Question

Tính:

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

2)\(\frac{1}{1-x}+\frac{x}{1+x}+\frac{z}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

Answer 1

1) dkxd:

\(x\ne0;x\ne-3\\ \frac{x-5}{x^2+3x}+\frac{6}{x+3}=\frac{x-5}{x\left(x+3\right)}+\frac{6}{x+3}\\ =\frac{x-5+6x}{x\left(x+3\right)}\\ =\frac{7x-5}{x\left(x+3\right)}\)

Answer 2

2) dkxd:

\(x\ne1;x\ne-1\\ \\ \frac{1}{1-x}+\frac{x}{x+1}+\frac{z}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\\ \\ =\frac{x+1+1-x^2}{1-x^2}+\frac{z}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\\ \\ =\frac{-\left(1+x^2\right)\left(x^2-x-2\right)+z-zx^2}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\\ \\ =-\frac{\left(1+x^4\right)\left(x^4+x^3+x^2+x+2+z-zx^2\right)+4-4x^4}{1-x^8}+\frac{8}{1+x^8}=...\)

Answer 3

1)

\(\frac{x-5}{x^2+3x}+\frac{6}{x+3}\)

\(=\frac{x-5}{x\left(x+3\right)}+\frac{6}{x+3}\)(MTC:x(x+3))

\(=\frac{x-5}{x\left(x+3\right)}+\frac{6x}{x\left(x+3\right)}\)

\(=\frac{x-5+6x}{x\left(x+3\right)}\)

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