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\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)
\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)
\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)
\(\Rightarrow x=3\)
\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Leftrightarrow25-5x=6x-8\)
\(\Leftrightarrow-5x-6x=-8-25\)
\(\Leftrightarrow-11x=-33\)
\(\Leftrightarrow x=3\)
Vậy x = 3
Từ đề bài ta có:
\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}\)
\(=50\).
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)
BÀI 1:
a)
\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}\\ =\dfrac{6}{12}-\dfrac{8}{12}+\dfrac{9}{12}\\ =\dfrac{6-8+9}{12}\\ =\dfrac{7}{12}\)
b)
\(-4\dfrac{1}{2}+1,2\cdot\left(-5\right)-30\%\\ =\dfrac{-9}{2}+\dfrac{6}{5}\cdot\left(-5\right)-\dfrac{3}{10}\\ =\dfrac{-9}{2}+\left(-6\right)-\dfrac{3}{10}\\ =\dfrac{-45}{10}+\dfrac{-60}{10}-\dfrac{3}{10}\\ =\dfrac{\left(-45\right)+\left(-60\right)-3}{10}\\ =\dfrac{-108}{10}\\ =\dfrac{54}{5}\)
c)
\(\dfrac{-7}{9}\cdot\dfrac{6}{13}+\dfrac{-7}{9}\cdot\dfrac{7}{13}+5\dfrac{7}{9}\\ =\dfrac{-7}{9}\cdot\left(\dfrac{6}{13}+\dfrac{7}{13}\right)+5\dfrac{7}{9}\\ =\dfrac{-7}{9}\cdot1+5\dfrac{7}{9}\\ =\dfrac{-7}{9}+5\dfrac{7}{9}\\ =5\)
BÀI 2
a)
\(\dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{2}{3}:\dfrac{3}{2}\\ \dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{4}{9}\\ \dfrac{3}{2}x=\dfrac{4}{9}+\dfrac{2}{3}\\ \dfrac{3}{2}x=\dfrac{10}{9}\\ x=\dfrac{10}{9}:\dfrac{3}{2}\\ x=\dfrac{20}{27}\)
b)
\(\left(\dfrac{9}{11}-x\right):\left(\dfrac{-10}{11}\right)=1-\dfrac{4}{5}\\ \left(\dfrac{9}{11}-x\right):\left(\dfrac{-10}{11}\right)=\dfrac{1}{5}\\ \dfrac{9}{11}-x=\dfrac{1}{5}\cdot\left(\dfrac{-10}{11}\right)\\ \dfrac{9}{11}-x=\dfrac{-2}{11}\\ x=\dfrac{9}{11}-\dfrac{-2}{11}\\ x=1\)
c)
\(\left(1,2x-\dfrac{4}{7}\right):\dfrac{4}{7}=75\%\\ \left(\dfrac{6}{5}x-\dfrac{4}{7}\right):\dfrac{4}{7}=\dfrac{3}{4}\\ \dfrac{6}{5}x:\dfrac{4}{7}-\dfrac{4}{7}:\dfrac{4}{7}=\dfrac{3}{4}\\ \dfrac{6}{5}x:\dfrac{4}{7}-1=\dfrac{3}{4}\\ \dfrac{6}{5}x:\dfrac{4}{7}=\dfrac{3}{4}+1\\ \dfrac{6}{5}x:\dfrac{4}{7}=\dfrac{7}{4}\\ \dfrac{6}{5}x=\dfrac{7}{4}\cdot\dfrac{4}{7}\\ \dfrac{6}{5}x=1\\ x=1:\dfrac{6}{5}\\ x=\dfrac{5}{6}\)
\(\left(4x-3\right)\left(\dfrac{3}{5}x+\dfrac{1}{2}\right)=0\)
\(=>4x-3=0\) hoặc \(\dfrac{3}{5}x+\dfrac{1}{2}=0\)
\(=>x=\dfrac{3}{4}\) hoặc x = -5/6
(4x - 3).(\(\dfrac{3}{5}\)x + \(\dfrac{1}{2}\)) = 0
=> TH1: 4x - 3 = 0
=> 4x =3
=> x = loại
=>TH2: (\(\dfrac{3}{5}\)x + \(\dfrac{1}{2}\)) = 0
=> \(\dfrac{3}{5}\) x = \(\dfrac{1}{2}\)
=> x = \(\dfrac{1}{2}\): \(\dfrac{3}{5}\)
=> x = \(\dfrac{5}{6}\)
a, \((\dfrac{-1}{2})\)2 -\(\dfrac{5}{6}\).\((\dfrac{-6}{7})-\dfrac{3}{4}:1\dfrac{2}{3}\)
=\(\dfrac{1}{4}+\dfrac{5}{7}-\dfrac{9}{20}\)
=\(\dfrac{35}{140}+\dfrac{100}{140}-\dfrac{63}{140}\)
=\(\dfrac{72}{140}\)= \(\dfrac{18}{35}\)
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