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a)(3x-2\(^4\)).7\(^3\)=2.7\(^4\)
3x-2\(^4\)=2.7\(^4:7^3\)
3x-16=2.7
3x-16=14
3x=30
=>x=10
Vậy x=10
a) Ta có:\(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
\(\Leftrightarrow3x-16=2\cdot\dfrac{7^4}{7^3}=2\cdot7=14\)
\(\Leftrightarrow3x=30\)
hay x=10
Vậy: x=10
a:x=3/5-7/8=24/40-35/40=-11/40
b: =>1/3:(2x-1)=-1/6
=>2x-1=-2
=>2x=-1
=>x=-1/2
c: =>x-3/4=17/2+7/4=34/4+7/4=41/4
=>x=11
d: =>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-7:2/5=-35/2
a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
3x - 7 = 2x + 5
3x - 2x = 5 + 7
x = 12
|3x - 2| = 7
\(\Rightarrow\left\{\begin{matrix}3x-2=7\\-\left(3x-2\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3x=9\\-3x+2=7\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=3\\-3x=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=3\\x=-\frac{5}{3}\end{matrix}\right.\)
4 - |x - 2| = -3
|x - 2| = 4 - (-3)
|x - 2| = 7
\(\Rightarrow\left\{\begin{matrix}x-2=7\\-\left(x-2\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=9\\-x+2=7\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=9\\-x=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=9\\x=-5\end{matrix}\right.\)
|2x - 3| = x - 1
\(\Rightarrow\left\{\begin{matrix}2x-3=x-1\\-\left(2x-3\right)=x-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}2x-x=-1+3\\-2x+3=x-1\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=2\\-2x-x=-1-3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2\\-3x=-4\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=2\\x=\frac{4}{3}\end{matrix}\right.\)
|3x + 1| = x + 3
\(\Rightarrow\left\{\begin{matrix}3x+1=x+3\\-\left(3x+1\right)=x+3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3x-x=3-1\\-3x-1=x+3\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}2x=2\\-3x-x=3+1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\-4x=4\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(a,x+\frac{3}{4}=\frac{1}{4}\)
\(\Rightarrow x=\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}\)
\(b,\frac{2}{3}x-\frac{3}{7}=\frac{1}{7}\)
\(\Rightarrow\frac{2}{3}x=\frac{1}{7}+\frac{3}{7}\)
\(\Rightarrow\frac{2}{3}x=\frac{4}{7}\)
\(\Rightarrow x=\frac{4}{7}:\frac{2}{3}=\frac{4}{7}\cdot\frac{3}{2}=\frac{6}{7}\)
\(c,1\frac{2}{3}x-50\%x=-1\frac{1}{6}\)
\(\Rightarrow\frac{5}{3}x-\frac{50}{100}x=-\frac{7}{6}\)
\(\Rightarrow\left[\frac{5}{3}-\frac{50}{100}\right]x=-\frac{7}{6}\)
\(\Rightarrow\left[\frac{5}{3}-\frac{1}{2}\right]x=-\frac{7}{6}\)
\(\Rightarrow\frac{7}{6}x=-\frac{7}{6}\Leftrightarrow x=-\frac{7}{6}:\frac{7}{6}=-\frac{7}{6}\cdot\frac{6}{7}=-1\)
a,x+\(\frac{3}{4}\)=\(\frac{1}{4}\)
=>x =\(\frac{1}{4}\)-\(\frac{3}{4}\)
=> x =\(\frac{-1}{2}\)
Vậy x=\(\frac{-1}{2}\)
b,\(\frac{2}{3}\)x-\(\frac{3}{7}\)=\(\frac{1}{7}\)
=>\(\frac{2}{3}\)x =\(\frac{1}{7}\)+\(\frac{3}{7}\)
=>\(\frac{2}{3}\)x =\(\frac{4}{7}\)
=> x =\(\frac{4}{7}\) :\(\frac{2}{3}\)
=> x =\(\frac{4}{7}\).\(\frac{3}{2}\)
=> x =\(\frac{6}{7}\)
Vậy x=\(\frac{6}{7}\)
c,\(\frac{12}{3}\)x-50%x =\(\frac{-11}{6}\)
=>\(\frac{12}{3}\)x-\(\frac{1}{2}\)x=\(\frac{-11}{6}\)
=>x(\(\frac{12}{3}\)-\(\frac{1}{2}\))=\(\frac{-11}{6}\)
=>x(\(\frac{24}{6}\)-\(\frac{3}{6}\))=\(\frac{-11}{6}\)
=>x\(\frac{21}{6}\) =\(\frac{-11}{6}\)
=>x =\(\frac{-11}{6}\):\(\frac{21}{6}\)
=>x =\(\frac{-11}{6}\).\(\frac{6}{21}\)
=>x =\(\frac{-11}{21}\)
Vậy x=\(\frac{-11}{21}\)
a) Ta có:\(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
\(\Leftrightarrow3x-16=2\cdot\dfrac{7^4}{7^3}=2\cdot7=14\)
\(\Leftrightarrow3x=30\)
hay x=10
Vậy: x=10
(3x - 24) . 73 = 2. 74
3x - 24 = 2.74 : 73
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 30 : 3 = 10