7(x - 3) - x(3 - x)

= (x - 3)(7 + x)

chỉ bt có v mà k bt có đúng k 

1 ) 7 ( x - 3 ) - x ( 3 - x ) 

= 7 ( x - 3 ) + x ( x - 3 )

= ( x - 3 ) ( 7 + x )

2 ) 4x² - 6x + 3 - 2x

= 4x² - 2x - 6x + 3

= 2x ( 2x - 1 ) - 3 ( 2x - 1 )

= ( 2x - 1 ) ( 2x - 3 )

3 ) ( 4 - x ) - 4x + x²

= ( 4 - x ) - x ( 4 - x )

= ( 4 - x ) (  1 - x )

4 ) x² - 2xy + y²

= ( x - y )²

1, \(7\left(x-3\right)-x\left(3-x\right)=7\left(x-3\right)+x\left(x-3\right)\)

\(=\left(x-3\right)\left(x+7\right)\)

2,\(4x^2-6x+\left(3-2x\right)=4x^2-8x+3\)

\(=4x^2-6x-2x+3=2x\left(2x-3\right)-\left(2x-3\right)\)

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

3,\(\left(4-x\right)-4x+x^2=x^2-5x+4\)

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

4,\(x^2-2xy+y^2=\left(x-y\right)^2\)

\(\sqrt{4x^2-4x+9}=3\\ \Rightarrow4x^2-4x+9=9\\ \Rightarrow4x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Ta có: \(\sqrt{4x^2-4x+9}=3\)

\(\Leftrightarrow4x^2-4x=0\)

\(\Leftrightarrow4x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

a) \(\left(3x-2^4\right).7^3=2.7^4\)\(\Leftrightarrow3x-2^4=2.7^4:7^3\)

\(\Leftrightarrow3x-16=2.7\)\(\Leftrightarrow3x-16=14\)\(\Leftrightarrow3x=30\)

\(\Leftrightarrow x=10\)

Vậy \(x=10\)

b) \(3x+4x=\left|-75\right|+23\)\(\Leftrightarrow7x=75+23\)

\(\Leftrightarrow7x=98\)\(\Leftrightarrow x=14\)

Vậy \(x=14\)

a) \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=> \(3x\cdot7^3-2^4\cdot7^3=2\cdot7\cdot7^3\)

=> \(3x\cdot7^3=14\cdot7^3+16\cdot7^3\)

=> \(3x\cdot7^3=\left(14+16\right)\cdot7^3\)

=> \(3x\cdot7^3=30\cdot7^3\)

=> \(3x=30\)(bỏ hai vế 7³)

=> \(x=10\)

Vậy x = 10

b) \(3x+4x=\left|-75\right|+23\)

=> \(7x=75+23\)

=> \(7x=98\)

=> \(x=14\)

Vậy x = 14

a) ( 3x - 2⁴ ). 7³ = 2. 7⁴

    ( 3x - 16 )       = 2 . (7⁴ : 7³ )

     ( 3x - 16)      = 2 . 7

      3x -16         = 14

        3x             = 14 + 16

         3x            = 30

           x            = 30 : 3

            x           = 10

Vậy x =10.

b) 3x + 4x = I - 75 I + 23

    3x + 4x = 75 + 23

     7x        = 98

       x        = 98 : 7

        x       = 14.

Vậy x = 14.

Hok tốt !

3x(12x-4)-(4x-3)(9x+4) = 9

36x² -12x-(36x² -16x-27x-12) = 9

36x² -12x-36x² -16x+27x+12 = 9

-x = 9-12

-x = -3

x= -3 : -1

x= 3

vậy x= 3

a) \(x^3-7x-6\)

\(=\left(x^3+2x^2\right)-\left(2x^2+4x\right)-\left(3x+6\right)\)

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

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

b)\(x^3-19x-30\)

\(=\left(x^3-5x^2\right)+\left(5x^2-25x\right)+\left(6x-30\right)\)

\(=\left(x-5\right)\left(x^2+5x+6\right)\)

\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)

c) \(a^3-6a^2+11a-6\)

\(=\left(a^3-a^2\right)-\left(5a^2-5a\right)+\left(6a-6\right)\)

\(=\left(a-1\right)\left(a^2-5a+6\right)\)

\(=\left(a-1\right)\left(a-2\right)\left(a-3\right)\)

\(m\left(x\right)+h\left(x\right)=g\left(x\right)-5\)

\(\Leftrightarrow m\left(x\right)=g\left(x\right)-h\left(x\right)-5\)

\(\Leftrightarrow m\left(x\right)=4x^2+3x+1-3x^2+2x+3-5\)

\(\Leftrightarrow m\left(x\right)=x^2+5x-1\)

\(2x-1-x^2=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\\ \left(1-3\right)^3-1=\left(-2\right)^3-1=-2-1=-3\\ \left(4x-1\right)^2-9x^2=\left(4x-1-3x\right)\left(4x-1+3x\right)=\left(x-1\right)\left(7x-1\right)\\ \left(x+2\right)^3+1=\left(x+2+1\right)\left[\left(x+2\right)^2+\left(x+2\right)+1\right]\\ =\left(x+3\right)\left(x^2+4x+4+x+2+1\right)\\ =\left(x+3\right)\left(x^2+5x+7\right)\)

Đúng đó. Nhưng ghi thêm: vậy đa thức trên vô nghiệm nha.

Ghi 3 > 0 hơi trẻ trâu tí !!!

Nhưng vẫn đúng

Thiếu kết luận

Tham khảo cách mik nè:

Ta có : 4x^2 >= 0

       => 4x^2 + 3 >= 3

       => 4x^2 + 3 >  0

Vậy đa thức trên vô nghiệm