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1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)
\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)
\(\Rightarrow2x-1=\frac{-4}{63}\)
\(\Rightarrow2x=-\frac{4}{63}+1\)
\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)
a: =>-3/2+x-7=5-1/3x+4/15
=>4/3x=413/30
hay x=413/40
b: \(\Leftrightarrow5-\dfrac{3}{2}x=-\dfrac{22}{3}\cdot\dfrac{-11}{8}=\dfrac{121}{12}\)
=>3/2x=-61/12
hay x=-61/18
c: (3x+2)2+|3x+2y|=0
=>3x+2=0 và 3x=-2y
=>x=-2/3 và -2y=-2
=>(x,y)=(-2/3;1)
1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
Vì \(2\ne3\) mà \(\left(3x+1\right)^2=\left(3x+1\right)^3\) nên
\(\left[{}\begin{matrix}3x+1=-1\\3x+1=0\\3x+1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-\dfrac{1}{3}\\x=0\end{matrix}\right.\)
Vậy...........
Chúc bạn học tốt!!!
Ta có : \(\left(3x+1\right)^2=\left(3x+1\right)^3\)
\(9x^2+1=27x^2+1\)
\(1-1=27x^2-9x^2\)
\(0=\left(27-9\right)x^2\)
\(16x^2=0\)
\(x^2=0\)
x = \(0\)
Vậy x = 0
\(\left(3x-1\right)^3=25\left(3x-1\right)\\ \Leftrightarrow\left(3x-1\right)^2=25\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\\ \left(3x-14\right)^3=2^5\cdot5^2+200\\ \Leftrightarrow\left(3x-14\right)^3=1000=10^3\\ \Leftrightarrow3x-14=10\Leftrightarrow x=8\)
\(\left(3x-1\right)^3=25\left(3x-1\right)\)
\(\Rightarrow\left(3x-1\right)\left(9x^2-6x+1-25\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(9x^2-6x-24\right)=0\)
\(\Rightarrow3\left(3x-1\right)\left(x-2\right)\left(3x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(\left(3x-14\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(3x-14\right)^3=1000\)
\(\Rightarrow3x-14=10\Rightarrow3x=24\Rightarrow x=8\)