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a/\(x:27=3,6\)
\(\Rightarrow x=97,2\)
b/\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
\(\Rightarrow\left(2x+1\right)^2=81\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{4;-5\right\}\)
Lời giải:
a.
$x:27=-2:3,6=\frac{-5}{9}$
$x=27.\frac{-5}{9}=-15$
b.
$\frac{2x+1}{-27}=\frac{-3}{2x+1}$
$\Rightarrow (2x+1)^2=(-27)(-3)=81=9^2=(-9)^2$
$\Rightarrow 2x+1=9$ hoặc $2x+1=-9$
$\Rightarrow x=4$ hoặc $x=-5$
\(\frac{2x-1}{3}=\frac{27}{2x-1}\)
\(\Rightarrow\left(2x-1\right)\left(2x-1\right)=27\cdot3\)
\(\Rightarrow\left(2x-1\right)^2=81\)
\(\Rightarrow\left(2x-1\right)^2=\left(\pm9\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
ko bt đề đúng ý bn chưa ?
\(\frac{2x-1}{3}=\frac{27}{2x-1}\)
\(\Leftrightarrow\left(2x-1\right)^2=27.3=81\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-4\end{cases}}}\)
Sửa đề: \(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)
ĐKXĐ: x<>1/2
\(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)
=>\(\left(2x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
\(7\left(x+1\right)+3x=27\)
\(7x+7+3x=27\)
\(10x=20\)
\(x=2\)
\(\left(x+2\right)\left(3-2x\right)+x=2x^2-3\)
\(3x+6-2x^2-4x+x=2x^2-3\)
\(-2x^2-2x^2=-3-6\)
\(-4x^2\)=\(=-9\)
\(x^2=\dfrac{9}{4}\)
\(=>x\in\left\{\dfrac{3}{2};\dfrac{-3}{2}\right\}\)
\(7\left(x+1\right)+3x=27\\ \Leftrightarrow7x+7+3x=27\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\)
Vậy x = 2
\(\left(x+2\right)\left(3-2x\right)+x=2x^2-3\\ \Leftrightarrow3x-4x-2x^2+6+x=2x^2-3\\ \Leftrightarrow-2x^2+6=2x^2-3\\ \Leftrightarrow4x^2=9\\ \Leftrightarrow x^2=\dfrac{9}{4}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\)
\(a,\left(1-2x\right)^3=-8\)
\(\Rightarrow1-2x=-2\)
\(\Rightarrow2x=3\)
\(\Rightarrow=\frac{3}{2}\)
\(b,\left(2x-1\right)^3=-27\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
\(\Rightarrow x=-1\)
\(\frac{2x-1}{3}=\frac{-27}{1-2x}\)
\(\Rightarrow\left(2x-1\right)\left(1-2x\right)=\left(-27\right).3\)
\(\Rightarrow2x-4x^2-1-2x=-81\)
\(\Rightarrow4x^2-1=-81\)
\(\Rightarrow4x^2=-81+1\)
\(\Rightarrow4x^2=-80\)
Vì \(x^2\ge0\)mà \(4x^2=-80\)
\(\Rightarrow x\in\theta\)
tíc mình nha
Đề là: \(2x+\dfrac{1}{-27}=\dfrac{-3}{2x}+1\)
Hay: \(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\) vậy em?
ĐKXĐ: x<>-1/2
\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
=>\(\left(2x+1\right)^2=\left(-27\right)\cdot\left(-3\right)=81\)
=>\(\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=8\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)