Dạng 3: Bài luyện tập có đáp án

$\begin{array}{l}MTC:2ab\\ \frac{1}{2a}=\frac{b}{2ab}\\ \frac{3}{2b}=\frac{3a}{2ab}\end{array}$

$\begin{array}{l}MTC:3x^4{y}^4\\ \frac{2}{x^4{y}^3}=\frac{6y}{3x^4{y}^4}\\ \frac{4}{3x^2{y}^4}=\frac{4}{3x^4{y}^4}\end{array}$

$\begin{array}{l}MTC:5(a-3)\\ \frac{2x}{5}=\frac{2x.(a-3)}{5(a-3)}\\ \frac{3x-2}{5a-15}=\frac{3x-2}{5(a-3)}\end{array}$

$\begin{array}{l}MTC:6x^2\\ \frac{5-x}{3x^2}=\frac{2(5-x)}{6x^2}\\ \frac{7}{6x}=\frac{7x}{6x^2}\end{array}$

$\begin{array}{l}MTC:\left(x-1\right)\left(x+1\right)\\ \frac{2}{x-1}=\frac{2(x+1)}{(x-1)(x+1)}\\ \frac{3}{x+1}=\frac{3(x-1)}{(x-1)(x+1)}\end{array}$

$\begin{array}{l}MTC:(x-2)(x+2)\\ \frac{4}{x-2}=\frac{4(x+2)}{(x-2)(x+2)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2x+3}{x^2-4}=\frac{2x+3}{(x-2)(x+2)}\end{array}$

$\begin{array}{l}MTC:2(x-1)(x+1)\\ \frac{x}{2x-2}=\frac{x}{2(x-1)}=\frac{x(x+1)}{2(x-1)(x+1)}\\ \frac{5}{x^2-1}=\frac{5}{(x-1)(x+1)}=\frac{10}{2(x-1)(x+1)}\end{array}$

$\begin{array}{c}MTC:18ab\\ \frac{b}{6b}=\frac{3a}{18ab}\\ \frac{c}{18ab}\\ \frac{x}{9b}=\frac{2ax}{18ab}\end{array}$

$\begin{array}{l}MTC:20ab\\ \frac{3}{4a}=\frac{15b}{20ab}\\ \frac{a-1}{10b}=\frac{2a.(a-1)}{20ab}\\ \frac{x+4}{20ab}\end{array}$

$\begin{array}{l}MTC:3a^2(b-1)\\ \frac{3}{b-1}=\frac{9a^2}{3a^2.(b-1)}\\ \frac{2}{3a}=\frac{2a.(b-1)}{3a^2.(b-1)}\\ \frac{3a+1}{a^2.(b-1)}=\frac{3.(3a+1)}{3a^2.(b-1)}\end{array}$

$\begin{array}{l}MTC:6a.(a-1)\\ \frac{2}{3a}=\frac{4.(a-1)}{6a.(a-1)}\\ \frac{1}{2a-2}=\frac{1}{2.(a-1)}=\frac{3a}{6a.(a-1)}\\ \frac{a-4}{6a^2-6a}=\frac{a-4}{6a.(a-1)}\end{array}$

$\begin{array}{l}MTC:2xy.(2x-3y)\\ \frac{5}{xy}=\frac{10.(2x-3y)}{2xy.(2x-3y)}\\ \frac{3b}{4x-6y}=\frac{3b}{2.(2x-3y)}=\frac{3bxy}{2xy.(2x-3y)}\\ \frac{-a}{2x^{2}y-3x{y}^2}=\frac{-a}{xy.(2x-3y)}=\frac{-2a}{2xy.(2x-3y)}\end{array}$

$\begin{array}{l}MTC:(a+3).(b-3)\\ \frac{-5b}{a+3}=\frac{-5b.(b-3)}{(a+3).(b-3)}\\ \frac{3a}{b-3}=\frac{3a.(a+3)}{(a+3).(b-3)}\\ \frac{2ab}{ab-3a+3b-9}=\frac{2ab}{(a+3).(b-3)}\end{array}$

$\begin{array}{l}MTC:3.{(a-2)}^2.\\ a+2=\frac{3.(a+2).{(a-2)}^2}{3.{(a-2)}^2}\\ \frac{a+1}{3a-6}=\frac{a+1}{3.(a-2)}=\frac{(a+1).(a-2)}{3.{(a-2)}^2}\\ \frac{a-1}{{(a-2)}^2}=\frac{3.(a-1)}{3.{(a-2)}^2}\end{array}$

