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Ta có: (X-3)2=(X-3)4
=>X-3)4-(X-3)2=0
=>(X-3)2.(X-3.X-3-1)=0
=>(X-3)2= 0 hoặc X-3.X-3-1=0
Suy ra x=3;4
b, 3.(2x+1)+2x+2
=3.2x+1+2x+1.2
=2x+1.(3+2)
=2x+1.5=40
=2x+1=40:5=8
=2x+1=23
=>x+1=3
=>x=3-1=2
=> là suy ra nha bạn
a) ( X - 3 ) 2 = ( X - 3 )4 <=> X - 3 =1
=> X = 1+3
=> X = 4
Chú thích <=> : khi và chỉ khi
=> : suy ra
b) Mình ko biết ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________-sorry-____________________________________________________________________________________________________________________________________-__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

\(\left(5^2+3^2\right).x+\left(5^2-3^2\right).x-40.x=10^2\)
\(\Rightarrow\left(25+9\right)x+\left(25-9\right)x-40x=100\)
\(\Rightarrow34x+16x-40x=100\Rightarrow\left(34+16-40\right)x=100\)
\(\Rightarrow10x=100\Rightarrow x=100\div10=10\)
Vậy \(x=10\)

1) x - 36 + 12 = - x+ 10
=> x + x = 10 + 24
=> 2x = 34
=> x = 34/2 = 17
2) (x + 15) - (11 - x) = (-2)2
=> x + 15 - 11 + x = 4
=> 2x = 4 - 4
=> 2x = 0
=> x = 0
3) 40 - 4x2 = (-6)2
=> 40 - 4x2 = 36
=> 4x2 = 40 - 36
=> 4x2 = 4
=> x2 = 1
=> x = \(\pm\)1
4) (-50) + 10x2 = (-25) x |-2|
=> -50 + 10x2 = -50
=> 10x2 = -50 + 50
=> 10x2 = 0
=> x2 = 0
=> x = 0
5) |x + 1| = 2020
=> \(\orbr{\begin{cases}x+1=2020\\x+1=-2020\end{cases}}\)
=> \(\orbr{\begin{cases}x=2019\\x=-2021\end{cases}}\)
6) (x + 1)5 + 8 = 0 (xem lại đề)
7) (-20) + x3 : 16 = -24
=> x3 : 16 = -24 + 20
=> x3 : 16 = -4
=> x3 = -4 . 16
=> x3 = -64 = (-4)3
=> x = -4
9) x14 = x17
=> x14 - x17 = 0
=> x14(1 - x3) = 0
=> \(\orbr{\begin{cases}x^{14}=0\\1-x^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
10) (-36) + (1 - x)2 = 0
=> (1 - x)2 = 36
=> (1 - x)2 = 62
=> \(\orbr{\begin{cases}1-x=6\\1-x=-6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)


a)2.|x-1|2-10=40
2.|x-1|2=40+10
2.|x-1|2=50
|x-1|2=50:2
|x-1|2=25
|x-1|2=52
|x-1|=5
\(\Rightarrow x-1=\pm5\)
*x-1=5 *x-1=-5
x=5+1 x=-5+1
x=6 x=-4

a, Đặt \(A=1+2+2^2+...+2^x\)
\(A=2^x+2^{x-1}+....+2^2+2+1\)(đảo lại số hạng để phục vụ tính bước sau )
\(2A=2^{x+1}+2^x+...+2^3+2^2+2\)
\(2A-A=2^{x+1}-1\)
Suy ra \(A=2^{x+1}-1\)
Khi đó \(2^{x+1}-1=1023\Rightarrow2^{x+1}=1024\Rightarrow2^{x+1}=2^{10}\Rightarrow x+1=10\Rightarrow x=9\)
Vậy x = 9
b ) Ta có \(A=1+3+3^2+3^3+...+3^{119}\)
\(A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(A=1.\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{116}.\left(1+3+3^2+3^3\right)\)
\(A=\left(1+3+3^2+3^3\right).\left(1+3^4+...+3^{116}\right)\)
\(A=40.\left(1+3^4+...+3^{116}\right)⋮40\)
Vậy A chia hết cho 40