1)\(\left(\frac{x^2-2}{x^2+2x}+\frac{1}{x+2}\right)\colon\frac{x+1}{x}\) 2)\(\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right)\colon\left(1-\frac{x}{x+2}\right)\) 3)\(\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right)\colon\frac{x+1}{x-2}\) 4)\(\left(\frac{2x}{x-3}+\frac{x}{x+3}+\frac{2x^2+3x+1}{9-x^2}\right)\colon\frac{x-1}{x+3}\) 5)\(\left(\frac{x}{x+3}-\frac{2x}{3-x}+\frac{3x^2+9}{9-x^2}\right)\colon\frac{3}{x-3}\)...
Đọc tiếp
1)\(\left(\frac{x^2-2}{x^2+2x}+\frac{1}{x+2}\right)\colon\frac{x+1}{x}\)
2)\(\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right)\colon\left(1-\frac{x}{x+2}\right)\)
3)\(\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right)\colon\frac{x+1}{x-2}\)
4)\(\left(\frac{2x}{x-3}+\frac{x}{x+3}+\frac{2x^2+3x+1}{9-x^2}\right)\colon\frac{x-1}{x+3}\)
5)\(\left(\frac{x}{x+3}-\frac{2x}{3-x}+\frac{3x^2+9}{9-x^2}\right)\colon\frac{3}{x-3}\)
6)\(\left(\frac{1}{x+2}+\frac{5}{x-2}+\frac{4}{x^2-4}\right)\colon\frac{6}{x+3}\)
1. đkxđ: \(x \neq 0;\ x \neq -2;\ x \neq -1\)
\(\left(\frac{x^2-2}{x^2+2x}+\frac{1}{x+2}\right):\frac{x+1}{x}=\left[\frac{x^2-2}{x(x+2)}+\frac{x}{x(x+2)}\right]\cdot\frac{x}{x+1}\)
\(=\frac{x^2+x-2}{x(x+2)}\cdot\frac{x}{x+1}=\frac{(x-1)(x+2)}{x(x+2)}\cdot\frac{x}{x+1}\)
\(= \frac{x-1}{x} \cdot \frac{x}{x+1} = \frac{x-1}{x+1}\)
2. đkxđ: \(x \neq \pm 2;\ x \neq -1\)
\(\left( \frac{x}{x^2-4} + \frac{1}{x+2} - \frac{2}{x-2} \right) : \left( 1 - \frac{x}{x+2} \right)\)
\(= \left[ \frac{x}{(x-2)(x+2)} + \frac{x-2}{(x-2)(x+2)} - \frac{2(x+2)}{(x-2)(x+2)} \right] : \left( \frac{x+2-x}{x+2} \right)\)
\(= \frac{x + x - 2 - 2x - 4}{(x-2)(x+2)} : \frac{2}{x+2}\)
\(=\frac{-6}{(x-2)(x+2)}\cdot\frac{x+2}{2}\)
\(= \frac{-3}{x-2}\)
3. đkxđ: \(x \neq 0;\ x \neq \pm 2;\ x \neq -1\)
\(\left( \frac{4x}{x^2+2x} + \frac{2}{x-2} - \frac{6-5x}{4-x^2} \right) : \frac{x+1}{x-2}\)
\(= \left[ \frac{4}{x+2} + \frac{2}{x-2} + \frac{6-5x}{(x-2)(x+2)} \right] : \frac{x+1}{x-2}\)
\(= \frac{4(x-2) + 2(x+2) + 6 - 5x}{(x-2)(x+2)} \cdot \frac{x-2}{x+1}\)
\(= \frac{4x - 8 + 2x + 4 + 6 - 5x}{(x-2)(x+2)} \cdot \frac{x-2}{x+1}\)
\(= \frac{x + 2}{(x-2)(x+2)} \cdot \frac{x-2}{x+1}\)
