Dạng 5 : bài tập tự luyện có đáp án

$\frac{x-4}{14}+\frac{-10-8x}{14}=\frac{x-4-10-8x}{14}=\frac{-7x-14}{14}=\frac{7\left(-x-2\right)}{14}=\frac{-x-2}{2}$

$\frac{7x-2}{24xy}+\frac{2-8xy}{24xy}=\frac{7x-2+2-8xy}{24xy}=\frac{7x-8xy}{24xy}=\frac{x\left(7-8y\right)}{24xy}=\frac{7-8y}{24y}$

$\frac{3x-y}{3x+y}+\frac{6x+4y}{3x+y}=\frac{3x-y+6x+4y}{3x+y}=\frac{9x+3y}{3x+y}=\frac{3\left(3x+y\right)}{3x+y}=3$

$\frac{3xy-4}{-25xy}+\frac{7-12x}{-25xy}+\frac{4x-3}{-25xy}=\frac{3xy-4+7-12x+4x-3}{-25xy}=\frac{3xy-8x}{-25xy}=\frac{x\left(3y-8\right)}{-25xy}=\frac{3y-8}{-25y}=\frac{8-3y}{25y}$

$\frac{3}{5-2a}+\frac{2a-2}{2a-5}=\frac{3}{5-2a}+\frac{2-2a}{5-2a}=\frac{3+2-2a}{5-2a}=1$

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

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

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

$\frac{2x}{3}+\frac{2ax+3x}{-3a}=\frac{2ax-2ax-3x}{3a}=\frac{-x}{a}$

$\frac{a+2b-2x}{ab-ax}+\frac{1}{x-b}=\frac{a+2b-2x-a}{a\left(b-x\right)}=\frac{2}{a}$

$\frac{3x-2y}{xy}+\frac{2}{x}+\frac{2}{y}=\frac{3x-2y+2y+2x}{xy}=\frac{5}{y}$

$\frac{3}{y-1}+\frac{-2}{x}+\frac{-3x^2+xy-x}{x^2\left(y-1\right)}=\frac{3x^2-2x\left(y-1\right)-3x^2+xy-x}{x^2\left(y-1\right)}=\frac{-2xy+x}{x^2\left(y-1\right)}=\frac{1-2y}{x\left(y-1\right)}$

$\frac{5}{x+2}+\frac{7-2x}{x^2-4}=\frac{5\left(x-2\right)+7-2x}{x^2-4}=\frac{3x-3}{x^2-4}$

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

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

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

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

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

$\frac{1}{9x-18}+\frac{22-7x}{72-18x^2}+\frac{5}{12x+24}=\frac{1}{9\left(x-2\right)}+\frac{22-7x}{18\left(4-x^2\right)}+\frac{5}{12\left(x+2\right)}$

$=\frac{4\left(x+2\right)+2\left(7x-22\right)+15\left(x-2\right)}{36\left(x^2-4\right)}=\frac{33x-66}{36\left(x^2-4\right)}=\frac{11}{12\left(x+2\right)}$

$\frac{5b-3}{9a^{2}b}+\frac{a+1}{5a{b}^3}+\frac{4a+2}{15a^{3}b}=\frac{5a{b}^2\left(5b-3\right)+9a^2\left(a+1\right)+3b^2\left(4a+2\right)}{45a^3{b}^3}=\frac{25a{b}^3+9a^3-3a{b}^2+9a^2+6b^2}{45a^3{b}^3}$

$\frac{5}{2x}+\frac{2x-3}{2x-1}+\frac{4x^2+3}{8x^2-4x}=\frac{10\left(2x-1\right)+4x\left(2x-3\right)+4x^2+3}{2x\left(2x-1\right)}=\frac{12x^2+8x-7}{4x\left(2x-1\right)}$

$=\frac{\left(6x+7\right)\left(2x-1\right)}{4x\left(2x-1\right)}=\frac{6x+7}{4x}$

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

$\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{{\left(x+3\right)}^2}+\frac{-1}{{\left(x-3\right)}^2}+\frac{x}{x^2-9}=\frac{x^2-6x+9-x^2-6x-9+x\left(x^2-9\right)}{{\left(x-3\right)}^2{\left(x+3\right)}^2}$

$=\frac{x^3-21x}{{\left(x-3\right)}^2{\left(x+3\right)}^2}$

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

$\frac{4}{\left(y-x\right)\left(z-x\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}=\frac{4\left(y-z\right)+3\left(z-x\right)-3\left(y-x\right)}{\left(y-x\right)\left(z-x\right)\left(y-z\right)}$$=\frac{y-z}{\left(y-x\right)\left(z-x\right)\left(y-z\right)}=\frac{1}{\left(y-x\right)\left(z-x\right)}$

$\frac{1}{x^2+3x+2}+\frac{2}{x^2+12x+35}+\frac{3}{x^2+7x+10}=\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}+\frac{3}{\left(x+2\right)\left(x+5\right)}$

$=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+5}-\frac{1}{x+7}+\frac{1}{x+2}-\frac{1}{x+5}=\frac{1}{x+1}-\frac{1}{x+7}=\frac{6}{\left(x+1\right)\left(x+7\right)}$

$\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{-ac}{\left(a-b\right)\left(b-c\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}$

$=\frac{bc\left(b-c\right)-ac\left(a-c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{bc\left(b-c\right)-ac\left(a-b+b-c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}$

$=\frac{\left(b-c\right)\left(bc-ac\right)+\left(a-b\right)\left(ab-ac\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1$