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^2 +1 =0 hoac x^2 -4 =0

x^2 = -1 (vo ly) hoac X^2 =4

suy ra x=2 hoac x=-2

 x²+1=0 hoặc x²-4=0

x² =-1=> vô lí

 x²-4=0=x^2=4=>x=2 hoặc x=-2

vậy x=2 hoặc -2

vì (x^2+1).(x^2-4)=0

\(\Rightarrow\)x^2+1=0 hoặc x^4-4=0

     x^2    =-1         x^4    =4

      x^2    =  -1^2      x^4    =  

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}

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

TH1:

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

=> x = \(\frac{-1}{2}\)

TH2:

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

=> x = \(\frac{3}{4}\)

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

\(\Leftrightarrow\left(x-1\right)^2-\left(x-3\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^2-\left[\left(x-3\right)^2\right]^2=0\)

\(\Leftrightarrow\left[\left(x-1\right)-\left(x-3\right)^2\right]\left[\left(x-1\right)+\left(x-3\right)^2\right]=0\)

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

\(\Leftrightarrow\left(-x^2+7x-10\right)\left(x^2-5x+8\right)=0\)

\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)\left(x^2-5x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy: ... 

(x-1)^2 =(x-3)^4=\(\left\{{}\begin{matrix}1+1\\2+2\\3+3\\4+4\end{matrix}\right.=2+4+6+8=\sqrt[]{251234=\Sigma\dfrac{2}{2}22\dfrac{2}{2}}\max\limits_{212}=\dfrac{21}{23}2123=\sum\limits1^{ }_{ }\text{(x-1)^2 =x=}\sum1\)

Bổ sung cho @ Huỳnh Thanh Phong.

(- \(x^2\) + 7\(x\)  - 10).(\(x^2\) - 5\(x\) + 8) = 0

(- \(x^2\) + 5\(x\) + 2\(x\) - 10).(\(x^2\) - \(\dfrac{5}{2}\)\(x\) - \(\dfrac{5}{2}\)\(x\) + \(\dfrac{25}{4}\) + \(\dfrac{7}{4}\)) = 0

[(- \(x^2\) + 5\(x\)) + (2\(x\) - 10)].[(\(x^2\) - \(\dfrac{5}{2}\)\(x\)) - (\(\dfrac{5}{2}\)\(x\) - \(\dfrac{25}{4}\)) + \(\dfrac{7}{4}\)] = 0

[ -\(x\)(\(x\) - 5) + 2.(\(x\) - 5)]. [\(x\)(\(x\) - \(\dfrac{5}{2}\)) - \(\dfrac{5}{2}\).(\(x\) - \(\dfrac{5}{2}\)) + \(\dfrac{7}{4}\)] = 0

(\(x\) - 5).(-\(x\) + 2).[(\(x-\dfrac{5}{2}\)).(\(x\) - \(\dfrac{5}{2}\)) + \(\dfrac{7}{4}\)] = 0

(\(x\) - 5).(-\(x\) + 2).[(\(x\) - \(\dfrac{5}{2}\))² + \(\dfrac{7}{4}\)] = 0 (1)

Vì (\(x\) - \(\dfrac{5}{2}\))² ≥ 0 ⇒ (\(x\) - \(\dfrac{5}{2}\))² + \(\dfrac{7}{4}\) ≥ \(\dfrac{7}{4}\)  (2)

Kết hợp (1) và (2) ta có:

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

\(\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy \(x\in\) {2; 5}