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 \(x\in\left\{4;-5\right\}\)

a) x=3,6\(\times\) 27=97,2

a,x:27=3,6

x=3,6:27

x\(\approx\)0,13

\(\Rightarrow\)x=0,1(3)

a) \(3\left(2x-5\right)+125=134\)

\(\Leftrightarrow3\left(2x-5\right)=9\)

\(\Leftrightarrow2x-5=3\)

\(\Leftrightarrow2x=8\Leftrightarrow x=4\)

b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)

\(\Leftrightarrow6x+9=27\)

\(\Leftrightarrow6x=18\Leftrightarrow x=3\)

d) \(27\left(x-27\right)-27=0\)

\(\Leftrightarrow27\left(x-27\right)=27\)

\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)

3.(x+5)-2x=27
3.x+3.5-2x=27
3x+15-2x=27
x.(3-2)+15=27
x.  1    +15=27
  x       +15=27
  x             =27-15
  x             =12

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 $x=-5$

`=(3x-1)^2+2(2x-1)(2x+5)+(2x+5)^2-(2x-3)(4x^2+6x+9):(2x-3)`

`=(3x-1)^2+2(2x-1)(2x+5)+(2x+5)^2-(2x-3)(4x^2+6x+9):(2x-3)`

`=(3x-1+2x+5)^2-(4x^2+6x+9)`

`=(5x+4)^2-(4x^2+6x+9)`

`=25x^2+40x+16-4x^2-6x-9`

`=21x^2+34x+7`

Ta có: \(\left(3x-1\right)^2-2\left(1-3x\right)\left(2x+5\right)+\left(5+2x\right)^2-\left(8x^3-27\right):\left(2x-3\right)\)

\(=\left(3x-1+2x+5\right)^2-\left(4x^2+6x+9\right)\)

\(=\left(5x+4\right)^2-\left(4x^2+6x+9\right)\)

\(=25x^2+40x+16-4x^2-6x-9\)

\(=21x^2+34x+7\)

a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.

ĐKXĐ: \(-\frac{5}{2}\le x\le2\)

Ta có \(VT=27\left(\sqrt{5+2x}+\sqrt{4-2x}\right)\ge27\sqrt{5+2x+4-2x}=81\)

Xét vế phải với hàm \(f\left(x\right)=\left(4x+1\right)^2\) trên \(\left[-\frac{5}{2};2\right]\)

\(f\left(-\frac{1}{4}\right)=0\) ; \(f\left(-\frac{5}{2}\right)=81\) ; \(f\left(2\right)=81\)

\(\Rightarrow0\le f\left(x\right)\le81\Rightarrow VP\le81\)

\(\Rightarrow VT\ge VP\) \(\forall x\in\left[-\frac{5}{2};2\right]\)

Vậy nghiệm của BPT là \(-\frac{5}{2}\le x\le2\)

 \(\frac{2x-1}{3}=\frac{27}{2x-1}\)

\(\Rightarrow\left(2x-1\right)\left(2x-1\right)=27\cdot3\)

\(\Rightarrow\left(2x-1\right)^2=81\)

\(\Rightarrow\left(2x-1\right)^2=\left(\pm9\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

ko bt đề đúng ý bn chưa ?

\(\frac{2x-1}{3}=\frac{27}{2x-1}\)

\(\Leftrightarrow\left(2x-1\right)^2=27.3=81\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-4\end{cases}}}\)

Mơn nhiều ạ...

Sửa đề: \(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

ĐKXĐ: x<>1/2

\(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

=>\(\left(2x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)