Quy đồng mẫu thức các phân thức Bài 13 trang 27 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 4: Quy đồng mẫu thức nhiều phân thức

Bài 13 trang 27 SBT Toán 8 Tập 1: 

a) $\frac{25}{14x^{2}y};\frac{14}{21x{y}^5}$;

b) $\frac{11}{102x^{4}y};\frac{3}{34x{y}^3}$;

c) $\frac{3x+\text{\hspace{0.17em}}1}{12x{y}^4};\frac{y-2}{9x^2{y}^3}$;

d) $\frac{1}{6x^3{y}^2};\frac{x+\text{\hspace{0.17em}}1}{9x^2{y}^4};\frac{x-1}{4x{y}^3}$;

e) $\frac{3+\text{}2x}{10x^{4}y};\frac{5}{8x^2{y}^2};\frac{2}{3x{y}^5}$;

f) $\frac{4x-4}{2x(x+\text{}3)};\frac{x-3}{3x(x+1)}$;

g) $\frac{2x}{{(x+\text{}2)}^3};\frac{x-2}{2x{(x+2)}^2}$;

h) $\frac{5}{3x-_{3}12x};\frac{3}{(2x+4).(x+\text{}3)}$.

Lời giải:

a) Mẫu thức chung: 42x²y⁵

Ta có

$\frac{25}{14x^{2}y}=\frac{25.3y^4}{14x^{2}y.3y^4}=\frac{75y^4}{42x^2{y}^5}$

$\frac{14}{21x{y}^5}=\frac{14.2x}{21x{y}^5.2x}=\frac{28x}{42x^2{y}^5}$

b) Mẫu thức chung: 102x⁴y³

Ta có

$\frac{11}{102x^{4}y}=\frac{11y^2}{102x^{4}y.y^2}=\frac{11y^2}{102x^4{y}^3}$

$\frac{3}{34x{y}^3}=\frac{3.3x^3}{34x{y}^3.3x^3}=\frac{9x^3}{102x^4{y}^3}$

c) Mẫu thức chung: 36x²y⁴

Ta có

$\frac{3x+\text{\hspace{0.17em}}1}{12x{y}^4}=\frac{(3x+\text{\hspace{0.17em}}1).3x}{12x{y}^4.3x}=\frac{9x^2+\text{\hspace{0.17em}}3x}{36x^2{y}^4};$

$\frac{y-2}{9x^2{y}^3}=\frac{(y-2).4y}{9x^2{y}^3.4y}=\frac{4y^2-8y}{36x^2{y}^4}$

d) Mẫu thức chung: 36x³y⁴

Ta có

$\frac{1}{6x^3{y}^2}=\frac{1.6y^2}{6x^3{y}^2.6y^2}=\frac{6y^2}{36x^3{y}^4};$

$\frac{x+\text{\hspace{0.17em}}1}{9x^2{y}^4}=\text{\hspace{0.17em}}\frac{(x+\text{\hspace{0.17em}}1).4x}{9x^2{y}^4.4x}=\frac{4x^2+\text{\hspace{0.17em}}4x}{36x^3{y}^4};$

$\frac{x-1}{4x{y}^3}=\frac{(x-1).9xy_2}{4x{y}^3.9x^{2}y}=\frac{9x^{3}y-9x^{2}y}{36x^3{y}^4}$

e) Mẫu thức chung: 120x⁴y⁵

Ta có

$\frac{3+\text{}2x}{10x^{4}y}=\frac{(3+\text{}2x).12y^4}{10x^{4}y.12y^4}=\frac{36y^4+\text{}24x{y}^4}{120x^4{y}^5};$

$\frac{5}{8x^2{y}^2}=\frac{5.15x^2{y}^3}{8x^2{y}^2.15x^2{y}^3}=\frac{75x^2{y}^3}{120x^4{y}^5};$

$\frac{2}{3x{y}^5}=\frac{2.40x^3}{3x{y}^5.40x^3}=\frac{80x^3}{120x^4{y}^5}$

f) Mẫu thức chung: 3x(x + 3)(x + 1)

Ta có

$\begin{array}{l}\frac{4x-4}{2x(x+\text{}3)}=\frac{2(x-1)}{x(x+3)}\\ =\frac{2(x-1).3(x+\text{}1)}{x(x+\text{}3).3(x+\text{}1)}=\frac{6(\text{}{x}^2-1)}{3x(x+3)(x+\text{}1)};\end{array}$

$\begin{array}{l}\frac{x-3}{3x(x+1)}=\frac{(x-3)(x+\text{}3)}{3x(x+1).(x+\text{}3)}\\ =\frac{x^2-9}{3x(x+1).(x+\text{}3)}\end{array}$

g) Mẫu thức chung: 2x(x + 2)³

Ta có

$\frac{2x}{{(x+\text{}2)}^3}=\frac{2x.2x}{{(x+\text{}2)}^3.2x}=\frac{4x^2}{2x{\left(x+2\right)}^3};$

$\begin{array}{l}\frac{x-2}{2x{(x+2)}^2}=\frac{(x-2).(x+\text{}2)}{2x{(x+2)}^2.(x+\text{}2)}\\ =\frac{x^2-4}{2x{(x+\text{}2)}^3}\end{array}$

h) Ta có: 3x³ – 12x = 3x(x² – 4) = 3x(x – 2)(x + 2)

(2x + 4)(x + 3) = 2(x + 2)(x + 3)

Suy ra mẫu thức chung là 6x(x – 2)(x + 2)(x + 3)

$\begin{array}{l}\frac{5}{3x-_{3}12x}=\frac{5}{3x(x+\text{}2).(x-2)}\\ =\frac{\mathrm{5.2.}(x+\text{}3)}{3x(x+\text{}2).(x-2).2(x+\text{}3)}\\ =\frac{10(x+3)}{6x(x+2)(x-2).(x+3)}\end{array}$

$\begin{array}{l}\frac{3}{(2x+4).(x+\text{}3)}=\frac{3}{2(x+2).(x+\text{}3)}\\ =\frac{3.3x(x-2)}{2(x+2).(x+\text{}3).3x(x-2)}\\ =\frac{9x(x-2)}{6x(x+2)(x-2)(x+3)}\end{array}$