right)2y}{2xy}\)

\(=\frac{2x^2-11x+10xy-2x^2+2xy+4y^2}{2xy}=\frac{12xy-11x+4y^2}{2xy}\)

\(x+y+\frac{3x^2}{2y}\)

\(=\frac{2xy}{2y}+\frac{2y^2}{2y}+\frac{3x^2}{2y}=\frac{2xy+2y^2+3x^2}{2y}=\frac{2y.\left(2x+y\right)+3x^2}{2y}=\frac{2x+y+3x^2}{2y}\)

p/s: mới lớp 7 ạ sai sót bỏ qua nha :>

aeeeeei nhầm nha :>

\(\frac{2xy}{2y}+\frac{2y^2}{2y}+\frac{3x^2}{2y}=\frac{2xy+2y^2+3x^2}{2y}=\frac{2y.\left(x+y\right)+3x^2}{2y}=\frac{x+y+2x^3}{2y}\)

tớ nghĩ tớ làm sai rồi

sorry bn nha :< 

lần sau nếu chắc 100% t mói tl 

ĐK: \(x\ne1\)

\(\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{1-2x}{x^2+x+1}-\frac{6}{x-1}\)

\(=\frac{4x^2-3x+5-\left(1-2x\right)\left(x-1\right)-6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{4x^2-3x+5+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{-12x}{\left(x-1\right)\left(x^2+x+1\right)}\)

Toán lớp 8 hở mày ?

ĐKXĐ : \(x\ne-1\)

\(\frac{1}{2x+2}-\frac{x-1}{3x^2+6x+3}\)

\(=\frac{1}{2\left(x+1\right)}-\frac{x-1}{3\left(x^2+2x+1\right)}\)

\(=\frac{1}{2\left(x+1\right)}-\frac{x-1}{3\left(x+1\right)^2}\)

\(=\frac{3\left(x+1\right)}{2\left(x+1\right)\cdot3\left(x+1\right)}-\frac{2\left(x-1\right)}{3\left(x+1\right)^2\cdot2}\)

\(=\frac{3x+3}{6\left(x+1\right)^2}-\frac{2x-2}{6\left(x+1\right)^2}\)

\(=\frac{3x+3-2x+2}{6\left(x+1\right)^2}\)

\(=\frac{x+5}{6\left(x+1\right)^2}\)

\(\frac{2x+y}{2x^2-xy}+\frac{8y}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)

\(=\frac{2x+y}{x\left(2x-y\right)}-\frac{8y}{\left(2x-y\right)\left(2x+y\right)}+\frac{2x-y}{x\left(2x+y\right)}\)

\(=\frac{\left(2x+y\right)^2-8xy+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{4x^2+4xy+y^2-8xy+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\frac{8x^2-8xy+2y^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{2\left(4x^2-4xy+y^2\right)}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\frac{2\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{2\left(2x-y\right)}{x\left(2x+y\right)}\)

\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{3x+2}{9x^2-4}-\frac{3x-2}{9x^2-4}+\frac{3x-6}{9x^2-4}\)

\(=\frac{3x+2-3x+2+3x-6}{9x^2-4}\)

\(=\frac{3x-2}{9x^2-4}\)

\(=\frac{1}{3x+2}\)

\(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x^2}{x^2-9}\)

\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\) \(-\frac{3\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)\(-\frac{x^2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}\)

\(=\frac{18-3x-9-x^3+3x^2}{\left(x-3\right)^2\left(x+3\right)}\)

\(=\frac{-x^3+3x^2-3x+9}{\left(x-3^2\right)\left(x+3\right)}\)

\(=\frac{\left(-x^2-3\right)\left(x-3\right)}{\left(x-3^2\right)\left(x+3\right)}\)

\(=\frac{-x^2-3}{\left(x-3\right)\left(x+3\right)}\)

học tốt

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