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

x(3-x)=0

Th1: x=0

Th2: 3-x=0     =>       x=3

Vậy x=0 và x=3

(x+2)(4x-8)=0

Th1: x+2= 0        =>         x=-2

Th2: 4x-8 =0           =>             4x =8

                                                x= 2

Vậy x= +- 2 (cộng trừ 2 nhé)

(x+1)+(x+2)+(x+3)+....+(x+199)=5750

199x +(1+2+3+...199) =5750

199x+ {(199+1)* [(199-1)+1] : 2} =5750

199x + 19900= 5750

199x = -14150

x= -14150/199

a)=>x-1;x-3 \(\in\)Ư(-5)={-1;-5;1;5}

còn lại thử từng TH nhé

b)\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

c)=>x²-4;x²-19 trái dấu

Ta có:x^2-4-(x^2-19)=x^2-4-x^2+19=15 >0

\(\Rightarrow\orbr{\begin{cases}x^2-4>0\\x^2-19< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x^2>4\\x^2< 19\end{cases}}\)

Ta có:4<x^2<19

=>x^2\(\in\){9;16}

=>x\(\in\){3;4}

\(\left(3x-4\right)\left(x-3\right)^3=0\)

\(=>\orbr{\begin{cases}3x-4=0\\x-3=0\end{cases}=>\orbr{\begin{cases}x=\frac{4}{3}\\x=3\end{cases}}}\)

Ủng hộ nha

a) \(\left(x+1\right)^2=\left(x+1\right)^0\)

\(=>\left(x+1\right)^2=1\)

\(=>\orbr{\begin{cases}x+1=-1\\x+1=1\end{cases}}\)

\(=>\orbr{\begin{cases}x=-2\\x=0\end{cases}}\)

\(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)

TH1:

\(x+\frac{1}{2}=0\)

=> x = \(\frac{-1}{2}\)

TH2:

\(x-\frac{3}{4}=0\)

=> x = \(\frac{3}{4}\)

a. \(\left(2x-3\right)\left(x+1\right)+\left(2x-3\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1+3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\4x-6=0\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2}\)

b. \(\left(x-4\right)\left(3x-2\right)+x^2-16=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{2}\end{matrix}\right.\)

(2x-3)(x+1)+(2x+3)(3x-7)=0

<=> (2x-3)(x+1)-(2x-3)(3x-7)=0

<=> (2x-3)(x+1-3x+7)=0

<=> (2x-3)(-2x+8)=0

<=> 2x-3=0 => x=3/2

Hoặc -2x+8=0 => x= 4

Vậy x thuộc{3/2;4}