\(\left(\dfrac{2}{3}x-1\right).\left(0,4x+5\right)=0\)
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29 tháng 10 2021
\(1,\Leftrightarrow3^x\left(1+3^2\right)=810\\ \Leftrightarrow3^x=\dfrac{810}{10}=81=3^4\\ \Leftrightarrow x=4\\ 2,\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=1\\0,4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{25}{2}\end{matrix}\right.\)
2 tháng 8 2018
\(a,\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=-\dfrac{64}{125}\)
\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)
\(\dfrac{3}{5}-\dfrac{2}{3}x=-\dfrac{4}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{4}{5}-\dfrac{3}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{7}{5}\)
\(x=\dfrac{21}{10}\)
\(b,\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
\(x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(x=\dfrac{2}{3}\)
\(c,\left(0,4x-1,3\right)^2=5,29\)
\(\left(0,4x-1,3\right)^2=2,3^2=\left(-2,3\right)^2\)
TH1: \(0,4x-1,3=2,3\)
\(0,4x=3,6\)
\(x=9\)
TH2: \(0,4x-1,3=-2,3\)
\(0,4x=-1\)
\(x=-\dfrac{5}{2}\)
=.= hok tốt!!
2 tháng 5 2017
a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
x=\(\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
b) \(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\)
\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)
\(x=\dfrac{2}{3}:-\dfrac{22}{3}\)
\(x=-\dfrac{1}{11}\)
Vậy \(x=-\dfrac{1}{11}\).
c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\)
\(60\%x=\dfrac{19}{9}\)
\(\dfrac{3}{5}x=\dfrac{19}{9}\)
\(x=\dfrac{19}{9}:\dfrac{3}{5}\)
\(x=\dfrac{95}{27}\)
Vậy \(x=\dfrac{95}{27}\).
d) \(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)
\(\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\)
\(\dfrac{2}{3}-x=\dfrac{3}{20}\)
\(x=\dfrac{2}{3}-\dfrac{3}{20}\)
\(x=\dfrac{31}{60}\)
Vậy \(x=\dfrac{31}{60}\).
e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\)
\(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\)
\(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\)
\(-2x=-\dfrac{73}{20}\)
\(x=-\dfrac{73}{20}:\left(-2\right)\)
\(x=\dfrac{73}{40}\)
Vậy \(x=\dfrac{73}{40}\).
28 tháng 8 2023
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
29 tháng 10 2021
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
5 tháng 1 2021
a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)
b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)
c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)
5 tháng 4 2017
1) \(-\dfrac{5}{9}+1\dfrac{5}{9}\cdot\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =-\dfrac{5}{9}+\dfrac{14}{9}\cdot\dfrac{7}{20}\cdot\dfrac{1}{49}\\ =-\dfrac{5}{9}+\dfrac{1}{90}=\dfrac{-49}{90}\)
2) \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{104}{195}+25\%\right)\cdot\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right)\cdot\dfrac{24}{47}-\dfrac{51}{13}\cdot\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =-\dfrac{4}{13}\)
3) \(1\dfrac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}\\ =\dfrac{7}{5}-\dfrac{2}{5}\\ =1\)
Tìm x : 1) \(60\%x+0,4x+x:3=2\\ \Leftrightarrow\dfrac{3}{5}x+\dfrac{2}{5}x+\dfrac{1}{3}x=2\\ \Leftrightarrow\dfrac{4}{3}x=2\\ \Leftrightarrow x=\dfrac{3}{2}\)
5 tháng 4 2017
Nốt nè bn
\(-2x-\dfrac{-3}{5}:\left(0,5\right)^2=-1\dfrac{1}{4}\\ \Leftrightarrow-2x+\dfrac{3}{5}:\dfrac{1}{4}=-\dfrac{5}{4}\\ \Leftrightarrow-2x+\dfrac{12}{5}=-\dfrac{5}{4}\\ \Leftrightarrow-2x=-\dfrac{73}{20}\\ x=-\dfrac{73}{40}\)
\(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{2}{3}-\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{31}{60}\)
26 tháng 12 2021
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
16 tháng 12 2022
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{2}{5}x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5:\dfrac{2}{5}=-\dfrac{25}{2}\end{matrix}\right.\)