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a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
=>x=3/5:2/3=3/5x3/2=9/10
b: \(\Leftrightarrow x\cdot2.8-50=34\)
=>2,8x=84
=>x=30
c: \(\Leftrightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)
hay x=5/2
d: \(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}=\dfrac{41}{4}\)
=>2x-3/4=41/4 hoặc 2x-3/4=-41/4
=>2x=44/4=11 hoặc 2x=-19/2
=>x=11/2 hoặc x=-19/4
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)
\(\Rightarrow-2\le x\le2\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)
\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)
\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)
\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)
\(\Rightarrow x\in\left\{11;12;13;14\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)
Bài 1: Tìm x biết:
a) \(\dfrac{6}{5}-2\left|1-3x\right|=1\dfrac{2}{3}\)
\(2\left|1-3x\right|=\dfrac{6}{5}-1\dfrac{2}{3}\)
\(2\left|1-3x\right|=\dfrac{-7}{15}\)
\(\left|1-3x\right|=\dfrac{-7}{15}:2\)
\(\left|1-3x\right|=\dfrac{-7}{30}\)
\(\left|1-3x\right|\in N\) nhưng \(\dfrac{-7}{30}\notin N\)
\(\Rightarrow x=\varnothing\)
b) \(\left(2,8x+50\right):\dfrac{-3}{2}=51\)
\(\left(2,8x+50\right)=51.\dfrac{-3}{2}\)
\(2,8x+50=\dfrac{-153}{2}\)
\(2,8x=\dfrac{-153}{2}-50\)
\(2,8x=\dfrac{-253}{2}\)
\(x=\dfrac{-253}{2}:2,8\)
\(x=\dfrac{-1265}{28}\)
c) \(\dfrac{x-2}{-2}=\dfrac{x+4}{3}\)
\(\Rightarrow\left(x-2\right).3=-2.\left(x+4\right)\)
\(x.3-2.3=\left(-2\right).x+\left(-2\right).4\)
\(3x-6=\left(-2\right)x+\left(-8\right)\)
\(3x-\left(-2\right)x=6+\left(-8\right)\)
\(5x=-2\)
\(x=\left(-2\right):5\)
\(x=\dfrac{-2}{5}\)
d) \(4\left(3-2x\right)-5\left(x-1\right)=12\)
\(4.3-4.2x-5x+5.1=12\)
\(12-8x-5x+5=12\)
\(12+\left(-8\right)x+\left(-5\right)x+5=12\)
\(12+\left(-13\right)x+5=12\)
\(\left(-13\right)x=12-12-5\)
\(\left(-13\right)x=-5\)
\(x=\left(-5\right):\left(-13\right)\)
\(x=\dfrac{5}{13}\)
Bài 2: Chứng minh:
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\) (đpcm)
bài này áp dụng quy tắc nhân chéo nha :vv
a) \(\dfrac{x}{5}=\dfrac{2}{5}\Leftrightarrow5x=2.5=10\Leftrightarrow x=\dfrac{10}{5}=2\)
b) \(\dfrac{3}{8}=\dfrac{6}{x}\Leftrightarrow3x=6.8=48\Leftrightarrow x=\dfrac{48}{3}=16\)
c)\(\dfrac{1}{9}=\dfrac{x}{27}\Leftrightarrow9x=27\Leftrightarrow x=\dfrac{27}{9}=3\)
d)\(\dfrac{4}{x}=\dfrac{8}{6}\Leftrightarrow8x=4.6=24\Leftrightarrow x=\dfrac{24}{8}=3\)
e) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+4x=-6+20\\ \Leftrightarrow7x=14\Leftrightarrow x=\dfrac{14}{7}=2\)
g) \(\dfrac{x}{-2}=\dfrac{-8}{x}\Leftrightarrow x^2=\left(-2\right)\left(-8\right)=16\\ \Rightarrow x=\pm4\)
\(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{2}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:\dfrac{-2}{3}\)
\(x=\dfrac{1}{10}.\dfrac{-3}{2}\)
\(x=\dfrac{-3}{20}\)