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a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\) ĐKXĐ: x \(\ne\) 0
<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)
<=> 15x + 28 = 3x
<=> 15x - 3x = -28
<=> 12x = -28
<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)
b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)
<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)
<=> -5x . 6 = 12 . 2
<=> -30x = 24
<=> x = \(-\dfrac{4}{5}\)
a) Ta có: \(\left|-5\right|+\left|x-1\right|=\left|7\right|\)
\(\Leftrightarrow\left|x-1\right|+5=7\)
\(\Leftrightarrow\left|x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-1\right\}\)
b) Ta có: \(2\cdot\left|2x-4\right|-\left|-4\right|=\left|-50\right|\)
\(\Leftrightarrow4\cdot\left|x-2\right|-4=50\)
\(\Leftrightarrow4\cdot\left|x-2\right|=54\)
\(\Leftrightarrow\left|x-2\right|=\dfrac{27}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\dfrac{27}{2}\\x-2=-\dfrac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{31}{2}\left(loại\right)\\x=-\dfrac{23}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(x\in\varnothing\)
a, | -5 | + | x-1 | = | 7 |
5 + | x - 1 | = 7
| x - 1 | = 2
TH1 x -1 = 2
x = 3
TH2 x -1 = -2
x= -1
a: =>3^x=3^4*3=3^5
=>x=5
b: =>\(2^{x+1}=2^5\)
=>x+1=5
=>x=4
c: \(\Leftrightarrow3^{x+2-3}=3\)
=>x-1=1
=>x=2
d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)
=>x=4 hoặc x=-4
e: (2x-1)^4=81
=>2x-1=3 hoặc 2x-1=-3
=>2x=4 hoặc 2x=-2
=>x=-1 hoặc x=2
f: (2x-6)^4=0
=>2x-6=0
=>x-3=0
=>x=3
a) \(3^x=81\cdot3\)
\(\Rightarrow3^x=3^4\cdot3\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
b) \(2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
c) \(3^{x+2}:27=3\)
\(\Rightarrow3^{x+2}:3^3=3\)
\(\Rightarrow3^{x+2-3}=3\)
\(\Rightarrow3^{x-1}=3\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
d) \(2x^2=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
e) \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f) \(\left(2x-6\right)^4=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)
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....
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: \(x=\dfrac{37}{70}\)
a: =>1/3(2x-5)=-2/3-3/2=-4/6-9/6=-13/6
=>2x-5=-13/6*3=-13/2
=>2x=-3/2
=>x=-3/4
b: =>2/5x=-3/4-1/2=-5/4
=>x=-5/4:2/5=-5/4*5/2=-25/8
a)
\(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{13}{6}\\ \Rightarrow2x-5=-\dfrac{13}{6}:\dfrac{1}{3}=-\dfrac{13}{2}\\ \Rightarrow2x=-\dfrac{13}{2}+5\\ \Rightarrow2x=-\dfrac{3}{2}\\ \Rightarrow x=-\dfrac{3}{2}:2\\ \Rightarrow x=-\dfrac{3}{4}\)
b)
\(\dfrac{2}{5}x+\dfrac{1}{2}=-\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{3}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{5}{4}\\ \Rightarrow x=-\dfrac{5}{4}:\dfrac{2}{5}=-\dfrac{25}{8}\)
Giải:
a)-3/10-(-1/5)+x)=-3/2
-1/5+x =-3/10-(-3/2)
-1/5+x =6/5
x =6/5-(-1/5)
x =7/5
b)-(-x+3/4)-12/8.(-32/15)=-(-1/2)
x-3/4+16/5 =1/2
x-3/4 =1/2-16/5
x =-27/10
x =-27/10+3/4
x =-39/20
c)x-3/x+5=4/3
=>(x-3).3=4.(x+5)
3x-9 =4x+20
3x-4x =20+9
-1x =29
x =-29
Câu b cậu nên tính lại cho kĩ nhé, ấn máy tính dễ nhầm lắm đấy!
Mk phải ấn: -(39/20+3/4)-12/8.-32/15=1/2
Vì x là số âm mà đằng trước x là dấu ''-'' nên -(-39/20)=39/20 ; -(-1/2)=1/2
Chúc bạn học tốt!