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\(x-0,27=\frac{\frac{73}{77}+\frac{73}{165}+\frac{73}{285}}{25.\left(\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{1}{77}+\frac{1}{165}+\frac{1}{285}\right)}{25.4.\left(\frac{5}{4.84}+\frac{3}{4.180}+\frac{4}{4.285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{5}{336}+\frac{1}{240}+\frac{1}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{2}{105}+\frac{1}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73}{100}\)
\(x=\frac{73}{100}+\frac{27}{100}\)
\(x=1\)
Vậy x=1
x-0,27=(73/77+73/165+73/185):[25.12/133)
x-0,27=(73/77+73/165+73/185):[25/1.12/133)
x-0,27=(73/77+73/165+73/185):300/133
x-0,27=1387/777:300/133
x-0,27=0,791381381
x=0,791381381-0,27
x=0,521381381
Tìm x :
x - 0,27 = \(\frac{73}{100}\)
x = \(\frac{73}{100}+0,27\)
x = 1
Cậu P khó quá mik chưa nghĩ ra cách tính nhanh nhất !
Cậu tự giải nhé !
Hok tốt
\(x-0,27=\frac{\frac{73}{77}+\frac{73}{165}+\frac{73}{285}}{28.\left(\frac{5}{84}+\frac{3}{285}+\frac{4}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{1}{77}+\frac{1}{165}+\frac{1}{285}\right)}{25.4.\left(\frac{5}{4.84}+\frac{3}{4.180}+\frac{4}{4.285}\right)}\)
\(x-\frac{27}{100}=\frac{73,\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{5}{366}+\frac{3}{240}+\frac{4}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{2}{105}+\frac{1}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73}{100}\)
\(x=\frac{27}{100}+\frac{73}{100}\)
\(\Rightarrow x=1\)
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
bài 1 :\(\frac{3}{8}-\frac{1}{5}+\frac{3}{40}=\frac{1}{4}\)
\(\frac{9}{7}\cdot\left(\frac{3}{7}-\frac{1}{2}\right)=-\frac{9}{98}\)
\(-\frac{3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot-\frac{3}{7}\cdot\frac{3}{7}=\left(\frac{4}{9}+\frac{5}{9}-1\right)\cdot-\frac{3}{7}=-1\cdot-\frac{3}{7}=\frac{3}{7}\)
bài 2: \(x+\frac{2}{5}=\frac{9}{10}\)
\(x=\frac{9}{10}-\frac{2}{5}\)
\(x=\frac{1}{2}\)
x = \(\frac{9}{10}\)- \(\frac{2}{5}\)
x =\(\frac{9}{10}\) - \(\frac{4}{10}\)
x = \(\frac{5}{10}\) = \(\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
ai thấy tớ đúng thì ủng hộ nha
tui đang âm
a,\(\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x};Đkxđ:x\ne1\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}\left(\frac{-9}{20}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{2-2x}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{-2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{7}{2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow70=-6\left(x-1\right)\)
\(\Rightarrow6x=6-70\)
\(\Rightarrow6x=-64\)
\(\Rightarrow x=\frac{-32}{3}x\ne1\)