$\begin{array}{l}MTC:(a-b).(a+b)\\ \frac{ab}{a^2-b^2}=\frac{ab}{(a-b).(a+b)}\\ \frac{a-b}{a^2-2ab+b^2}=\frac{a-b}{{(a-b)}^2}=\frac{1}{a-b}=\frac{a+b}{(a-b).(a+b)}\\ \frac{a+b}{a^2+2ab+b^2}=\frac{a+b}{{(a+b)}^2}=\frac{1}{a+b}=\frac{a-b}{(a-b).(a+b)}\end{array}$

$\begin{array}{l}MTC:2.(x-1).(x+1)\\ \frac{2x}{x^2-1}=\frac{2x}{(x-1).(x+1)}=\frac{4x}{2.(x-1).(x+1)}\\ \frac{-3x}{2x+2}=\frac{-3x}{2.(x+1)}=\frac{-3x.(x-1)}{2.(x-1).(x+1)}\\ \frac{4}{x-1}=\frac{8.(x+1)}{2.(x-1).(x+1)}\end{array}$

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

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

$\begin{array}{l}MTC:(3-2x).(3+2x)\\ \frac{3x^2}{9-4x^2}=\frac{3x^2}{(3-2x).(3+2x)}\\ \frac{2-x}{2x-3}=\frac{x-2}{3-2x}=\frac{(x-2).(3+2x)}{(3-2x).(3+2x)}\\ \frac{x-3}{2x+3}=\frac{(x-3).(3-2x)}{(3-2x).(3+2x)}\end{array}$

$\begin{array}{l}MTC:30.(x-1).(x+1)\\ \frac{x}{10x-10}=\frac{x}{10.(x-1)}=\frac{3x.(x+1)}{30.(x-1).(x+1)}\\ \frac{9x-11}{30x^2-30}=\frac{9x-11}{30.(x^2-1)}=\frac{9x-11}{30.(x-1).(x+1)}\\ \frac{1}{3x+3}=\frac{1}{3.(x+1)}=\frac{10.(x-1)}{30.(x-1).(x+1)}\end{array}$

$\begin{array}{l}MTC:x.(2a-x).(2a+x)\\ \frac{1}{x-2a}=\frac{-x.(2a+x)}{x.(2a-x).(2a+x)}\\ \frac{8a^2}{4a^{2}x-x^3}=\frac{8a^2}{x.(2a-x).(2a+x)}\\ \frac{1}{x+2a}=\frac{x.(2a-x)}{x.(2a-x).(2a+x)}\end{array}$

$\begin{array}{l}MTC:2.(a-1).(a^2+a+1)\\ \frac{1}{2a-2}=\frac{1}{2.(a-1)}=\frac{a^2+a+1}{2.(a-1).(a^2+a+1)}\\ \frac{a+1}{a^2+a+1}=\frac{2.(a-1).(a+1)}{2.(a-1).(a^2+a+1)}\\ \frac{a^3+2a}{a^3-1}=\frac{2.(a^3+2a)}{2.(a-1).(a^2+a+1)}\end{array}$

$\begin{array}{l}MTC: {(2a+1)}^2.(4a^2+2a+1)\\ \frac{-a}{4a^2+4a+1}=\frac{-a}{{(2a+1)}^2}=\frac{-a.(4a^2+2a+1)}{{(2a+1)}^2.(4a^2+2a+1)}\\ \frac{1-2a}{8a^3+1}=\frac{1-2a}{(2a+1).(4a^2+2a+1)}=\frac{(1-2a).(1+2a)}{{(2a+1)}^2.(4a^2+2a+1)}\\ \frac{4a^2-4a+1}{4a^2+2a+1}=\frac{{(2a-1)}^2}{4a^2+2a+1}=\frac{{(2a-1)}^2.{(1+2a)}^2}{{(2a+1)}^2.(4a^2+2a+1)}\end{array}$

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

$\begin{array}{l}MTC:2.(3a-1).(5a+2)\\ \frac{a}{2-6a}=\frac{a}{2.(1-3a)}=\frac{-a}{2.(3a-1).(5a+2)}\\ \frac{a-1}{2+5a}=\frac{2.(3a-1)}{2.(3a-1).(5a+2)}\\ \frac{a^2-3a}{15a^2+a-2}=\frac{a^2-3a}{(3a-1).(5a+2)}=\frac{2.(a^2-3a)}{2.(3a-1).(5a+2)}\end{array}$