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\(a,\Rightarrow\dfrac{\left(-3\right)^x}{\left(-3\right)^4}=\left(-3\right)^3\\ \Rightarrow\left(-3\right)^{x-4}=\left(-3\right)^3\\ \Rightarrow x-4=3\Rightarrow x=7\\ b,Sửa:\left(x-\dfrac{1}{2}\right)^2=25\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=5\\x-\dfrac{1}{2}=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{5}\\x=-\dfrac{9}{5}\end{matrix}\right.\)
a) Ta có: f(x)=-3
<=>x5-2x2+x4-x5+3x2-x4-3+2x=-3
<=>(x5-x5)+(-2x2+3x2)+(x4-x4)+2x-3=-3
<=>x2+2x-3=-3
<=>x2+2x=0
<=>x(x+2)=0
<=>x=0 hoặc x+2=0
<=>x=0 hoặc x=-2
Vậy..........
b)đa thức f(x) có nghiệm
<=>f(x)=0
<=>x2+2x-3=0
<=>x2+3x-x-3=0
<=>x(x+3)-(x+3)=0
<=>(x-1)(x+3)=0
<=>x-1=0 hoặc x+3=0
<=>x=1 hoặc x=-3
Vậy nghiệm của đa thức f(x) là x=-3;x=1
a: =>5x-2=0 hoặc 2x+1/3=0
=>x=-1/6 hoặc x=2/5
b: Đặt x/2=y/3=k
=>x=2k; y=3k
xy=54
=>6k^2=54
=>k^2=9
=>k=3 hoặc k=-3
TH1: k=3
=>x=6; y=9
TH2: k=-3
=>x=-6; y=-9
c: =>5050x=-213
=>x=-213/5050
\(\left(2x-2\right)^3=\left(2x-2\right)^{12}\)
\(\Rightarrow\left(2x-2\right)^{15}\)
\(\Rightarrow2x-2=0\Rightarrow2x=2\Rightarrow x=1\)
\(\left(2x-2\right)^3=\left(2x-2\right)^{12}\)
\(\Rightarrow\left(2x-2\right)^9=1\)
\(\Rightarrow2x-2=1\)
\(\Rightarrow2x=3\Rightarrow x=\dfrac{3}{2}\)
a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)
\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)
\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)
\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)
\(x=\dfrac{-1}{3}\)
b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)
\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)
\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)
\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)
\(2x=\dfrac{-17}{24}\)
\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)
\(x=\dfrac{-17}{48}\)
c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)
a, 1/4 + 3/4 : x = -2
3/4 : x = -2 - 1/4
3/4 : x = -9/4
x = 3/4 : -9/4
x = -1/3
\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)