Hình như bạn ghi sai đề. Đề đúng: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x+10=30\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

\(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)

\(25x^2+10x+1-\left(25x^2-9\right)=30\)

\(25x^2+10x+1-25x^2+9=30\)

\(10x+10=30\)

\(10x=20\)

\(x=2\)

T_T tôi chưa hc lp 8

(5x+1)²-(5x+3)(5x-3)=30

=>25x²+10x+1-25x²+9=30

=>(25x²-25x²)+10x+1+9=30

=>10x+10=30

=>10x=20

=>x=2

Ta có:  (5x+1)^2 - (25x^2 - 9)= 30

             (5x+1 - 5x)(5x+1+5x) + 9 = 30

                  10x +10 = 30 

=>  x=2

 (x+2)^2= 9 
=> (x+2)^2= 3^2=(-3)^2
TH1: x+2=3
=> x=3-2=1
TH2: x+2=-3
=> x=(-3)-2=-5

Bài làm :

\(a,\left(x+2\right)^2-9=0\)

\(\Leftrightarrow\left(x+2\right)^2=9\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy x = 1 hoặc x = -5 .

\(b,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-\left(25x^2-3^2\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow\left(25x^2-25x^2\right)+10x=30-9-1\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

Vậy x = 2 .

\(c,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)

\(\Leftrightarrow x^3+x^2+x-x^2-x-1+\left(x^2+2x\right)\left(2-x\right)=5\)

\(\Leftrightarrow x^3-1+2x^2-x^3+4x-2x^2=5\)

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

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\frac{3}{2}\)

Vậy x = 3/2 .

Học tốt nhé

( x + 2 )² - 9 = 0

<=> ( x + 2 )² - 3² = 0

<=> ( x + 2 - 3 )( x + 2 + 3 ) = 0

<=> ( x - 1 )( x + 5 ) = 0

<=> x = 1 hoặc x = -5

( 5x + 1 )² - ( 5x - 3 )( 5x + 3 ) = 30

<=> 25x² + 10x + 1 - ( 25x² - 9 ) = 30

<=> 25x² + 10x + 1 - 25x² + 9 = 30

<=> 10x + 10 = 30

<=> 10x = 20

<=> x = 2

( x - 1 )( x² + x + 1 ) + x( x + 2 )( 2 - x ) = 5

<=> x³ - 1 + x( 4 - x² ) = 5

<=> x³ - 1 + 4x - x³ = 5

<=> 4x - 1 = 5

<=> 4x = 6

<=> x = 6/4 = 3/2

c: =>(x+2)(x+3)(x-5)(x-6)=180

=>(x^2-3x-10)(x^2-3x-18)=180

=>(x^2-3x)^2-28(x^2-3x)=0

=>x(x-3)(x-7)(x+4)=0

=>\(x\in\left\{0;3;7;-4\right\}\)

c: =>(x-3)(x+2)(2x+1)(3x-1)=0

=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

a) (5x+1)²-(5x+3).(5x-3)=30

\(\Leftrightarrow25x^2+10x+1-25x^2+9-30=0\)

\(\Leftrightarrow10x-20=0\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

b) (x-3).(x²+3x+9)+x.(x+2).(2-x)=1

\(\Leftrightarrow x^3-3^3+x\left(4-x^2\right)-1=0\)

\(\Leftrightarrow x^3-27+4x-x^3-1=0\)

\(\Leftrightarrow4x-28=0\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

câu a cái ngoặc của 5x+1

1 , <=>  25x^2 + 10x + 1 - ( 25x^2 - 9) = 30 

      <=> 25x^2 + 10x + 1 - 25x^2 + 9 = 30 

      <=> 10x + 10                 = 30 

     <=> 10 ( x + 1)                 = 30

       <=>  x + 1                      = 3 

        <=>  x                           = 2 

2, ( x + 3)(x^2 - 3x + 9 ) - x(x+2)(x-2) = 15 

<=> x^3 - 27 - x(x^2 - 4)                      = 15

<=> x^3 - 27 - x^3 + 4x                        = 15 

<=> 4x -27                                          = 15 

<=> 4x                                                = 15 + 27

<=> 4x                                                 =42

<=> x                                                     = 42/4 = 21/2 

******************

\(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(=25x^2+1+10x-\left(25x^2-9\right)=30\)

\(25x^2+1+10x-25x^2+9=30\)

\(10x+10=30\)

\(10x=30-10=20\)

\(x=\frac{20}{10}=2\)

a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

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

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)