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bạn ơi tại sao lại bằng 2x +6, bạn có thể giải đáp cho mình đc ko
a) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\\ =\dfrac{-5x-10}{8-4x}.\dfrac{4-2x}{x+2}\\ =\dfrac{-5\left(x+2\right)}{4\left(2-x\right)}.\dfrac{2\left(2-x\right)}{x+2}=\dfrac{-5}{2}\)
b)\(\dfrac{x^2-36}{2x+10}.\dfrac{3}{6-x}\\ =\dfrac{\left(x-6\right).\left(x+6\right)}{2\left(x+5\right)}.\dfrac{-3}{x-6}\\ =\dfrac{-3x-18}{2x+10}\)
a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)
b. =\(\frac{2.\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x.\left(x+3\right)}=\frac{-2}{x^2}\)
Chắc chắn đúng, mik nhaaaaaa
\(\frac{x+6}{x^2-4}-\frac{2}{x^2+2x}\)
\(=\frac{x+6}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x\left(x+2\right)}\)
\(=\frac{x\left(x+6\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+6x-2x+4}{x\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+4x+4}{x\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2}{x\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x+2}{x\left(x-2\right)}\)
\(\frac{x+6}{x^2+4}-\frac{2}{x^2+2x}\)
\(=\frac{x+6}{\left(x+2\right)^2}-\frac{2}{x\left(x+2\right)}\)
\(=\frac{x\left(x+6\right)}{x\left(x+2\right)^2}-\frac{2\left(x+2\right)}{x\left(x+2\right)^2}\)
\(=\frac{x^2+6x-2x-4}{x\left(x+2\right)^2}\)
\(=\frac{x^2+4x-4}{x\left(x+2\right)^2}\)
\(\frac{3}{x-3}-\frac{x-6}{x^2+3x}=\frac{3}{x-3}-\frac{x-6}{x\left(x+3\right)}=\frac{3}{x-3}+\frac{x-6}{x\left(x-3\right)}=\frac{3x}{x\left(x-3\right)}+\frac{x-6}{x\left(x-3\right)}=\frac{3x+x-6}{x\left(x-3\right)}=\frac{4x-6}{x\left(x-3\right)}\)
\(\frac{3\left(x+1\right)}{x+2}-\frac{3x-6}{x^2-4}\)
\(=\frac{3\left(x+1\right)}{x+2}-\left(\frac{3x-6}{x^2-4}\right)\)
\(=\frac{3x^2-6x^2-12x+24}{x^3+2x^2-4x-8}\)
\(=\frac{3\left(x+2\right)\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x+2\right)\left(x-2\right)}\)
\(=\frac{3x-6}{x+2}\)
\(\frac{x^2+4x+4}{1-x}.\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\)
\(=\frac{x^2+4x+4}{1-x}.\left[\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\right]\)
\(=\frac{x^4+2x^3-3x^2-4x+4}{-3x^4-15x^3-18x^2+12x+24}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x+2\right)\left(x+2\right)}{3\left(-x+1\right)\left(x+2\right)\left(x+2\right)\left(x+2\right)}\)
\(=\frac{-x+1}{3x+6}\)
3 2 x + 6 - x - 6 2 x 2 + 6 x