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(x+3)5/(x+3)2=64/27
=>(x+3)3=64/27
=>(x+3)3=(4/3)3
=>\(\orbr{\begin{cases}x+3=\frac{4}{3}\\x+3=-\frac{4}{3}\end{cases}=>\orbr{\begin{cases}x=-\frac{5}{3}\\x=-\frac{13}{3}\end{cases}}}\)
a)\(\left(x+1\right)^3=-27\)
\(\left(x+1\right)^3=\left(-3\right)^3\)
x+1=-3
x=(-3)-1
x=-4
b)6-3x=8
3x=6-8
3x=(-2)
x=\(-\frac{2}{3}\)
a) \(\left(x+1\right)^3=-27\)
\(\Rightarrow\left(x+1\right)^3=\left(-3\right)^3\)
\(\Rightarrow x-1=-3\)
\(\Rightarrow x=-4\)
Vậy \(x=-4\)
b) \(\sqrt{36}-\sqrt{9}.x=\sqrt{64}\)
\(\Rightarrow6-3.x=8\)
\(\Rightarrow3x=-2\)
\(\Rightarrow x=\frac{-2}{3}\)
Vậy \(x=\frac{-2}{3}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}+\dfrac{1}{2}\)
Vậy : \(x=\dfrac{5}{6}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-x=3\sqrt{3}\\\dfrac{2}{3}-x=-3\sqrt{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2-9\sqrt{3}}{3}\\x=\dfrac{2+9\sqrt{3}}{3}\end{matrix}\right.\)
Sửa đề: \(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)
ĐKXĐ: x<>1/2
\(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)
=>\(\left(2x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)
\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)
\(\Rightarrow3x-12=-2x+4\)
\(\Rightarrow3x+2x=4+12\)
\(\Rightarrow5x=16\)
\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)
\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)
\(\dfrac{x}{3}-2=\dfrac{1}{15}\)
=>\(\dfrac{x}{3}=2+\dfrac{1}{15}=\dfrac{31}{15}\)
=>\(x=\dfrac{31}{15}\cdot3=\dfrac{31}{5}\)
\(ĐK:x\ge0\\ PT\Leftrightarrow\left(x-\dfrac{3}{4}\right)\left(x^2+\dfrac{3}{4}x+\dfrac{9}{16}\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(n\right)\\\sqrt{x}=3\left(n\right)\\x^2+2\cdot\dfrac{3}{8}x+\dfrac{9}{64}+\dfrac{27}{64}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=9\\\left(x+\dfrac{3}{8}\right)^2+\dfrac{27}{64}=0\left(\text{vô nghiệm}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=9\end{matrix}\right.\)