Question

[Kaijo]

7,Thực hiện phép tính

a,\(\frac{4x+1}{2}-\frac{3x+2}{3}\) 

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

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

d,\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\) 

e,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\) 

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

Answer 1

[Ánh Tuyết]

a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)

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

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

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

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

Answer 2

[๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ]

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

\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)

tương tự đến hết nha a hay cj gì đps !