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

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

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!!!

`2//(5x-8)-3(4x-5)=4(3x-4)`

`<=>5x-8-12x+15=12x-16`

`<=>-19x=-23`

`<=>x=23/19`     Vậy `x=23/19`

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

`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`

`<=>4x=8`

`<=>x=2`     Vậy `x=2`

`A`

3(x+5) -3 = 2(x+1) +7

⇒3x+15-3=2x+2+7

⇒3x+12=2x+9

⇒x=-3

15-(x+2)² = -1

⇒(x+2)²=16

⇒\(\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

75-(2x+1)³ =11

a) ta có: 3(x+5)-3=2(x+1)+7

nên 3x-2x=2+7-15+3

hay x=-3

b) Ta có: \(15-\left(x+2\right)^2=-1\)

nên \(\left(x+2\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

c) Ta có: \(75-\left(2x+1\right)^3=11\)

\(\Leftrightarrow2x+1=4\)

hay \(x=\dfrac{3}{2}\)

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) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)

\(\Leftrightarrow3x+15-3=2x+2+7\)

\(\Leftrightarrow3x+12=2x+9\)

hay x=-3

b) Ta có: \(15-\left(x+2\right)^2=-1\)

\(\Leftrightarrow\left(x+2\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

c) Ta có: \(75-\left(2x+1\right)^3=11\)

\(\Leftrightarrow\left(2x+1\right)^3=64\)

\(\Leftrightarrow2x+1=4\)

hay \(x=\dfrac{3}{2}\)

a) Ta có: $3\left(x+5\right)-3=2\left(x+1\right)+7$

$\iff 3x+15-3=2x+2+7$

$\iff 3x+12=2x+9$

hay x=-3

b) Ta có: $15-{}^{}=-1$

$\iff {}^{}=16$

⇔[x+2=4x+2=−4⇔[x=2x=−6⇔[x+2=4x+2=−4⇔[x=2x=−6

c) Ta có: $75-{}^{}$

`3(x+5)-3=2(x+1)+7`

`3x+15-3=2x+2+7`

`x=-3`

.

`15-(x+2)^2=-1`

`(x+2)^2=16`

`(x+2)^2=4^2=(-4)^2`

`[(x+2=4),(x+2=-4):}`

`[(x=2),(x=-6):}`

.

`75-(2x+1)^3=11`

`(2x+1)^3=64`

`(2x+1)^2=4^3`

`2x+1=4`

`x=3/2`

$3(x+5)-3=2(x+1)+7$

$3x+15-3=2x+2+7$

$x=-3$

.

$15-{}^{}=-1$

${}^{}=16$

${}^{}={}^{}={}^{}$

$[\begin{array}{c}x+2=4\\ x+2=-4\end{array}$

$[\begin{array}{c}x=2\\ x=-6\end{array}$

a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)

\(\Leftrightarrow3x-2x=2+7-15+3\)

hay x=-3

b) Ta có: \(15-\left(x+2\right)^2=-1\)

\(\Leftrightarrow\left(x+2\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

- Mình có thể giúp bài 1 :)

 5^x +  5^ ( x + 2 ) = 650

5^x +  5^x . 5² = 650

 5^x .( 1 + 25 ) = 650

 5^x . 26 = 650 

5^x = 650 : 26

 5^x = 25

 5^x = 5² 

=> x = 2 

Vậy x = 2 

The Reflection Of Light~ Ủa đề  là 5^x-5^x+2= 650 sao ông lại đổi thành 5^x-5^(x+2) zậy