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Bài 1:
\(\Leftrightarrow-\dfrac{5}{7}:x=-\dfrac{7}{18}-\dfrac{1}{6}=\dfrac{-7}{18}-\dfrac{3}{18}=\dfrac{-10}{18}=\dfrac{-5}{9}\)
=>x=5/9:5/7=7/9
Bài 2:
a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}=\dfrac{1000}{2}=500\)
b: \(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\)
\(=-\dfrac{1}{1000}\)
Bài 2:
a) \(\frac{4}{9}+x=\frac{-5}{3}\)
\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)
\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)
Vậy: \(x=\frac{-19}{9}\)
b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)
\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)
\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)
c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)
\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-5\right\}\)
Bài 1:
a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)
\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)
\(=\frac{2+3}{12}=\frac{5}{12}\)
b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{-7}{10}\)
\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)
Bài 2:
a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)
\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)
\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)
\(\Leftrightarrow-2x=-6-1=-7\)
hay \(x=\frac{7}{2}\)
Vậy: \(x=\frac{7}{2}\)
\(=\left(1-\frac{1}{10}\right).\left(1-\frac{2}{10}\right)...\left(1-\frac{10}{10}\right)...\left(1-\frac{15}{10}\right)=\left(1-\frac{1}{10}\right).\left(1-\frac{2}{10}\right)...0...\left(1-\frac{15}{10}\right)=0\)
a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
\(\frac{\frac{25}{108}.\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\frac{-41}{21}}\)=\(\frac{\frac{5755}{108}+\frac{187}{4}}{\frac{-973}{410}}\)=\(\frac{\frac{8531}{84}}{\frac{-973}{410}}\)=-241,0180
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(=\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right)\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\left(-\frac{41}{21}\right)}\)
\(=\frac{\frac{25}{108}\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}\cdot\left(-\frac{21}{41}\right)}=\frac{\frac{2701}{27}}{-\frac{973}{410}}\)
Tính nốt vì số dữ quá , lần sau để số ít thôi
