3x+2 / 5x+7  = 3x-1 / 5x + 1

=> (3X+2)( 5X+1) = ( 3x-1) ( 5x+7) 

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

15x^2 + 3x + 10x + 2 = 15x^2 + 21x - 5x - 7

15x^2+ 3x + 10x - 15x^2 - 21x + 5x = -7-2

13x - 16x = -9

-3x = -9

x = 3

Vậy x=3

Đúng 1000%

đề bài là j

3x+2 / 5x+7  = 3x-1 / 5x + 1

=> (3X+2)( 5X+1) = ( 3x-1) ( 5x+7) 

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

15x^2 + 3x + 10x + 2 = 15x^2 + 21x - 5x - 7

15x^2+ 3x + 10x - 15x^2 - 21x + 5x = -7-2

13x - 16x = -9

-3x = -9

x = 3

\(\left(3x-1\right)^5=32\\ \Rightarrow\left(3x-1\right)^5=2^5\\ \Rightarrow3x-1=2\\ \Rightarrow x=1\)

(3x - 1)^5 = 2^5

3x -1 =2

3x= 2+1

3x =3

x =1

`(3x - 1)^5 = 32`

`<=> (3x - 1)^5 = 2^5`

`<=> 3x - 1 = 2`

`<=> 3x = 3`

`<=> x = 1`

a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)

=>-4x=5

hay x=-5/4

b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

=>42x=41

hay x=41/42

`a)(10x+9)x-(5x-1)(2x+3)=8`

`<=>10x^2+9x-10x^2-15x+2x+3=8`

`<=>-4x=5`

`<=>x=-5/4`     Vậy `S={-5/4}`

`b)(3x-5)(7-5x)+(5x+2)(3x-2)-2=0`

`<=>21x-15x^2-35+25x+15x^2-10x+6x-4-2=0`

`<=>42x=41`

`<=>x=41/42`       Vậy `S={41/42}`

a: 

=>-4x=5

hay x=-5/4

b: 

=>42x=41

hay x=41/42

= 3x^2-2x+2

k mình nha

a) 4x + 3x = 217

  x( 4 + 3 ) = 217

  7x = 217

   x = 217 : 7 = 31

Vậy x = 31
b) 9x - 3x = 216

    ( 9 -3)x = 216

     6x = 216

   x = 216:6 = 36

Vậy x = 36

c) 6x - 3x + 23 = 230

( 6 - 3 )x = 230 - 23

    3x = 207

     x = 207 : 3 = 69

Vậy x = 69

d) 5x + 3x + x = 72

     5x + 3x + 1x = 72

    ( 5 + 3 + 1 )x = 72

                9x = 72

               x = 72 : 9 = 8

Vậy x = 8

Chúc bạn học tốt nhé

a) \(4x+3x=217\)

\(\Rightarrow x\cdot\left(3+4\right)=217\)

\(\Rightarrow7x=217\)

\(\Rightarrow x=\dfrac{217}{7}\)

\(\Rightarrow x=31\)

b) \(9x-3x=216\)

\(\Rightarrow x\cdot\left(9-3\right)=216\)

\(\Rightarrow6x=216\)

\(\Rightarrow x=\dfrac{216}{6}\)

\(\Rightarrow x=36\)

c) \(6x-3x+23=230\)

\(\Rightarrow x\cdot\left(6-3\right)=230-23\)

\(\Rightarrow3x=207\)

\(\Rightarrow x=\dfrac{207}{3}\)

\(\Rightarrow x=69\)

d) \(5x+3x+x=72\)

\(\Rightarrow x\cdot\left(5+3+1\right)=72\)

\(\Rightarrow9x=72\)

\(\Rightarrow x=\dfrac{72}{9}\)

\(\Rightarrow x=8\)

c.on nhìu a

là tính x hả bn

nếu đúng vậy thì ok để mình giải co

ko số​ nhân đó là ích

   \(3x+5x=10\)

\(\left(5+3\right)x=10\)

               \(8x=10\)

                  \(x=\frac{10}{8}=\frac{5}{4}\)

\(3x-32>-5x+1\)

\(\Leftrightarrow3x+5x>32+1\)

\(\Leftrightarrow8x>33\)

\(\Leftrightarrow x>\frac{33}{8}\)

Đề bài yêu cầu gì vậy?? 

Ko co' câu hỏi hao tu làm?? :v

\(3x-32>-5x+1\)

\(\Leftrightarrow3x+5x>1+32\)

\(\Leftrightarrow8x>33\)

\(\Leftrightarrow x>\frac{33}{8}\)

Vậy \(x>\frac{33}{8}\)thì \(3x-32>-5x+1\)

\(\Leftrightarrow9x+3=-20x+16\)

=>29x=13

hay x=13/29

\(\sqrt{-3x^3+5x+14}+\sqrt{-5x^3+6x+28}=\left(4-2x-x^2\right)\sqrt{2-x}\) (ĐKXĐ: \(x\in R,x\le2\))

\(\Leftrightarrow\sqrt{\left(2-x\right)\left(3x^2+6x+7\right)}+\sqrt{\left(2-x\right)\left(5x^2+10x+14\right)}-\left(4-2x-x^2\right)\sqrt{2-x}=0\)

\(\Leftrightarrow\sqrt{2-x}\left(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}-4+2x+x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\left(1\right)\end{cases}}\)

Pt \(\left(1\right)\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-\left(x+1\right)^2+5\left(2\right)\)

Ta có: \(\left(x+1\right)^2\ge0\Rightarrow\sqrt{2\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

Tương tự: \(\sqrt{5\left(x+1\right)^2+9}\ge3\). Từ đó: \(VT_{\left(2\right)}\)\(\ge2+3=5\)

Mà \(VP_{\left(2\right)}=-\left(x+1\right)^2+5\le5\) nên dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)(tm)

Vậy tập nghiệm của pt cho là \(S=\left\{2;-1\right\}.\)