\(= \frac{1}{x-2} \cdot \frac{x-2}{x+1} = \frac{1}{x+1}\)
4. đkxđ: \(x \neq \pm 3;\ x \neq 1\)
\(\left( \frac{2x}{x-3} + \frac{x}{x+3} + \frac{2x^2+3x+1}{9-x^2} \right) : \frac{x-1}{x+3}\)
\(= \left[ \frac{2x}{x-3} + \frac{x}{x+3} - \frac{2x^2+3x+1}{(x-3)(x+3)} \right] : \frac{x-1}{x+3}\)
\(= \frac{2x(x+3) + x(x-3) - (2x^2+3x+1)}{(x-3)(x+3)} \cdot \frac{x+3}{x-1}\)
\(= \frac{2x^2 + 6x + x^2 - 3x - 2x^2 - 3x - 1}{(x-3)(x+3)} \cdot \frac{x+3}{x-1}\)
\(= \frac{x^2 - 1}{(x-3)(x+3)} \cdot \frac{x+3}{x-1}\)
\(= \frac{(x-1)(x+1)}{(x-3)(x+3)} \cdot \frac{x+3}{x-1} = \frac{x+1}{x-3}\)
5. đkxđ: \(x \neq \pm 3\)
\(\left( \frac{x}{x+3} - \frac{2x}{3-x} + \frac{3x^2+9}{9-x^2} \right) : \frac{3}{x-3}\)
\(= \left[ \frac{x}{x+3} + \frac{2x}{x-3} - \frac{3x^2+9}{(x-3)(x+3)} \right] : \frac{3}{x-3}\)
\(= \frac{x(x-3) + 2x(x+3) - (3x^2+9)}{(x-3)(x+3)} \cdot \frac{x-3}{3}\)
\(= \frac{x^2 - 3x + 2x^2 + 6x - 3x^2 - 9}{(x-3)(x+3)} \cdot \frac{x-3}{3}\)
\(= \frac{3x - 9}{(x-3)(x+3)} \cdot \frac{x-3}{3}\)
\(= \frac{3(x-3)}{(x-3)(x+3)} \cdot \frac{x-3}{3} = \frac{x-3}{x+3}\)
6. đkxđ: \(x \neq \pm 2;\ x \neq -3\)
\(\left( \frac{1}{x+2} + \frac{5}{x-2} + \frac{4}{x^2-4} \right) : \frac{6}{x+3}\)
\(= \left[ \frac{1}{x+2} + \frac{5}{x-2} + \frac{4}{(x-2)(x+2)} \right] : \frac{6}{x+3}\)
\(= \frac{(x-2) + 5(x+2) + 4}{(x-2)(x+2)} \cdot \frac{x+3}{6}\)
\(= \frac{x - 2 + 5x + 10 + 4}{(x-2)(x+2)} \cdot \frac{x+3}{6}\)
\(= \frac{6x + 12}{(x-2)(x+2)} \cdot \frac{x+3}{6}\)
\(= \frac{6(x+2)}{(x-2)(x+2)} \cdot \frac{x+3}{6} = \frac{x+3}{x-2}\)
Dưới đây là lời giải rút gọn cho từng câu.
1.
\(\left(\right. \frac{x^{2} - 2}{x^{2} + 2 x} + \frac{1}{x + 2} \left.\right) : \frac{x + 1}{x}\)Ta có
\(x^{2} + 2 x = x \left(\right. x + 2 \left.\right) .\)Quy đồng:
\(\frac{x^{2} - 2}{x \left(\right. x + 2 \left.\right)} + \frac{x}{x \left(\right. x + 2 \left.\right)} = \frac{x^{2} + x - 2}{x \left(\right. x + 2 \left.\right)} = \frac{\left(\right. x + 2 \left.\right) \left(\right. x - 1 \left.\right)}{x \left(\right. x + 2 \left.\right)} = \frac{x - 1}{x} .\)Chia cho \(\frac{x + 1}{x}\):
\(\frac{x - 1}{x} \cdot \frac{x}{x + 1} = \boxed{\frac{x - 1}{x + 1}} .\)2.
