Bài 19 trang 41 Toán 8 Tập 1 Chân trời sáng tạo

Giải Toán 8 Bài tập cuối chương 1 - Chân trời sáng tạo

Bài 19 trang 41 Toán 8 Tập 1: Thực hiện các phép tính sau:

a) $\frac{8y}{3x^2}.\frac{9x^2}{4y^2}$

b) $\frac{3x+x^2}{x^2+x+1}.\frac{3x^3-3}{x+3}$

c) $\frac{2x^2+4}{x-3}.\frac{3x+1}{x-1}:\frac{x^2+2}{6-2x}$

d) $\frac{2x^2}{3y^3}:\left(-\frac{4x^3}{21y^2}\right)$

e) $\frac{2x+10}{x^3-64}:\frac{{\left(x+5\right)}^2}{2x-8}$

g) $\frac{1}{x+y}\left(\frac{x+y}{xy}-x-y\right)-\frac{1}{x^2}:\frac{y}{x}$

Lời giải:

a) $\frac{8y}{3x^2}.\frac{9x^2}{4y^2}=\frac{8y.9x^2}{3x^2.4y^2}=\frac{2.4y\mathrm{.3.3}{x}^2}{3x^2.y.4y}=\frac{6}{y}$

b) $\frac{3x+x^2}{x^2+x+1}.\frac{3x^3-3}{x+3}$

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

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

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

c) $\frac{2x^2+4}{x-3}.\frac{3x+1}{x-1}:\frac{x^2+2}{6-2x}$

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

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

$=\frac{2.\left(3x+1\right).\left(-2\right)}{x-1}=\frac{-4\left(3x+1\right)}{x-1}$

d) $\frac{2x^2}{3y^3}:\left(-\frac{4x^3}{21y^2}\right)$

$=\frac{2x^2}{3y^3}.\frac{-21y^2}{4x^3}$

$=\frac{2x^2.\left(-21\right)y^2}{3y^3.4x^3}$

$=\frac{2x^2.\left(-7\right).3y^2}{3y^2.y.2x^2.2x}$

$=\frac{-7}{2xy}$;

e) $\frac{2x+10}{x^3-64}:\frac{{\left(x+5\right)}^2}{2x-8}$

$=\frac{2\left(x+5\right)}{x^3-4^3}.\frac{2x-8}{{\left(x+5\right)}^2}$

$=\frac{2\left(x+5\right).2\left(x-4\right)}{\left(x-4\right)\left(x^2+4x+16\right).{\left(x+5\right)}^2}$

$=\frac{2.2}{\left(x^2+4x+16\right).\left(x+5\right)}$

$=\frac{4}{\left(x+5\right)\left(x^2+4x+16\right)}$;

g) $\frac{1}{x+y}\left(\frac{x+y}{xy}-x-y\right)-\frac{1}{x^2}:\frac{y}{x}$

$=\frac{1}{x+y}\left[\frac{x+y}{xy}-\left(x+y\right)\right]-\frac{1}{x^2}.\frac{x}{y}$

$=\frac{1}{x+y}.\frac{x+y}{xy}-\frac{1}{x+y}.\left(x+y\right)-\frac{1}{xy}$

$=\frac{1}{xy}-1-\frac{1}{xy}$

$=\left(\frac{1}{xy}-\frac{1}{xy}\right)-1=-1.$

Lời giải bài tập Toán 8 Bài tập cuối chương 1 hay, chi tiết khác: