Dạng 2: Quy đồng các mẫu thức

$BCNN\left(2;4\right)=4$

MTC: $4a$

$\frac{3-a}{2a}=\frac{2\left(3-a\right)}{4a}$

$\frac{-3}{4}=\frac{-3}{4a}$

MTC: $12x$

$\frac{2x}{3}=\frac{2x.4x}{3.4x}=\frac{8x^2}{12x};\frac{2ax+3}{4x}=\frac{\left(2ax+3\right).3}{4x.3}=\frac{6ax+9}{12x}$

Ta có: $4a=2^2.a$

$20ab=2^2\mathrm{.5.}ab$

$10b^2=2.5b^2$

MTC: $2^2\mathrm{.3.}a{b}^2=20a{b}^2$

$\frac{x}{4a}=\frac{x.5b^2}{4a.5b^2}=\frac{5b^{2}x}{20a{b}^2}$

$\frac{6a-5bx}{20ab}=\frac{\left(6a-5bx\right).b}{20ab}=\frac{6ab-5b^{2}x}{20a{b}^2}$

$\frac{a-1}{10b}=\frac{2ab.\left(a-1\right)}{10b.ab}=\frac{2a^{2}b-2ab}{20a{b}^2}$

Ta có: $63x^2{y}^3={7.3}^2.x^2{y}^3$

$15x{z}^2=\mathrm{3.5.}x{z}^2$

$9y^{2}z=3^2{y}^{2}z$

MTC: $3^2\mathrm{.5.7}{x}^2{y}^3{z}^2=315x^2{y}^3{z}^2$

$\frac{13z}{63x^2{y}^3}=\frac{13z.5z^2}{63x^2{y}^3.5z^2}=\frac{65z^3}{315x^2{y}^3{z}^2}$

$\frac{-y}{15x{z}^2}=\frac{-y.21x{y}^3}{15x{z}^2.21x{y}^3}=\frac{-21x{y}^4}{315x^2{y}^3{z}^2}$

$\frac{2x}{9y^{2}z}=\frac{2x.35x^{2}yz}{9y^{2}z.35x^{2}yz}=\frac{70x^{3}yz}{315x^2{y}^3{z}^2}$

MTC: $6\left(x-1\right)$

$\frac{5}{6}=\frac{5\left(x-1\right)}{6\left(x-1\right)}=\frac{5x-5}{6\left(x-1\right)}$

$\frac{x-2}{3\left(x-1\right)}=\frac{\left(x-2\right).2}{3\left(x-1\right).2}=\frac{2x-4}{6\left(x-1\right)}$

MTC: $10x\left(x+1\right)$

$\frac{1}{2x}=\frac{5\left(x+1\right)}{2x.5\left(x+1\right)}=\frac{5x+5}{10x\left(x+1\right)}$

$\frac{5}{10x+10}=\frac{5}{10\left(x+1\right)}=\frac{5.x}{10\left(x+1\right).x}=\frac{5.x}{10x.\left(x+1\right)}$

$\frac{x+7}{5x^2+5x}=\frac{x+7}{5x\left(x+1\right)}=\frac{\left(x+7\right).2}{5x\left(x+1\right).2}=\frac{2x+14}{10x\left(x+1\right)}$

MTC: $15\left(x+1\right)$

$\frac{2}{5x+5}=\frac{2}{5\left(x+1\right)}=\frac{2.3}{5\left(x+1\right).3}=\frac{6}{15\left(x+1\right)}$

$\frac{2+x}{3x+3}=\frac{2+x}{3\left(x+1\right)}=\frac{\left(2+x\right).5}{3\left(x+1\right).5}=\frac{10+5x}{15\left(x+1\right)}$

MTC: $2(x^2-4)$

$\frac{1}{2x+4}=\frac{x-2}{2(x^2-4)}$

$\frac{x}{2x-4}=\frac{x+2}{2(x^2-4)}$

$\frac{3}{4-x^2}=\frac{-6}{2(x^2-4)}$

MTC: $x\left(4x^2-1\right)$

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

$\frac{20}{4x^3-x}=\frac{20}{x\left(4x^2-1\right)}$

$\frac{7}{2x^2+x}=\frac{7}{x\left(2x+1\right)}=\frac{7\left(2x-1\right)}{x\left(4x^2-1\right)}=\frac{14x-7}{x\left(4x^2-1\right)}$

MTC: $x(x^3+1)$

$\frac{x}{x^3+1}=\frac{x^2}{x(x^3+1)}$

$\frac{x+1}{x^2+x}=\frac{x+1}{x(x+1)}=\frac{1}{x}=\frac{x^3+1}{x(x^3+1)}$

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

MTC: ${(x+1)}^2{(x+2)}^2$

$\frac{1}{x^2+3x+2}=\frac{x^2+3x+2}{{(x+1)}^2{(x+2)}^2}$

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

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

Vì phân thức A có tử thức là $5>0$và mẫu thức là $x^2+4x+5={\left(x+2\right)}^2+1>0$nên phân thức A có GTLN khi $x^2+4x+5={\left(x+2\right)}^2+1$ có GTNN.

Vì ${\left(x+2\right)}^2\ge 0$ nên $x^2+4x+5={\left(x+2\right)}^2+1\ge 1$ có GTNN bằng 1 khi $x=-2$.

Vậy GTLN của $A=\frac{7}{x^2+4x+5}$ bằng 7 khi $x=-2$.

Ta có: $-\left|2x-4\right|\le 0\Rightarrow 3-\left|2x-4\right|\le 3$

$\Rightarrow B=\frac{6}{3-\left|2x-4\right|}\ge \frac{6}{3}=2$

Vậy B đạt GTNN bằng 2 khi $x=2$

Ta có: $-x^2+6x=-{\left(x-3\right)}^2+9\le 9$

$\Rightarrow \frac{-x^2+6x}{4}=\frac{-{\left(x-3\right)}^2+9}{4}\le \frac{9}{4}$

$\Rightarrow C=\frac{-x^2+6x}{4}\le \frac{9}{4}$

Vậy C đạt GTLN bằng $\frac{9}{4}$ khi $x=3$