\(\frac{\left(1x3x5x...x99\right)x\left(2x2x2x2...2x2\left\{5\:thưso2\right\}\right)}{51x52x53x...x100}\)
Rút gọn biểu thức (51x52x53x.....x100)/ (1x3x5x...x99)
Tìm x trong biểu thức sau:
\(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{8x9}+\frac{1}{9x10}\right)x100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
Tìm y :
\(\frac{9}{10}x100-\left(\frac{5}{2}:\left(y+\frac{206}{100}\right)\right):\frac{1}{2}=89\)
\(\frac{9}{10}.100-\left(\frac{5}{2}\left(y+\frac{206}{100}\right)\right):\frac{1}{2}=89\)
\(90-\left(\frac{5}{2}\left(y+\frac{103}{50}\right)\right)=89.\frac{1}{2}\)
\(90-\left(\frac{5}{2}\left(y+\frac{103}{50}\right)\right)=\frac{89}{2}\)
\(\frac{5}{2}\left(y+\frac{103}{50}\right)=90-\frac{89}{2}\)
\(\frac{5}{2}\left(y+\frac{103}{50}\right)=\frac{180}{2}-\frac{89}{2}\)
\(\frac{5}{2}\left(y+\frac{103}{50}\right)=\frac{91}{2}\)
\(y+\frac{103}{50}=\frac{91}{2}.\frac{2}{5}\)
\(y+\frac{103}{50}=\frac{91}{5}\)
\(y=\frac{91}{5}-\frac{103}{50}\)
\(y=\frac{910}{50}-\frac{103}{50}\)
\(y=\frac{807}{50}\)
Bài 2 :
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
b, \(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{\left(x-20\right)}=\frac{-3}{4}\)
\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(\frac{2}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
bằng 13,34590301 ( mình bấm máy tính bạn nhé :) )
Tính A = \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+5\right)}+\frac{8}{\left(x+5\right)\left(x+6\right)}\)
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
Tìm x biết: \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Ta có: \(\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{1}{x+2}-\frac{1}{x+5}\); \(\frac{5}{\left(x+5\right)\left(x+10\right)}=\frac{1}{x+5}-\frac{1}{x+10}\)
\(\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{1}{x+10}-\frac{1}{x+17}\);
=> Phương trình tương đương:
\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)<=> \(\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
<=> \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
=> x=15
Đáp số: x=15