Bài 1: Rút gọn
1) \(x^2-y^2 \over 6x^2y^2
\)÷ \(x+y \over 12xy\)
2) \(5x \over 2x+1
\) ÷ \(3x(x-1) \over 4x^2-1\)
3)( \(2x-1\over 2x+1
\)-\(2x-1\over 2x+1
\)) ÷ \(4x \over 10x-5
\)
4) \(2\over 9x^2+6x+1
\)- \(3x \over 9x^2-1
\)
5) (\(5\over x^2+2x+1 \)+\(2x \over x^2-1
\)) ÷ \(2x^2+7x-5 \over 3x-3\)
6) (\(3\over x-3 \)+ \(2x \over x^2-9
\) + \(x\over x+3
\)) ÷ \(2x\over x+3\)
7) (\(3\over x^2-9
\)+\(1\over x^2+3x
\)-\(1\over x^2-3x
\)) ÷ \(x-2\over...
Đọc tiếp
Bài 1: Rút gọn
1) \(x^2-y^2 \over 6x^2y^2 \)÷ \(x+y \over 12xy\)
2) \(5x \over 2x+1 \) ÷ \(3x(x-1) \over 4x^2-1\)
3)( \(2x-1\over 2x+1 \)-\(2x-1\over 2x+1 \)) ÷ \(4x \over 10x-5 \)
4) \(2\over 9x^2+6x+1 \)- \(3x \over 9x^2-1 \)
5) (\(5\over x^2+2x+1 \)+\(2x \over x^2-1 \)) ÷ \(2x^2+7x-5 \over 3x-3\)
6) (\(3\over x-3 \)+ \(2x \over x^2-9 \) + \(x\over x+3 \)) ÷ \(2x\over x+3\)
7) (\(3\over x^2-9 \)+\(1\over x^2+3x \)-\(1\over x^2-3x \)) ÷ \(x-2\over 2x^2+6x\)
1)
ĐK: \(x,y\neq 0\); \(x+y\neq 0\)
\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)
\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)
2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)
\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)
\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)
3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)
\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)
4) ĐK: \(x\neq \frac{\pm 1}{3}\)
\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)
\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)
\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)
5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)
\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)
\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{3}{(x+1)^2}\)