\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-\left(x^2+5x\right)=0\\ 2\left(x+5\right)-\left[x\left(x+5\right)\right]=0\\ \left(2-x\right)\left(x+5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

2(x + 5) - x^2 - 5x = 0

=> 2(x+5)-(2x+5x) = 0

=> 2(x+5) - x(x+5) = 0

=> (x+5).(2-x) = 0

=> x+5=0 => x = -5

     2-x=0 => x = 2

Vậy x=-5 và x=2

Ta có

\(5^x.5^{x+1}.5^{x+2}\le10^{18}:2^{18}=>5^{3x+3}\le5^{18}=>3x+3\le18=>x\le5\)

 20 . 2^x + 1 = 10.4^2 + 1 

 20 . 2^x + 1 = 10 . 16 + 1

20 . 2^x + 1 = 161

20 . 2^x = 161 - 1

20 . 2^x = 160

2^x = 8

2^x = 2^3

=> x = 3

Ban co giai dc phan 2 ko ?

( 4 - x : 2 )^3 - 1 = 2 . ( 2^3 - 5 : 2^0 )

( 4 - x : 2 )^3 - 1 = 2 . ( 8 - 5 : 1 )

( 4 - x : 2 )^3 - 1 = 2 . 3

( 4 - x : 2 )^3 - 1 = 6

( 4 - x : 2 )^3  = 7

=> ko tìm đc x

nhầm tí mjk sửa lại nè

(x+2)(x+3)-(x-2)(x+5​)=0

<=>x²+5x+6-(x²+3x-10)=0

<=>x²+5x+6-x²-3x+10=0

<=>2x+16=0

<=>2x=-16

<=>x=-8

nhầm tí

(x+2)(x+3)-(x-2)(x+5​)=0

<=>x²+5x+6-(x²+3x-10)=0

<=>x²+5x+6-x²-3x+10=0

<=>2x+16=0

<=>2x=-16

<=>x=-8

Ta có : x² + 8x - 7 = 0

=> x² + 8x + 16 - 9 = 0

=> (x + 4)² - 9 = 0

=>  (x + 4)² = 9

=> \(\orbr{\begin{cases}x+4=3\\x+4=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-7\end{cases}}\)

a) (3-x)²-(5+x)²=0 => x=-1

b) (x-3)²-x(x+2)=7 => x=0,25

c) x²+8x-7=0   (ko biết làm)

d) x²+4x-5=0  => x=1

thank you 2 ban nhieu !!!