`(5x-1)^2-(5x-4)(5x+4)=7`

`\Leftrightarrow 25x^2-10x+1-25x^2+16-7=0`

`\Leftrightarrow  -10x+10=0`

`\Leftrightarrow  x=1`

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

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

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

\(10x+10=30\)

\(10x=20\)

\(x=2\)

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

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

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 

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

5\(^{x+1}\) - 5\(^x\) = 2.2⁸ + 8

5\(^x\).(5 - 1) = 520

5\(^x\).4        = 520

5\(^x\)           = 520 : 4

5\(^x\)           = 130

Với \(x\) = 0 ⇒ 5\(^x\) = 5⁰ = 1 < 130 (loại)

Với \(x\) > 0 ⇒ 5\(^x\) = \(\overline{...5}\) \(\ne\) 130 (loại)

Vậy \(x\) \(\in\) \(\varnothing\) 

\(5^{x+1}-5^x=2.2^8+8\\ 5^x\left(5-1\right)=512+8\\ 5^x.4=520\\ 5^x=\dfrac{520}{4}=130\)

Em xem lại đề

- Help Me :((

\(25\left(x+3\right)^2+\left(1-5x\right)\left(1+5x\right)=0\)

\(25\left(x^2+6x+9\right)+1-25x^2=0\)

\(25x^2+150x+225+1-25x^2=0\)

\(150x=-226\)

\(x=-\frac{113}{75}\)

25 ( x + 3 )² + ( 1 - 5x )( 1 + 5x ) = 0

25 ( x² + 6x + 9 ) + 1 + 5x - 5x - 25x² = 0

25x² + 150x + 225 + 1 + 5x - 5x - 25x² = 0

150x + 226 = 0

150x = -226

x = -226/150

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

`\rightarrow 5x^2-15x= [x(5x-1)-2(5x-1)-5]`

`\rightarrow 5x^2-15x=(5x^2-x-10x+2-5)`

`\rightarrow 5x^2-15x=5x^2-11x-3`

`\rightarrow 5x^2-15x-5x^2+11x+3=0`

`\rightarrow -4x+3=0`

`\rightarrow 4x=3`

`\rightarrow x=`\(\dfrac{3}{4}\)

Vậy, `x=`\(\dfrac{3}{4}\)

Còn biến `y` thì mình k thấy bạn nhé!

Cho mk sửa lại từ dòng thứ 6 (tính cả đề)

`\rightarrow -4x+3=0`

`\rightarrow -4x=-3`

`\rightarrow x=-3/-4`

`\rightarrow x=3/4`

Vậy, `x=3/4`

a) \(\left(5x-2\right)\left(5x+2\right)-\left(5x+3\right)\left(5x-4\right)=0\)

\(\Leftrightarrow5x+8=8\)

\(\Leftrightarrow5x=8-8\)

\(\Leftrightarrow x=5.0\)

\(\Leftrightarrow x=0\)

b) 

câu b sao bạn

\(\left(4x-3\right)\left(4x+2\right)+\left(4x+5\right)\left(1-4x\right)=2.5^2\)

\(16x^2+8x-12x-6+4x-16x^2+5-20x=50\)

\(-20x-1=50\)

\(-20x=51\)

\(x=\frac{-51}{20}\)

Vậy \(x=\frac{-51}{20}\)

b: \(\Leftrightarrow x\left(x-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\)

c: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(b,\Leftrightarrow x\left(x-25\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)