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a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
Giải:
a) \(\dfrac{12}{16}=\dfrac{-x}{4}=\dfrac{21}{y}=\dfrac{z}{80}\)
\(\Rightarrow x=\dfrac{12.-4}{16}=-3\)
\(\Rightarrow y=\dfrac{16.21}{12}=28\)
\(\Rightarrow z=\dfrac{12.80}{16}=60\)
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)\) =0
\(\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(x.\left(\dfrac{1}{3}+\dfrac{2}{5}\right)\) \(=0+\dfrac{2}{5}\)
\(x.\dfrac{11}{15}\) \(=\dfrac{2}{5}\)
x \(=\dfrac{2}{5}:\dfrac{11}{15}\)
x \(=\dfrac{6}{11}\)
c) (2x-3)(6-2x)=0
⇒2x-3=0 hoặc 6-2x=0
x=3/2 hoặc x=3
d) \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}\)
\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}\)
\(2x-5=\dfrac{-13}{2}\)
\(2x=\dfrac{-13}{2}+5\)
\(2x=\dfrac{-3}{2}\)
\(x=\dfrac{-3}{2}:2\)
\(x=\dfrac{-3}{4}\)
e) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}:2\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{8}\) hoặc \(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-1}{8}\)
\(x=\dfrac{11}{12}\) hoặc \(x=\dfrac{5}{12}\)
a/ => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)
=> \(\dfrac{1}{x}=\dfrac{2}{5}\)
=> x = 5/2
b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)
=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)
=> \(x=\dfrac{2}{5}\)
c/ => | x + 1| = 10/21
=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)
d/ => \(5x+5=6x-3\)
=> x = 8
1,
a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)
b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)
=\(\dfrac{-2}{15}\)
2,
a, 2x+19=25
=>x=3
b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)
=>x=\(\dfrac{-3}{2}\)
Bài 1:
a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)
\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)
\(=\dfrac{-5}{12}\)
b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)
\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)
4-x=5 hoặc 4-x=-5
x=-1 hoặc x=9
e: =>2/7-x=2/5
=>7-x=5
=>x=2
f: =>2x+3/3=10/3
=>2x+3=10
=>2x=7
=>x=7/2
g: =>(14+x)/7=15/7
=>x+14=15
=>x=1
h: =>(2x+3)/x=13/x
=>2x+3=13
=>2x=10
=>x=5
\(\dfrac{2x-1}{12}=\dfrac{5}{3}\)
\(\Rightarrow\left(2x-1\right)3=12\cdot5\)
\(\Rightarrow6x-3=60\)
\(\Rightarrow6x=63\)
\(\Rightarrow x=\dfrac{21}{2}\)
\(\dfrac{2x-3}{15}=\dfrac{3}{5}\)
\(\Rightarrow\left(2x-3\right)5=15\cdot3\)
\(\Rightarrow10x-15=45\)
\(\Rightarrow10x=60\)
\(\Rightarrow x=6\)
#Đạt Đang Bận Thở
\(\dfrac{2x-1}{12}=\dfrac{20}{12}\)
\(\Leftrightarrow2x-1=20\)
\(\Leftrightarrow2x=21\)
\(\Leftrightarrow x=\dfrac{21}{2}\)
b. \(\dfrac{2x-3}{15}=\dfrac{9}{15}\)
\(\Leftrightarrow2x-3=9\)
\(\Leftrightarrow x=6\)
b, 92 : 4 - 27 = \(\dfrac{x+350}{x}\) + 315
23 - 27 = 1 + \(\dfrac{350}{x}\) + 315
316 + \(\dfrac{350}{x}\) = -4
\(\dfrac{350}{x}\) = - 316 - 4
\(\dfrac{350}{x}\) = -320
-320 \(x\) = 350
\(x\) = 350: (-320)
\(x\) = - \(\dfrac{35}{32}\) (loại)
Vậy \(x\) \(\in\) \(\varnothing\)
c, 720 : [ 41 - (2\(x\) - 5)] = 23.5
41 - (2\(x\) - 5) = 720 : (23.5)
41 - 2\(x\) + 5 = 18
46 - 2\(x\) =18
2\(x\) = 46 - 18
2\(x\) = 28
\(x\) = 28: 2
\(x\) = 14
Vậy \(x\) = 14
\(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)
\(\Rightarrow\dfrac{2x+19}{21}-\dfrac{2x+17}{23}-\dfrac{2x+7}{33}+\dfrac{2x+5}{35}=0\)
\(\Rightarrow\left(\dfrac{2x+19}{21}+1\right)-\left(\dfrac{2x+17}{23}+1\right)-\left(\dfrac{2x+7}{33}+1\right)+\left(\dfrac{2x+5}{35}+1\right)=0\)
\(\Rightarrow\dfrac{2x+40}{21}-\dfrac{2x+40}{23}-\dfrac{2x+40}{33}+\dfrac{2x+40}{35}=0\)
\(\Rightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)
\(\Rightarrow2x+40=0\Rightarrow x=-20\)( do \(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}>0\))
Ta có: \(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)
\(\Leftrightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow2x+40=0\)
hay x=-20