Dạng 5:bài luyện tập rút gọn phân thức có đá án

$\frac{14x{y}^5\left(2x-3y\right)}{21x^{2}y{\left(2x-3y\right)}^2}=\frac{2y^4}{3x\left(2x-3y\right)},\left(x,y\ne 0;2x\ne 3y\right)$

$\frac{8xy{(3x-1)}^3}{12x^3(1-3x)}=\frac{-2y{\left(3x-1\right)}^2}{3x^2},\left(x\ne 0;x\ne \frac{1}{3}\right)$

$\frac{20x^2-45}{{\left(2x+3\right)}^2}=\frac{5\left(4x^2-9\right)}{{\left(2x+3\right)}^2}=\frac{5\left(2x-3\right)\left(2x+3\right)}{{\left(2x+3\right)}^2}=\frac{5\left(2x-3\right)}{2x+3},\left(x\ne \frac{-3}{2}\right)$

$\frac{5x^2-10xy}{2{(2y-x)}^3}=\frac{5x\left(x-2y\right)}{2{\left(2y-x\right)}^3}=-\frac{5x}{2{\left(2y-x\right)}^2},\left(x,y\ne 0;x\ne 2y\right)$

$\frac{80x^3-125x}{3(x-3)-(x-3)(8-4x)}=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}=\frac{5x\left(4x+5\right)}{\left(x-3\right)},\left(x\ne 3;x\ne \frac{5}{4}\right)$

$\frac{9-{(x+5)}^2}{x^2+4x+4}=\frac{\left(3-x-5\right)\left(3+x+5\right)}{{\left(x+2\right)}^2}=\frac{\left(-x-2\right)\left(x+8\right)}{{\left(x+2\right)}^2}=-\frac{x+8}{x+2},\left(x\ne -2\right)$

$\frac{32x-8x^2+2x^3}{x^3+64}=\frac{2x\left(16-4x+x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4},\left(x\ne -4\right)$

$\frac{5x^3+5x}{x^4-1}=\frac{5x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}=\frac{5x}{x^2-1},\left(x\ne \pm 1\right)$

$\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{{\left(x+2\right)}^2}=\frac{x+3}{x+2},\left(x\ne -2\right)$

$\frac{10x{y}^2(x+y)}{15xy{(x+y)}^3}=\frac{2y}{3{\left(x+y\right)}^2},\left(x\ne -y\right)$

$\frac{x^2-xy-x+y}{x^2+xy-x-y}=\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\frac{x-y}{x+y},\left(x\ne 1,x\ne -y\right)$

$\frac{3x^2-12x+12}{x^4-8x}=\frac{3{\left(x-2\right)}^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)};\left(x\ne 0,x\ne 2\right)$

$\frac{7x^2+14x+7}{3x^2+3x}=\frac{7{\left(x+1\right)}^2}{3x\left(x+1\right)}=\frac{7\left(x+1\right)}{3x},\left(x\ne 0,x\ne -1\right)$

$\frac{2a^2-2ab}{ac+ad-bc-bd}=\frac{2a\left(a-b\right)}{\left(a-b\right)\left(c+d\right)}=\frac{2a}{c+d},\left(a\ne b,c\ne -d\right)$

$\frac{x^2-xy}{y^2-x^2}=\frac{x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}=-\frac{x}{x+y},\left(x\ne \pm y\right)$

$\frac{2x-2y}{x^2-2xy+y^2}=\frac{2\left(x-y\right)}{{\left(x-y\right)}^2}=\frac{2}{x-y},\left(x\ne y\right)$

$\frac{2-2a}{a^3-1}=\frac{2\left(1-a\right)}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-2}{a^2+a+1},\left(a\ne 1\right)$

$\frac{x^2-6x+9}{x^2-8x+15}=\frac{{\left(x-3\right)}^2}{\left(x-3\right)\left(x-5\right)}=\frac{x-3}{x-5},\left(x\ne 3,x\ne 5\right)$

$\frac{x^4-2x^3}{2x^4-x^3}=\frac{x^3\left(x-2\right)}{x^3\left(2x-1\right)}=\frac{x-2}{2x-1},\left(x\ne 0,x\ne \frac{1}{2}\right)$

$\frac{x^7-x^4}{x^6-1}=\frac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\frac{x^4}{x^3+1},\left(x\ne \pm 1\right)$

$\frac{{(x+2)}^2-{(x-2)}^2}{16x}=\frac{8x}{16x}=\frac{1}{2},\left(x\ne 0\right)$

$\frac{24,5x^2-0,5y^2}{3,5x^2-0,5xy}=\frac{0,5\left(7x-y\right)\left(7x+y\right)}{0,5x\left(7x-y\right)}=7x+y,\left(x\ne 0,y\ne 7x\right)$