\(\left(\right. \frac{x}{x^{2} - 4} + \frac{1}{x + 2} - \frac{2}{x - 2} \left.\right) : \left(\right. 1 - \frac{x}{x + 2} \left.\right)\)Ta có
\(x^{2} - 4 = \left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right) .\)Quy đồng:
\(\frac{x}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} + \frac{x - 2}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} - \frac{2 \left(\right. x + 2 \left.\right)}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)}\) \(= \frac{x + x - 2 - 2 x - 4}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} = \frac{- 6}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} .\)Mặt khác
\(1 - \frac{x}{x + 2} = \frac{2}{x + 2} .\)Do đó
\(\frac{- 6}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} \cdot \frac{x + 2}{2} = \boxed{- \frac{3}{x - 2}} .\)3.
\(\left(\right. \frac{4 x}{x^{2} + 2 x} + \frac{2}{x - 2} - \frac{6 - 5 x}{4 - x^{2}} \left.\right) : \frac{x + 1}{x - 2}\)Ta có
\(\frac{4 x}{x \left(\right. x + 2 \left.\right)} = \frac{4}{x + 2} ,\)và
\(4 - x^{2} = - \left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right) .\)Suy ra
\(- \frac{6 - 5 x}{4 - x^{2}} = \frac{6 - 5 x}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} .\)Quy đồng:
\(\frac{4 \left(\right. x - 2 \left.\right) + 2 \left(\right. x + 2 \left.\right) + 6 - 5 x}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} = \frac{x + 2}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} = \frac{1}{x - 2} .\)Chia:
\(\frac{1}{x - 2} \cdot \frac{x - 2}{x + 1} = \boxed{\frac{1}{x + 1}} .\)4.
\(\left(\right. \frac{2 x}{x - 3} + \frac{x}{x + 3} + \frac{2 x^{2} + 3 x + 1}{9 - x^{2}} \left.\right) : \frac{x - 1}{x + 3}\)Ta có
\(9 - x^{2} = - \left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right) .\)Quy đồng:
\(\frac{2 x \left(\right. x + 3 \left.\right) + x \left(\right. x - 3 \left.\right) - \left(\right. 2 x^{2} + 3 x + 1 \left.\right)}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)}\) \(= \frac{x^{2}}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} .\)Chia:
\(\frac{x^{2}}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} \cdot \frac{x + 3}{x - 1} = \boxed{\frac{x^{2}}{\left(\right. x - 3 \left.\right) \left(\right. x - 1 \left.\right)}} .\)5.
\(\left(\right. \frac{x}{x + 3} - \frac{2 x}{3 - x} + \frac{3 x^{2} + 9}{9 - x^{2}} \left.\right) : \frac{3}{x - 3}\)Đổi dấu:
\(\frac{1}{3 - x} = - \frac{1}{x - 3} , 9 - x^{2} = - \left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right) .\)Nên
\(- \frac{2 x}{3 - x} = \frac{2 x}{x - 3} ,\) \(\frac{3 x^{2} + 9}{9 - x^{2}} = - \frac{3 \left(\right. x^{2} + 3 \left.\right)}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} .\)Quy đồng:
\(\frac{x \left(\right. x - 3 \left.\right) + 2 x \left(\right. x + 3 \left.\right) - 3 \left(\right. x^{2} + 3 \left.\right)}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{9 \left(\right. x - 1 \left.\right)}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} .\)Chia:
\(\frac{9 \left(\right. x - 1 \left.\right)}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} \cdot \frac{x - 3}{3} = \boxed{\frac{3 \left(\right. x - 1 \left.\right)}{x + 3}} .\)6.
\(\left(\right. \frac{1}{x + 2} + \frac{5}{x - 2} + \frac{4}{x^{2} - 4} \left.\right) : \frac{6}{x + 3}\)Ta có
\(x^{2} - 4 = \left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right) .\)Quy đồng:
\(\frac{x - 2 + 5 \left(\right. x + 2 \left.\right) + 4}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} = \frac{6 \left(\right. x + 2 \left.\right)}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)} = \frac{6}{x - 2} .\)Chia:
\(\frac{6}{x - 2} \cdot \frac{x + 3}{6} = \boxed{\frac{x + 3}{x - 2}} .\)Kết quả cuối cùng