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a)\(123-5:\left(x+4\right)=38\)
\(5:\left(x+4\right)=123-38\)
\(5:\left(x+4\right)=85\)
\(x+4=5:85\)
\(x=\dfrac{1}{17}-4\)
\(x=-\dfrac{67}{17}\)
b)\(70-5.\left(x-3\right)=45\)
\(5.\left(x-3\right)=70-45\)
\(5.\left(x-3\right)=35\)
\(x-3=35:5\)
\(x-3=7\)
\(x=7+3\)
\(x=10\)
a. \(6^2:4.3+2.5^2\)
= \(36:12+2.25\)
= \(3+50\)
=\(53\)
b. \(2.\left(5.4^2-18\right)\)
= \(2.\left(5.16-18\right)\)
= \(2.\left(80-18\right)\)
= \(2.62\)
= \(124\)
c. \(80:\left\{\left[\left(11-2\right).2\right]+2\right\}\)
\(=80:\left\{\left[9.2\right]+2\right\}\)
\(=80:\left\{18+2\right\}\)
\(=80:20\)
\(=4\)
\(\dfrac{3}{1}+\dfrac{3}{3}+\dfrac{3}{6}+...+\dfrac{3}{x\cdot\left(x+1\right):2}=\dfrac{2015}{336}\\ \dfrac{6}{2}+\dfrac{6}{6}+\dfrac{6}{12}+...+\dfrac{6}{x\cdot\left(x+1\right)}=\dfrac{2015}{336}\\ 6\cdot\dfrac{1}{2}+6\cdot\dfrac{1}{6}+6\cdot\dfrac{1}{12}+...+6\cdot\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{2015}{336}\\ =6\cdot\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2015}{336}\\ 6\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2015}{336}\\ 6\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2015}{336}\\ 6\cdot\left(1-\dfrac{1}{x+1}\right)=\dfrac{2015}{336}\\ 1-\dfrac{1}{x+1}=\dfrac{2015}{336}:6\\ 1-\dfrac{1}{x+1}=\dfrac{2015}{2016}\\ \dfrac{1}{x+1}=1-\dfrac{2015}{2016}\\ \dfrac{1}{x+1}=\dfrac{1}{2016}\\ \Rightarrow x+1=2016\\ x=2016-1\\ x=2015\)
\(1,3.\dfrac{15}{39}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(=\dfrac{13}{10}.\dfrac{15}{39}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\)
Câu 2:
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
c: \(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{4}-\dfrac{1}{11}=\dfrac{7}{44}\)
\(4\dfrac{1}{3}\cdot\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\\ \dfrac{13}{3}\cdot\dfrac{-1}{3}\le x\le\dfrac{2}{3}\cdot\dfrac{-11}{12}\\ \dfrac{-13}{9}\le x\le\dfrac{-11}{18}\\ \dfrac{-26}{18}\le x\le\dfrac{-11}{18}\\ \Rightarrow x=-1\)
a: \(=105-96=9\)
b: =225+108=333
c: =-8x9-8x(-27)
\(=-8\left(9-27\right)=144\)
d: \(=1\cdot5+\left(-8\right)\cdot6-\left(-27\right)\cdot7=5-48+189=146\)
a: \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)\)
\(=\dfrac{37}{3}-3-4-\dfrac{3}{2}\)
\(=\dfrac{74-9}{6}-7=\dfrac{65}{6}-7=\dfrac{65-42}{7}=\dfrac{23}{7}\)
b: \(3\dfrac{5}{6}+2\dfrac{1}{6}\cdot6\)
\(=3+\dfrac{5}{6}+\dfrac{13}{6}\cdot6\)
\(=16+\dfrac{5}{6}=\dfrac{101}{6}\)
c: \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
\(=3+\dfrac{1}{2}+4+\dfrac{5}{7}-5-\dfrac{5}{14}\)
\(=2+\dfrac{7+10-5}{14}=2+\dfrac{12}{14}=2+\dfrac{6}{7}=\dfrac{20}{7}\)
d: \(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99+2}{22}=\dfrac{101}{22}\)
các bn giải mk nha