(12x-5)(4x-1)+(3x-7)(1-16x)=81

48x^2-12x-20x+5+3x-48x^2-7+112x=81

83x-2=81

83x=83

x=1

(12x-5)(4x-1)+(3x-7)(1-16x)=81
12x ( 4x- 1) - 5(4x-1) + 3x(1-16x) -7(1-16x) =81
12x . 4x -12x - 20x + 5 + 3x - 3x. 16x - 7 + 7.16x = 81
48x^2 -12x - 20x + 5 + 3x - 48x^2 - 7 + 112x = 81
(48x^2 - 48x^2) + (112x - 12x - 20x +3x) - ( 7 - 5) =81
83x = 81 + 2
83x =83
x=1

ht nha bạn

ai mà biết

đề thiếu nha

`x=49`

`->x+1=49+1 ->x+1=50` (1)

`F (x) = x^3 +50x^2 - 50x+50`

Thay (1) vào ta được :

`-> F (x) =x^3 + (x+1)x^2 - (x+1)x+50`

`->F (x)=x^3 +x^3+x^2-x^2 -x+50`

`-> F (x) = 2x^3 - x+50`

`-> F (49)=2 . 49^3 -49+50 =235299`

Vậy `F (49)=235299`

Bài 1 

ta có a+3+b-3 =a +b chia hết cho 4

nên (b-a )(a+b) cũng chia hết cho 4

bài 2.

ta có: \(M=6x^2-5x-6-12xy+6y^2+6y-3x+2y+2027\)

\(=6\left(x-y\right)^2-8\left(x-y\right)+2021=24-16+2021=2029\)

 x( x+8) + x (3 - x ) = -22

=> x( x+8 + 3 - x ) = -22

=>  11x   =  -22

=>  x   =  -2

Vậy x  = -2

`x (x+8) +x(3-x)=-22`

`->x^2 +8x+3x-x^2=-22`

`-> 11x=-22`

`->x=-2`

Vậy `x=-2`

(x-4)(x+4)^2 + x(x+5)(5-x) = 12

=>( x^2 - 16 )( x+4) + x( 25 - x^2) = 12

=> x^3 + 4x^2 - 16x - 64 + 25x - x^3 - 12 = 0

=> 4x^2 + 9x - 76 = 0 

=> x = \(\frac{-9\pm\sqrt{1297}}{8}\)

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

\(\Leftrightarrow16-2x+9x^2-\left(9x^2-4\right)=0\)

\(\Leftrightarrow16-24x+9x^2-9x^2+4=0\)

\(\Leftrightarrow20-24x=0\)

\(\Leftrightarrow4\left(5-6x\right)=0\)

\(\Leftrightarrow5-6x=0\)

\(\Leftrightarrow-6x=-5\)

\(\Leftrightarrow x=\frac{5}{6}\)