Question

[Trần Hữu Minh Trí]

Quy đồng mẫu thức a) 74x+1674�+16 và 11x2−1611�2−16b) 6(x+3)36(�+3)3 và x−32x(x+3)2�−32�(�+3)2 c) −61−x−61−� ; 3xx2+x+13��2+�+1 và x2−3x+5x3−1�2−3�+5�3−1 d)175x175� ; 24x−2y24�−2� và x−y8y2−2x2

Answer 1

[Nguyễn Lê Phước Thịnh]

a: ĐKXĐ: \(x\notin\left\{4;-4\right\}\)

\(\dfrac{7}{4x+16}=\dfrac{7}{4\left(x+4\right)}=\dfrac{7\left(x-4\right)}{4\left(x+4\right)\left(x-4\right)}\)

\(\dfrac{11}{x^2-16}=\dfrac{11\cdot4}{4\left(x^2-16\right)}=\dfrac{44}{4\left(x-4\right)\left(x+4\right)}\)

b: \(\dfrac{6}{x\left(x+3\right)^2};\dfrac{x-3}{2x\left(x+3\right)^2}\)

ĐKXĐ: \(x\notin\left\{0;-3\right\}\)

\(\dfrac{6}{x\left(x+3\right)^2}=\dfrac{6\cdot2}{2x\left(x+3\right)^2}=\dfrac{12}{2x\left(x+3\right)^2}\)

\(\dfrac{x-3}{2x\left(x+3\right)^2}=\dfrac{x-3}{2x\left(x+3\right)^2}\)

c: \(\dfrac{-6}{1-x};\dfrac{3x}{x^2+x+1};\dfrac{x^2-3x+5}{x^3-1}\)

ĐKXĐ: \(x\ne1\)

\(-\dfrac{6}{1-x}=\dfrac{6}{x-1}=\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

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

d: \(\dfrac{17}{5x};\dfrac{24}{x-2y};\dfrac{x-y}{8y^2-2x^2}\)

ĐKXĐ: \(x\ne0;x\ne\pm2y\)

\(\dfrac{17}{5x}=\dfrac{17\cdot2\left(x-2y\right)\left(x+2y\right)}{5x\cdot2\cdot\left(x-2y\right)\left(x+2y\right)}=\dfrac{34\left(x^2-4y^2\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{24}{x-2y}=\dfrac{24\cdot10x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{240x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{-5x\left(x-y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{-5x^2+5xy}{10x\left(x-2y\right)\left(x+2y\right)}\)