\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

\(a.x=-0,6\)

\(c.x=-11,6\)

Pt nhju ak!!!

a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\) 

    \(x\)        = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)

    \(x\)        = \(\dfrac{52}{63}\)

b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)

     2\(x\).(2\(x\) + 1) = 30

     4\(x^2\)+ 2\(x\)  - 30 = 0

  4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0

   (4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0

   4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0

        2 (\(x\) + 3).(2\(x\) -  5) = 0

            \(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)

             \(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}

c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\) = \(\dfrac{85}{x}\)

     \(\dfrac{85}{x}\)         = \(\dfrac{85}{24}\)

       \(x\)           = 24

a) \(\left(2x+3\right)^3=\left(2x+3\right)^8\)

TH1 \(2x+3=1\)

\(2x=1-3=-2\)

\(x=-1\)

TH2 \(2x+3=0\)

\(2x=-3\Rightarrow x=-\frac{3}{2}\)

b) ? sai đề

c) \(\left|5-3\right|=\left|11+2x\right|\Rightarrow\left|2\right|=\left|11+2x\right|\)

\(\hept{\begin{cases}11+2x=-2\\11+2x=2\end{cases}\Rightarrow}\hept{\begin{cases}2x=13\\2x=9\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{13}{2}\\x=\frac{9}{2}\end{cases}}\)

d) \(\left(x-5\right)^4=\left(x-5\right)^6\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=6\end{cases}}\)

Cảm ơn bạn. nhưng đúng ko z bn

Dũng Lê Trí giúp mik bài này cái

b) (6-2x)⁹ = (6-2x)³¹

^{Giài nhanh giùm mik nha}

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)

\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)

\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)

\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)

\(-x^3=27\)

\(x=-3\)

Bài 1:

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(6x-9+4-2x=-3\)

\(4x=-2\)

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

b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)

\(2x^3-4x+x^2-2x^3-x^2=-12\)

\(-4x=-12\)

\(x=\frac{1}{3}\)

Bài 2:

a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(6x^2+9x-6x^2-15x+4x+10=8\)

\(-2x=8\)

\(x=-4\)

b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)

\(4x^2-4x-3x^2+15-x^2=-7\)

\(-4x=-22\)

\(x=\frac{11}{2}\)

c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)

\(6x-4x-10-12x+6x+12=-6\)

\(-4x=-8\)

\(x=2\)