MA
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
BL
19 tháng 12 2016
A = (\(\frac{x}{x+1}\) + \(\frac{1}{x-1}\) ) : (\(\frac{2x+2}{x-1}\) - \(\frac{4x}{x^2-1}\) )
A = (\(\frac{x}{x+1}\) + \(\frac{1}{x-1}\) ) : ( \(\frac{2x+2}{x-1}\) - \(\frac{4x}{\left(x+1\right)\left(x-1\right)}\) )
\(\Rightarrow\) MTC: (x+1)(x-1)
A = ( \(\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\) + \(\frac{x+1}{\left(x+1\right)\left(x-1\right)}\) ) : (\(\frac{2\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\) - \(\frac{4x}{\left(x+1\right)\left(x-1\right)}\) )
A = \(\frac{x^2+1}{\left(x+1\right)\left(x-1\right)}\) : \(\frac{2x^2+2}{\left(x+1\right)\left(x-1\right)}\)
A = \(\frac{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{2\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}\)
A = \(\frac{1}{2}\)
mệt rồi :v ngủ =)))
1 tháng 8 2018
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{264}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{3}{4}\left(1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}\right)}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)\(+\frac{5}{8}\)
\(\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
TN
14 tháng 1 2017
\(VT=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+3}-\frac{1}{x+4}\)
\(=\frac{1}{x}-\frac{1}{x+4}=\frac{x+4-x}{x\left(x+4\right)}=\frac{4}{x\left(x+4\right)}\)
\(\Rightarrow\frac{4}{x\left(x+4\right)}=\frac{m}{x\left(x+4\right)}=VP\Rightarrow m=4\)
26 tháng 2 2020
a/\(=\frac{6x^2+8x+7}{x^3-1}+\frac{x}{x^2+x+1}-\frac{6}{x-1}\)=\(\frac{6x^2+8x+7+x\left(x-1\right)-6\left(x^2+x+1\right)}{x^3-1}=\frac{x^2+x+1}{x^3-1}=\frac{1}{x-1}\)
b/\(=\frac{1-x+x^2-1}{\left(x-1\right)^2}=\frac{x\left(x-1\right)}{\left(x-1\right)^2}=\frac{x}{x-1}\)
CT
8 tháng 12 2019
vt sai đề nâk
từ gt=> xy+yz+xz=0
áp dụng bdt bunhia
=> A>=0
dấu= xr khi x=y=z
-> dấu = k xr
..........
hoặc:
8 tháng 12 2019
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\frac{\Rightarrow1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(\Rightarrow xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)
\(=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{x+1}{x}}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{x}{x+1}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{\frac{x+1+x}{x+1}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{\frac{2x+1}{x+1}}}}=\frac{1}{1+\frac{1}{1+\frac{x+1}{2x+1}}}\)
\(=\frac{1}{1+\frac{1}{\frac{2x+1+x+1}{2x+1}}}=\frac{1}{1+\frac{1}{\frac{3x+2}{2x+1}}}=\frac{1}{1+\frac{2x+1}{3x+2}}=\frac{1}{\frac{3x+2+2x+1}{3x+2}}=\frac{1}{\frac{5x+3}{3x+2}}=\frac{3x+2}{5x+3}\)