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\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
a: =>2(2x-3)-9=5-3x-2
=>4x-6-9=-3x+3
=>4x-15=-3x+3
=>7x=18
=>x=18/7
b: =>\(\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\dfrac{21}{3x}+2\)
=>\(\dfrac{23}{3x}=\dfrac{4}{5}+2+\dfrac{1}{4}=\dfrac{61}{20}\)
=>3x=460/61
=>x=460/183
4: \(\Leftrightarrow3^{x+4}\cdot\dfrac{1}{3}-4\cdot3^x=3^{16}\left(1-4\cdot3^3\right)\)
=>\(3^x\cdot27-4\cdot3^x=3^{16}\cdot\left(-107\right)\)
=>3^x*23=3^16*(-107)
=>\(x\in\varnothing\)
2: \(\Leftrightarrow2^x\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)=2^{10}\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)\)
=>2^x=2^10
=>x=10
3: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>8^x=8^9
=>x=9
1: \(\Leftrightarrow3^x\cdot\left(4\cdot\dfrac{1}{9}+2\cdot3\right)=3^4\left(4+2\cdot3^3\right)\)
=>3^x=3^4*3^2
=>x=4+2=6
a)Ta có: \(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}x\right)\)
\(\Leftrightarrow\dfrac{x}{2}-\dfrac{3x-13}{5}=\dfrac{-7}{5}-\dfrac{7x}{10}\)
\(\Leftrightarrow\dfrac{5x}{10}-\dfrac{2\left(3x-13\right)}{10}=\dfrac{-14}{10}-\dfrac{7x}{10}\)
\(\Leftrightarrow5x-6x+26=-14-7x\)
\(\Leftrightarrow-x+26+14+7x=0\)
\(\Leftrightarrow6x=-40\)
hay \(x=-\dfrac{20}{3}\)
d) Ta có: \(\dfrac{2x-3}{2}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}+\dfrac{-9}{6}=\dfrac{5-3x}{6}-\dfrac{2}{6}\)
\(\Leftrightarrow6x-9-9=5-3x-2\)
\(\Leftrightarrow6x-18-3+3x=0\)
\(\Leftrightarrow x=\dfrac{7}{3}\)
\(4.3^x+3^{x+1}=63\)
\(\Rightarrow4.3^x+3.3^x=63\)
\(\Rightarrow7.3^x=63\Rightarrow3^x=9=3^2\Rightarrow x=2\)
\(9.\left(\dfrac{2}{3}\right)^{x+2}-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\left(\dfrac{2}{3}\right)^2\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\dfrac{4}{9}^{ }.\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\left(4-1\right)=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\dfrac{1}{3}=\dfrac{4}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=4\)
mà \(0< \left(\dfrac{2}{3}\right)^x< 1;4>0;x>0\)
\(\Rightarrow x\in\varnothing\)
Ta có: \(\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(\Leftrightarrow2\left(2x-3\right)-9=5-3x-2\)
\(\Leftrightarrow4x-15=-3x+3\)
\(\Leftrightarrow4x+3x=3+15\)
\(\Leftrightarrow x=\dfrac{18}{7}\)
\(\Leftrightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{7}{6}\)
\(\Leftrightarrow\dfrac{4x-6}{6}-\dfrac{5-3x}{6}=\dfrac{7}{6}\)
\(\Leftrightarrow4x-6-5+3x=7\)
\(\Leftrightarrow7x=7+5+6=18\)
\(\Leftrightarrow x=\dfrac{18}{7}\)
Vậy ...
a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10
=1/2x^5-3x^4-5x^3
b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x
=-15x^5+12x^4-9x^3+9x^2
c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)
d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
\(\Leftrightarrow2\left(2x-3\right)-9=5-3x-2\)
=>4x-6-9=-3x+3
=>7x=18
hay x=18/7
\(-3\left(\dfrac{x}{3}-\dfrac{1}{4}\right)-\dfrac{1}{3}\left(3x+\dfrac{1}{2}\right)=7x\\ \Rightarrow-\dfrac{3x}{3}+\dfrac{3}{4}-\dfrac{3x}{3}-\dfrac{1}{6}=7x\\ \Rightarrow-x+\dfrac{3}{4}-x-\dfrac{1}{6}=7x\\ \Rightarrow-x+\dfrac{3}{4}-x-\dfrac{1}{6}-7x=0\\ \Rightarrow\left(-x-x-7x\right)+\left(\dfrac{3}{4}-\dfrac{1}{6}\right)=0\\ \Rightarrow\left[x\left(-1-1-7\right)\right]+\left(\dfrac{3.3}{4.3}-\dfrac{1.2}{6.2}\right)=0\\ \Rightarrow-9x+\left(\dfrac{9}{12}-\dfrac{2}{12}\right)=0\\ \Rightarrow-9x+\dfrac{7}{12}=0\\ \Rightarrow-9x=\dfrac{-7}{12}\\ \Rightarrow x=-\dfrac{7}{12}:\left(-9\right)\\ \Rightarrow x=\dfrac{-7}{12}\cdot\dfrac{-1}{9}\\ \Rightarrow x=\dfrac{7}{108}\)
|3x-3/2|=3
3x-3/2=3 hoặc 3x-3/2=-3
TH1: 3x-3/2=3
3x=3+3/2
3x=9/2
x=9/2:3
x=3/2
TH2: 3x-3/2=-3
3x=-3+3/2
3x=-3/2
x=-3/2:3
x=-1/2
Vậy x=3/2 hoặc x=-1/2
|3x|=3+3/2 , |3x|=9/2 ,|x|=9/2 : 3, |x|=3/2,[x=3/2
[x=(-3/2)