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MP
22 tháng 7 2017
\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)
\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)
\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)
\(\Rightarrow x=3\)
Q
22 tháng 7 2017
\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Leftrightarrow25-5x=6x-8\)
\(\Leftrightarrow-5x-6x=-8-25\)
\(\Leftrightarrow-11x=-33\)
\(\Leftrightarrow x=3\)
Vậy x = 3
giải giùm tớ nha
DH
14 tháng 5 2017
Từ đề bài ta có:
\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}\)
\(=50\).
JH
15 tháng 5 2017
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)
27 tháng 8 2017
Ta có: \(\dfrac{1}{2}\cdot y+\dfrac{2}{3}\cdot y=\dfrac{7}{6}\Rightarrow y\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{7}{6}\Rightarrow\dfrac{7}{6}y=\dfrac{7}{6}\Rightarrow y=\dfrac{7}{6}:\dfrac{7}{6}=1\)
Vậy \(D=\left\{1\right\}\)
c
2 tháng 8 2017
a, \((\dfrac{-1}{2})\)2 -\(\dfrac{5}{6}\).\((\dfrac{-6}{7})-\dfrac{3}{4}:1\dfrac{2}{3}\)
=\(\dfrac{1}{4}+\dfrac{5}{7}-\dfrac{9}{20}\)
=\(\dfrac{35}{140}+\dfrac{100}{140}-\dfrac{63}{140}\)
=\(\dfrac{72}{140}\)= \(\dfrac{18}{35}\)
Bài 1:
1: \(A=2+2^2+\cdots+2^{2022}\)
=>\(2A=2^2+2^3+\cdots+2^{2023}\)
=>2A-A=\(2^2+2^3+\cdots+2^{2023}-2-2^2-\cdots-2^{2022}\)
=>\(A=2^{2023}-2\)
2: \(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2020\cdot2022}\right)\)
\(=\left(1+\frac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\frac{1}{\left(3-1\right)\left(3+1\right)}\right)\cdot\ldots\cdot\left(1+\frac{1}{\left(2021-1\right)\left(2021+1\right)}\right)\)
\(=\frac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\frac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot\ldots\cdot\frac{2021^2-1+1}{\left(2021-1\right)\left(2021+1\right)}\)
\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{2021^2}{2020\cdot2022}\)
\(=\frac{2\cdot3\cdot\ldots\cdot2021}{1\cdot2\cdot\ldots\cdot2020}\cdot\frac{2\cdot3\cdot\ldots\cdot2021}{3\cdot4\cdot\ldots\cdot2022}=\frac{2021}{1}\cdot\frac{2}{2022}=\frac{2021}{1011}\)
3: \(C=\frac{14+\frac{7}{15}+\frac74}{2+\frac{1}{15}+\frac14}:\frac{5+\frac{5}{17}+\frac{5}{13}}{6+\frac{6}{17}+\frac{6}{13}}+\frac{5858}{505}\)
\(=\frac{7\left(2+\frac{1}{15}+\frac14\right)}{2+\frac{1}{15}+\frac14}:\frac{5\left(1+\frac{1}{17}+\frac{1}{13}\right)}{6\left(1+\frac{1}{17}+\frac{1}{13}\right)}+\frac{58}{5}\)
\(=7\cdot\frac65+\frac{58}{5}=\frac{42+58}{5}=\frac{100}{5}=20\)