A=ba số hạng đầu
\(A=\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+6}=\frac{1}{x}-\frac{1}{x+6}\\ \)
B=3 số hạng tiếp theo
\(2B=\frac{1}{x+6}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}+\frac{1}{x+10}=\frac{1}{x+6}\)
\(A+B=\frac{1}{x}-\frac{1}{x+6}+\frac{1}{2\left(x+6\right)}=\frac{1}{x}-\frac{1}{2\left(x+6\right)}=\frac{12+x}{2x\left(x+6\right)}\)
1. \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
2 . \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
3 . \(4\left(3x-2\right)-3\left(x-4\right)=7x+10\)
4. \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
Tính A = \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+5\right)}+\frac{8}{\left(x+5\right)\left(x+6\right)}\)
tính(rút gọn)
a,\(\left(x+3-\frac{1}{x+3}\right)\left(x+\frac{3}{x+4}\right)\)
b,\(\left(2x-4-\frac{x-12}{3x+4}\right)\left(3x-2-\frac{10}{2x+1}\right)\)
c,\(\left(2x-8-\frac{x+10}{3x+1}\right)\left(x-6-\frac{x-6}{3x+2}\right)\)
d,\(\left(1+\frac{1}{x}\right):\left(1-\frac{1}{x^2}\right)\)
Tìm x biết \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+4\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
Giải phương trình:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{^{x^2}}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
1 . \(\left(3x-2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
2 . \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
3 . \(4\left(0,5-1,5x\right)=\frac{5x-6}{3}\)
4 . \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
5 . \(\frac{8x^2}{3\left(1-4x^2\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
A) \(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
B)\(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)
Giải các phương trình trên :
ĐS: a) x=8 b) x= -9
Tính nhanh tổng sau:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\)
Chứng minh: \(\frac{2}{x^2-1}+\frac{4}{x^2-4}+...+\frac{20}{x^2-100}=\frac{11}{\left(x-10\right)\left(x+1\right)}+\frac{11}{\left(x-9\right)\left(x+2\right)}+...+\frac{11}{\left(x-1\right)\left(x+10\right)}\)
