tìm x,biết
a)0,(31)+x=0,3(7) b)0,(4).x=\(\dfrac{5}{6}\)
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23 tháng 3 2022
a, \(x=-\dfrac{5}{6}-\dfrac{7}{12}=\dfrac{-10-7}{12}=-\dfrac{17}{12}\)
b, \(\dfrac{2}{9}-x=-\dfrac{4}{3}.\dfrac{5}{6}=-\dfrac{24}{18}=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{2}{9}+\dfrac{4}{3}=\dfrac{14}{9}\)
c, \(-3=x-1\Leftrightarrow x=-2\)
d, \(\dfrac{3}{5}x-\dfrac{2}{3}=\dfrac{4}{5}:\dfrac{1}{5}=4\Leftrightarrow\dfrac{3}{5}x=4+\dfrac{2}{3}=\dfrac{14}{3}\Leftrightarrow x=\dfrac{14}{3}:\dfrac{3}{5}=\dfrac{70}{9}\)
TM
6 tháng 3 2021
a/ \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{37}{70}\)
Vậy....
b/ \(x+\dfrac{4}{5}=-\dfrac{5}{12}\cdot\dfrac{3}{25}\)
\(x+\dfrac{4}{5}=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}-\dfrac{4}{5}=-\dfrac{17}{20}\)
Vậy....
c/ \(\dfrac{x}{182}=-\dfrac{6}{12}\cdot\dfrac{35}{91}\)
\(\dfrac{x}{182}=-\dfrac{5}{26}\)
\(=>x\cdot26=-5\cdot182\)
\(26x=-910\)
\(x=-910:26=-35\)
Vậy....
6 tháng 3 2021
a) Ta có: \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{7}{70}+\dfrac{30}{70}\)
hay \(x=\dfrac{37}{70}\)
Vậy: \(x=\dfrac{37}{70}\)
20 tháng 11 2021
a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
N
20 tháng 11 2021
\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
22 tháng 7 2023
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
CD
1
AH
Akai Haruma
Giáo viên
31 tháng 10 2023
Lời giải:
a.
$0< x< \frac{1}{4}+\frac{4}{5}$
$\Rightarrow 0< x< \frac{21}{20}$ hay $0< x< 1,05$
$\Rightarrow x=1$
b.
$\frac{4}{7}+\frac{3}{7}< x< \frac{5}{3}+\frac{2}{3}$
$\Rightarrow 1< x< \frac{7}{3}$
$\Rightarrow x=2$
CD
1
24 tháng 12 2023
`#3107.101107`
`1.`
`a,`
`(2x - 3)^2 = |3 - 2x|`
`=> (2x - 3)^2 = |2x - 3|`
`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
Vậy, `x \in {3/2; 2; 1}`
`b,`
`(x - 1)^2 + (2x - 1)^2 = 0`
`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
`c,`
`5 - x^2 = 1`
`=> x^2 = 4`
`=> x^2 = (+-2)^2`
`=> x = +-2`
Vậy, `x \in {-2; 2}`
`d,`
`x - 2\sqrt{x} = 0`
`=> x^2 - (2\sqrt{x})^2 = 0`
`=> x^2 - 4x = 0`
`=> x(x - 4) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x \in {0; 4}`
`g,`
`(x - 1) + 1/7 = 0`
`=> x - 1 + 1/7 = 0`
`=> x - 6/7 = 0`
`=> x = 6/7`
Vậy, `x = 6/7.`
KK
5 tháng 9 2021
a. (x - 22) - 1 = 0
<=> x - 4 - 1 = 0
<=> x = 5
b. 4 - (x - 2)2 = 0
<=> 22 - (x - 2)2 = 0
<=> (2 - x + 2)(2 + x - 2) = 0
<=> x(4 - x) = 0
<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
KK
5 tháng 9 2021
d. (3x - 2)2 - (2x + 3)2 = 5(x + 4)(x - 4)
<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x2 - 16)
<=> (x - 5)(5x + 1) = 5x2 - 80
<=> 5x2 + x - 25x - 5 = 5x2 - 80
<=> 5x2 - 5x2 + x - 25x = -80 + 5
<=> -24x = -75
<=> x = \(\dfrac{25}{8}\)
LO
1
18 tháng 6 2023
a: =>x+5>0 và x-2<0
=>-5<x<2
=>x thuộc {-4;-3;...;1}
b: =>(x-5)(x+5)>0
=>x>5 hoặc x<-5
=>x thuộc Z\{-5;-4;-3;...;3;4;5}
c: =>(x+6)(x-7)>0
=>x>7 hoặc x<-6
a) \(0,\left(31\right)+x=0,3\left(7\right)\\ \Rightarrow\dfrac{31}{99}+x=\dfrac{17}{45}\\ \Rightarrow x=\dfrac{17}{45}-\dfrac{31}{99}=\dfrac{32}{495}=0,0\left(64\right)\)
Vậy \(x=0,0\left(64\right)\)
b) \(0,\left(4\right)\cdot x=\dfrac{5}{6}\\ \Rightarrow\dfrac{4}{9}\cdot x=\dfrac{5}{6}\\ \Rightarrow x=\dfrac{5}{6}:\dfrac{4}{9}\\ \Rightarrow x=\dfrac{5}{6}\cdot\dfrac{9}{4}\\ \Rightarrow x=\dfrac{15}{8}=1,875\)
Vậy \(x=1,875\)