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1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)
\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)
\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)
\(15-20x+6x-12=0\)
\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn
a) ĐKXĐ: \(x\ne-1;x\ne2\)
Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
⇔\(\frac{1}{x+1}-\frac{5}{x-2}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(x-2-5x-5+15=0\)
⇔\(-4x+8=0\)
⇔\(-4x=-8\)
⇔\(x=\frac{-8}{-4}=2\)(loại)
Vậy: x không có giá trị
b) ĐKXĐ: \(x\ne0;x\ne\frac{3}{2}\)
Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
⇔\(\frac{x}{\left(2x-3\right)\cdot x}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)
⇔\(x-3-10x+15=0\)
⇔\(-9x+12=0\)
⇔\(-9x=-12\)
⇔\(x=\frac{-12}{-9}=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
c) ĐKXĐ:\(x\ne3;x\ne1\)
Ta có: \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2\left(x-3\right)}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{4}{x-3}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}-\frac{4}{x-3}=0\)
⇔\(\frac{6}{x-1}-\frac{8}{x-3}=0\)
⇔\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=0\)
⇔\(6\left(x-3\right)-8\left(x-1\right)=0\)
⇔6x-18-8x+8=0
⇔-2x-10=0
⇔-2(x+5)=0
Vì 2≠0 nên x+5=0
hay x=-5
Vậy: x=-5
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
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