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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) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
d: =>-x-5/6=7/12-4/12=3/12=1/4
=>-x=1/4+5/6=13/12
hay x=-13/12
e: =>x+3=-5
hay x=-8
f: =>4,5-2x=-1/2
=>2x=5
hay x=5/2
a: =>11(x-3)=6(x-5)
=>11x-33=6x-30
=>5x=3
=>x=3/5
b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3
=>-9/4x+11/12=8/3
=>-9/4x=32/12-11/12=21/12=7/4
=>x=-7/9
c: =>1/2x-1/3-2/3x-1=x
=>-1/6x-4/3=x
=>-7/6x=4/3
=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7
d: =>1-2x-3x+1=7/2
=>-5x=3/2
=>x=-3/10
`a)1/2+[-1]/[-3]-5/12 < 2x < 12/[-31]+136/31`
`186/372+124/372-155/372 < [744x]/372 < [-144]/372+1632/372`
`186+124-155 < 744x < -144+1632`
`155 < 744x < 1488`
`155:744 < 744x:744 < 1488:744`
`5/24 < x < 2`
Vậy `5/24 < x < 2`
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`b)[-2]/5 < x/15 < 1/6`
`[-12]/30 < [2x]/30 < 5/30`
`-12 < 2x < 5`
`-12:2 < 2x:2 < 5:2`
`-6 < x < 5/2`
Vậy `-6 < x < 5/2`
Giải:
a) x - \(\dfrac{9}{25}\)= \(\dfrac{16}{25}\)
x = \(\dfrac{16}{25}\)+\(\dfrac{9}{25}\)
x = \(\dfrac{25}{25}\)
x = 1
b) \(\dfrac{-12}{30}\)<\(\dfrac{x}{30}\)<\(\dfrac{5}{30}\)
=> x có thể bằng \(\dfrac{-11}{30}\) đến \(\dfrac{4}{30}\)
=> x bằng -5; -4; -3; -2; -1;0;1;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 1/3 - 25% - 5/12) + 2x=1,6 : 3/5`
`(4/3 - 1/4 - 5/12) +2x=8/5 : 3/5`
`2/3 + 2x = 8/3`
`2x=2`
`x=1`
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: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)
\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)
\(\dfrac{2x-1}{12}=\dfrac{5}{3}\)
\(\Rightarrow\left(2x-1\right)3=12\cdot5\)
\(\Rightarrow6x-3=60\)
\(\Rightarrow6x=63\)
\(\Rightarrow x=\dfrac{21}{2}\)
\(\dfrac{2x-3}{15}=\dfrac{3}{5}\)
\(\Rightarrow\left(2x-3\right)5=15\cdot3\)
\(\Rightarrow10x-15=45\)
\(\Rightarrow10x=60\)
\(\Rightarrow x=6\)
#Đạt Đang Bận Thở
\(\dfrac{2x-1}{12}=\dfrac{20}{12}\)
\(\Leftrightarrow2x-1=20\)
\(\Leftrightarrow2x=21\)
\(\Leftrightarrow x=\dfrac{21}{2}\)
b. \(\dfrac{2x-3}{15}=\dfrac{9}{15}\)
\(\Leftrightarrow2x-3=9\)
\(\Leftrightarrow x=6\)