(\(\frac{3}{4}\)-x)\(^3\)=\(\frac{1}{64}\)
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\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{4}{14}-\frac{2}{13}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{2}{6}+\frac{2}{14}-\frac{2}{26}}{\frac{4}{6}+\frac{4}{14}-\frac{4}{26}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{356}}{\frac{4}{4}-\frac{4}{16}+\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)
\(=\frac{2\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}{4\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}\times\frac{3\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{356}\right)}{4\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
\(=\frac{2}{4}\times\frac{3}{4}+\frac{5}{8}\)
\(=\frac{1}{2}\times\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}\)
\(=\frac{8}{8}=1\)
\(\frac{\frac{109}{3.7.13}}{\frac{361}{3.14.13}}\)\(\frac{\frac{153}{256}}{\frac{51}{64}}\)+5/8
=\(\frac{327}{722}\)+5/8
=\(\frac{3113}{2888}\)
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{264}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{3}{4}\left(1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}\right)}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)\(+\frac{5}{8}\)
\(\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
\(B=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
=>\(B=\frac{1.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}{3.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{14}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{\frac{4}{4}-\frac{4}{16}-\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)
=>\(B=\frac{1}{3}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
=>\(B=\frac{1}{3}.\frac{3}{4}+\frac{5}{8}\)
=>\(B=\frac{1}{4}+\frac{5}{8}\)
=>\(B=\frac{2}{8}+\frac{5}{8}\)
=>\(B=\frac{7}{8}\)
l-i-k-e cho mình nhé bạn.
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)
\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)
\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)
\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)
Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)
\(\Rightarrow x+66=0\)
\(\Rightarrow x=0-66=-66\)
Auto làm nốt câu b
a, Cộng cả 2 vế với 2
Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)
\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)
=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)
Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)
=> \(x=-65\)
b , Lm tương tự như Câu a
Chúc bn hok tốt
mình ko bt làm
\(\left(\frac{3}{4}-x\right)^3=\frac{1}{64}\)
\(\Leftrightarrow\left(\frac{3}{4}-x\right)^3=\frac{1}{4}^3\)
\(\Leftrightarrow\frac{3}{4}-x=\frac{1}{4}\)
\(\Leftrightarrow x=\frac{1}{2}\)