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

$\frac{(a-b)(c-d)}{(b^2-a^2)(d^2-c^2)}=\frac{\left(a-b\right)\left(c-d\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)},\left(a\ne \pm b,c\ne \pm d\right)$

$\frac{x^{2}y+2x{y}^2+y^3}{2x^2+xy-y^2}=\frac{xy+y^2}{2x-y}$

$\frac{x^2+3xy+2y^2}{x^3+2x^{2}y-x{y}^2-2y^3}=\frac{1}{x-y}$

$\frac{45x(3-x)}{15x{(x-3)}^3}=\frac{-45x\left(x-3\right)}{15x{\left(x-3\right)}^3}=\frac{-3}{{\left(x-3\right)}^2},\left(x\ne 0,x\ne 3\right)$

$\frac{y^2-x^2}{x^3-3x^{2}y+3x{y}^2-y^3}=-\frac{\left(x-y\right)\left(x+y\right)}{{\left(x-y\right)}^3}=-\frac{x+y}{{\left(x-y\right)}^2},\left(x\ne y\right)$

$\frac{a{x}^4-a^{4}x}{a^2+ax+x^2}=\frac{ax\left(x^3-a^3\right)}{a^2+ax+x^2}=\frac{ax\left(x-a\right)\left(x^2+ax+a^2\right)}{a^2+ax+x^2}=ax\left(x-a\right)$

$\frac{x^3+x^2-6x}{x^3-4x}=\frac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\frac{x\left(x+3\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}=\frac{x+3}{x-2},\left(x\ne 0,x\ne \pm 2\right)$

$\frac{x^3+3x}{3x^3+x^5}=\frac{x\left(x^2+3\right)}{x^3\left(3+x^2\right)}=\frac{1}{x^2},\left(x\ne 0\right)$

$\frac{x^4-2x^3}{2x^2-x^3}=\frac{x^3\left(x-2\right)}{x^2\left(2-x\right)}=\frac{x^3\left(x-2\right)}{x^2\left(2-x\right)}=-x,\left(x\ne 0,x\ne 2\right)$

$\frac{10ab-5a^2}{16b^2-8ab}=\frac{5a\left(2b-a\right)}{8b\left(2b-a\right)}=\frac{5a}{8b},\left(b\ne 0,a\ne 2b\right)$

$\frac{a^7+1}{a^{15}+a^8}=\frac{a^7+1}{a^8\left(a^7+1\right)}=\frac{1}{a^8},\left(a\ne 0,a\ne -1\right)$

$\frac{2x-4y}{0,2x^2-0,8y^2}=\frac{2\left(x-2y\right)}{0,2\left(x-2y\right)\left(x+2y\right)}=\frac{10}{x+2y},\left(x\ne \pm 2y\right)$

$\frac{x^2-9y^2}{1,5x+4,5y}=\frac{\left(x-3y\right)\left(x+3y\right)}{1,5\left(x+3y\right)}=\frac{x-3y}{1,5}$

$3a^2+3b^2=10ab\iff 3.\frac{a}{b}+3\frac{b}{a}=10\iff 3x+\frac{3}{x}=10\iff \left[\begin{array}{l}x=3\\ x=\frac{1}{3}\end{array}\right.$ với $x=\frac{a}{b}$

$\Rightarrow \frac{a}{b}=\frac{1}{3}\iff 3a=b$ vì $b>a>0$

$\frac{x^2-y^2}{(x+y)(ay-ax)}=\frac{\left(x-y\right)\left(x+y\right)}{a\left(x+y\right)\left(y-x\right)}=\frac{-1}{a},\left(x\ne \pm y\right)$

$\frac{2ax-2x-3y+3ay}{4ax+4x+6y+6ay}=\frac{\left(a-1\right)\left(2x+3y\right)}{\left(2a+2\right)\left(2x+3y\right)}=\frac{a-1}{2a+2},\left(2x\ne -3y\right)$

Điều kiện để phân thức M có nghĩa là mẫu thức khác 0.

Xét:$\begin{array}{l}{\left(a+b+c\right)}^2-\left(ab+bc+ca\right)=0 \iff a^2+b^2+c^2+ab+bc+ca=0.\\ \iff 2a^2+2b^2+2c^2+2ab+2bc+2ca=0\iff {\left(a+b\right)}^2+{\left(b+c\right)}^2+{\left(c+a\right)}^2=0 \\ \iff a+b=b+c=c+a \iff a=b=c.\end{array}$

Vậy điều kiện để phân thức M có nghĩa là a,b,c không đồng thời bằng 0.

tức là $a^2+b^2+c^2\ne 0$.