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 

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

a) ( 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

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

⇔ x² + 6x + 9 + x² - 4 - 2( x² - 2x + 1 ) = 7

⇔ 2x² + 6x + 5 - 2x² + 4x - 2 = 7

⇔ 10x + 3 = 7

⇔ 10x = 4

⇔ x = 4/10 = 2/5

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

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

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

\(10x+10=30\)

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

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

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

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

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

\(\Leftrightarrow-14x-9=5\)

\(\Leftrightarrow-14x=14\)

\(\Leftrightarrow x=-1\)

Vậy....

b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)

\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(\Leftrightarrow-10+17x=-44\)

\(\Leftrightarrow17x=-34\)

\(\Leftrightarrow x=-2\)

Vậy....

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

\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)

\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)

\(\Leftrightarrow10x+10=30\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

Vậy....

d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)

\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)

\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)

\(\Leftrightarrow14x-3=7\)

\(\Leftrightarrow14x=10\)

\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)

Vậy...

câu hỏi đây

bn Nhật Hạ lm đúng r đấy

conkia

a) \(x^2-2x+1=25\)

\(\Rightarrow x^2-2x+1-25=0\)

\(\Rightarrow x^2-2x.1+1^2-5^2=0\)

\(\Rightarrow\left(x-1\right)^2-5^2=0\)

\(\Rightarrow\left(x-1-5\right)\left(x-1+5\right)=0\)

\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=\left(-4\right)\end{cases}}\)

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

\(\Rightarrow\left(5x\right)^2+2.5x.1+1^2-\left(25x^2-9\right)-30=0\)

\(\Rightarrow25x^2+10x+1-25x^2+9-30=0\)

\(\Rightarrow\left(25x^2-25x^2\right)+10x+\left(1+9-30\right)=0\)

\(\Rightarrow10x-20=0\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

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