Ngày 11/12/2017 bạn Thảo Mi gửi bài tập
Bài 1. Thực hiện các phép tính:
a) $\frac{3x - y}{4}$ + $\frac{2y + x}{6}$ + $\frac{y - x}{9}$
b) $\frac{3x^2y + 5}{15x^3y^4}$ + $\frac{5 - 3x^2}{15x^4y^3}$
Bài giải:
a) MTC = 36
$\frac{3x - y}{4}$ + $\frac{2y + x}{6}$ + $\frac{y - x}{9}$ = $\frac{9(3x - y)}{4.9}$ + $\frac{6(2y + x)}{6.6}$ + $\frac{4(y - x)}{9.4}$ = $\frac{27x - 9y}{36}$ + $\frac{12y + 6x}{36}$ + $\frac{4y - 4x}{36}$
= $\frac{27x - 9y + 12y + 6x + 4y - 4x}{36}$ = $\frac{29x + 7y}{36}$
b) MTC = 15$x^4y^4$
$\frac{3x^2y + 5}{15x^3y^4}$ + $\frac{5 - 3x^3}{15x^4y^3}$ = $\frac{x(3x^2y + 5)}{15x^3y^4.x}$ + $\frac{y(5 - 3x^3)}{15x^4y^3.y}$ = $\frac{3x^3y + 5x}{15x^4y^4}$ + $\frac{5y - 3x^3y}{15x^4y^4}$ = $\frac{3x^3y - 3x^3y + 5x + 5y}{15x^4y^4}$ = $\frac{5(x + y)}{15x^4y^4}$ = $\frac{x + y}{3x^4y^4}$
Bài 2. Làm tính cộng
a) $\frac{x}{2x + 2}$ + $\frac{x}{3x - 3}$ + $\frac{-x}{1 - x^2}$
b) $\frac{x^2 - 5x}{x^3 + 1}$ + $\frac{x + 2}{x^2 - x + 1}$ + $\frac{1}{x + 1}$
Bài giải:
a) Ta có $\frac{x}{2x + 2}$ + $\frac{x}{3x - 3}$ + $\frac{-x}{1 - x^2}$ = $\frac{x}{2(x + 1)}$ + $\frac{x}{3(x - 1)}$ + $\frac{x}{x^2 - 1}$
Khi đó MTC = 6(x + 1)(x - 1)
$\frac{x}{2x + 2}$ + $\frac{x}{3x - 3}$ + $\frac{-x}{1 - x^2}$ = $\frac{x}{2(x + 1)}$ + $\frac{x}{3(x - 1)}$ + $\frac{x}{(x + 1)(x - 1)}$
= $\frac{x.3(x - 1)}{2(x + 1).3(x - 1)}$ + $\frac{x.2(x + 1)}{3(x - 1).2(x + 1)}$ + $\frac{6x}{6(x + 1)(x - 1)}$ = $\frac{3x^2 - 3x + 2x^2 + 2x + 6x}{6(x + 1)(x - 1)}$
= $\frac{5x^2 + 5x}{6(x + 1)(x - 1)}$ = $\frac{5x(x + 1)}{6(x + 1)(x - 1)}$ = $\frac{5x}{6(x - 1)}$
b) Ta có $\frac{x^2 - 5x}{x^3 + 1}$ + $\frac{x + 2}{x^2 - x + 1}$ + $\frac{1}{x + 1}$ = $\frac{x^2 - 5x}{(x + 1)(x^2 - x + 1)}$ + $\frac{x + 2}{x^2 - x + 1}$ + $\frac{1}{x + 1}$
MTC = (x + 1)($x^2$ - x + 1)
$\frac{x^2 - 5x}{(x + 1)(x^2 - x + 1)}$ + $\frac{x + 2}{x^2 - x + 1}$ + $\frac{1}{x + 1}$ = $\frac{x^2 - 5x}{(x + 1)(x^2 - x + 1)}$ + $\frac{(x + 2)(x + 1)}{(x^2 - x + 1)(x + 1)}$ + $\frac{x^2 - x + 1}{(x + 1)(x^2 - x + 1)}$
= $\frac{x^2 - 5x}{(x + 1)(x^2 - x + 1)}$ + $\frac{x^2 + 2x + x + 2}{(x^2 - x + 1)(x + 1)}$ + $\frac{x^2 - x + 1}{(x + 1)(x^2 - x + 1)}$ = $\frac{x^2 - 5x + x^2 + 3x + 2 + x^2 - x + 1}{(x + 1)(x^2 - x + 1)}$
= $\frac{3x^2 - 3x + 3}{(x + 1)(x^2 - x + 1)}$ = $\frac{3(x^2 - x + 1)}{(x + 1)(x^2 - x + 1)}$ = $\frac{3}{x + 1}$
