\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)
tương tự đến hết nha a hay cj gì đps !
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, \(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{x\left(x-1\right)}\)
a) \(\frac{4.x+1}{2}-\frac{3.x+2}{3}=\frac{3.\left(4.x+1\right)-2.\left(3.x+2\right)}{6}\)
\(=\frac{12.x+3-6.x-4}{6}\)
\(=\frac{6.x-1}{6}\)
b)\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)
\(=\frac{x+3}{\left(x-1\right).\left(x+1\right)}-\frac{1}{x.\left(x+1\right)}\)
\(=\frac{x.\left(x+3\right)-\left(x-1\right)}{x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{x^2+3.x-x+1}{x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{x^2+2.x+1}{x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{x+1}{x.\left(x-1\right)}\)
\(=\frac{x+1}{x^2-x}\)
c)\(\frac{3}{2.x^2+2.x}+\frac{2.x-1}{x^2-1}-\frac{1}{2}\)
\(=\frac{3}{2.x.\left(x+1\right)}+\frac{2.x-1}{\left(x-1\right).\left(x+1\right)}-\frac{1}{2}\)
\(=\frac{3.\left(x-1\right)+2.x.\left(2.x-1\right)-x.\left(x-1\right).\left(x+1\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{3.x-3+4.x^2-2.x-x.\left(x^2-1\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{3.x-3+4.x^2-2.x-x^3+x}{2.x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{2.x-3+4.x^2-x^3}{2.x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{-x^3+4.x^2+2.x-3}{2.x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{-x^3-x^2+5.x^2+5.x-3.x-3}{2.x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{-x^2.\left(x+1\right)+5.x.\left(x+1\right)-3.\left(x+1\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{-\left(x+1\right).\left(x^2-5.x+3\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)
\(=\frac{-\left(x^2-5.x+3\right)}{2.x.\left(x-1\right)}\)
\(=-\frac{x^2-5.x+3}{2.x^2-2.x}\)