gọi 2021-x = a

2023-x=b

2x-4044=c

ta có a + b + c=2021-x+2023-x+2x-4044=0

suy ra a + b = -c

suy ra (a+b)^3 =-c^3

ta có a^3 + b^3 + c^3=(a+b)^3 -3ab(a+b) + c^3 = -c^3 +3abc +c^3 = 3abc 

ta có (2021-x)^3 + (2023-x)^3 + (2x-4044)^3 = 0

=> 3(2021-x)(2023-x)(2x-4044)=0

=> th 1 x = 2021,  th 2 x = 2023; th3 x = 2022

\(5^x-2-3^2=2^4-\left(2^8x2^4-2^{10}x2^2\right)\)

\(5^x-2-3^2=2^4-\left(2^{8+4}-2^{10+2}\right)\)

\(5^x-2-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(5^x-2-3^2=2^4-0\)

\(5^x-2-3^2=2^4\)

\(5^x-2-9=16\)

\(5^x-2=16+9\)

\(5^x-2=25\)

\(5^x=25+2\)

\(5^x=27\)

Bởi vì 27 không phân tích được 1 số có số mũ là 2

\(\Rightarrow\) Không tồn tại x

\(5^{x-2}-9=16-\left(256.16-1024.4\right)\)

\(\Rightarrow5^{x-2}-9=16-\left(4096-4096\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}-9=16\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow x-2=25:5\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=5\)

\(5^{x-2}-3^2=2^4-\left(2^8\cdot2^4-2^{10}\cdot2^2\right)\)

\(\Rightarrow5^{x-2}-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}=16+9\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow5^{x-2}=5^2\)

\(\Rightarrow x-2=2\)

\(\Rightarrow x=2+2\)

\(\Rightarrow x=4\)

Vậy: x=4

a)

(x+4)(3x-5) = 0

=> x + 4 = 0 hoặc 3x-5 = 0

     x = -4                 x = 5/3

b)

  2x² + 7x + 3 = 0

  2x² + 6x + x + 3= 0

  (2x+1)(x+3) = 0

=> 2x+1 = 0 hoặc x + 3 = 0

    x = -1/2              x = -3

-3x+(-9)+5x-5=-10

(-3x+5x)+(-9-5)=-10

-2x+(-14)=-10

-2x=-10-(-14)

-2x=24

x=24:(-2)

x=-12. chúc bạn học tối nha

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

\(\Rightarrow2x\left(x-3\right)=x\left(x-3\right)\)

\(\Rightarrow2x=x\)

\(\Rightarrow x=0\)

\(2x.\left(x-3\right)=x^2-3x\)

\(\left(x-3\right)=x^2-3x:2x\)

Ta có: \(2x\left(x-3\right)=x^2-3x\)

\(\Leftrightarrow2x\left(x-3\right)-x\left(x-3\right)=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

(2x+1)(y+2)=4

⇒(2x+1) và (y+2) ∈ Ư (4) = { 1,-1,2,-2,4,-4 }

⇒2x+1=1        ⇒2x=1-1=0            ⇒x=0:2=0

   y+2=4             y=4-2=2                  y=2

⇒2x+1=-1      ⇒2x=-1-1=-2        ⇒x=-2:2=-1

   y+2=-4           y=-4-2=-6             y=-6

⇒2x+1=2        ⇒2x=2-1=1          ⇒x=1:2=0,5

   y+2=-2           y=-2-2=-4             y=-4

\(\left(2x-1\right)\left(y-2\right)=4\)

\(\Rightarrow2x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Mà \(2x+1\) lẻ

\(\Rightarrow2x+1=\pm1\)

Xét \(2x+1=1\Rightarrow x=0\)

\(\Rightarrow y-2=4\Rightarrow y=6\)

Xét \(2x+1=-1\Rightarrow x=-1\)

\(\Rightarrow y-2=-4\Rightarrow y=-2\)

(2x+1)(y+2)=4

⇒(2x+1) và (y+2) ∈ Ư (4) = { 1,-1,2,-2,4,-4 }

⇒2x+1=1        ⇒2x=1-1=0            ⇒x=0:2=0

   y+2=4             y=4-2=2                  y=2

⇒2x+1=-1      ⇒2x=-1-1=-2        ⇒x=-2:2=-1

   y+2=-4           y=-4-2=-6             y=-6

⇒2x+1=2        ⇒2x=2-1=1          ⇒x=1:2=0,5

   y+2=-2           y=-2-2=-4             y=-4

=>2x+1/2=0 hoặc 2x-3=0

=>x=-1/4 hoặc x=3/2

\(\left[{}\begin{matrix}\dfrac{1}{2}+2x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)

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

( 1/2 + 2x ) = 0

( 2x - 3 ) = 0

( 1/2 + 2x ) = 0

\(\Leftrightarrow\) ( 2x + 1/2 ) = 0

2x + 1/2 - 1/2 = 0 - 1/2

2x = -1/2

2x/2 = -1/2 : 2

x = -1/4

2x - 3 = 0

2x - 3 + 3 = 0 + 3

2x = 3

2x/2 = 3/2

x = 3/2

\(\Rightarrow\) x₁ = -1/4 ; x₂ = 3/2