Quy đồng mẫu thức a) <span class="mjx-numerato...

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: \(x\ne0;x\ne\pm2y\) \(\dfrac{17}{5x}=\dfrac{17\cdot2\left(x-2y\right)\left(x+2y\right)}{5x\cdot2\cdot\left(x-2y\right)\left(x+2y\right)}=\dfrac{34\left(x^2-4y^2\right)}{10x\left(x-2y\right)\left(x+2y\right)}\) \(\dfrac{24}{x-2y}=\dfrac{24\cdot10x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{240x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\) \(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}\) \(=\dfrac{-5x\left(x-y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{-5x^2+5xy}{10x\left(x-2y\right)\left(x+2y\right)}\) Câu trả lời: 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: \(x\ne0;x\ne\pm2y\) \(\dfrac{17}{5x}=\dfrac{17\cdot2\left(x-2y\right)\left(x+2y\right)}{5x\cdot2\cdot\left(x-2y\right)\left(x+2y\right)}=\dfrac{34\left(x^2-4y^2\right)}{10x\left(x-2y\right)\left(x+2y\right)}\) \(\dfrac{24}{x-2y}=\dfrac{24\cdot10x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{240x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\) \(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}\) \(=\dfrac{-5x\left(x-y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{-5x^2+5xy}{10x\left(x-2y\right)\left(x+2y\right)}\)

Quy đồng mẫu thức a)

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Trần Hữu Minh Trí

18 tháng 11 2023

Quy đồng mẫu thức a)

\frac{7}{4 x + 16}

74�+16 và

\frac{11}{x^{2} - 16}

11�2−16b)

\frac{6}{(x + 3)^{3}}

6(�+3)3 và

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

�−32�(�+3)2 c)

\frac{- 6}{1 - x}

−61−� ;

\frac{3 x}{x^{2} + x + 1}

3��2+�+1 và

\frac{x^{2} - 3 x + 5}{x^{3} - 1}

�2−3�+5�3−1 d)

\frac{17}{5 x}

175� ;

\frac{24}{x - 2 y}

24�−2� và

\frac{x - y}{8 y^{2} - 2 x^{2}}

#Hỏi cộng đồng OLM #Toán lớp 8

Nguyễn Lê Phước Thịnh

18 tháng 11 2023

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: \(x\ne0;x\ne\pm2y\)

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

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

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

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