\(a,=x\left(y-3\right)+y\left(y-3\right)=\left(x+y\right)\left(y-3\right)\\ b,=\left(x+2\right)^2-16y^2=\left(x+4y+2\right)\left(x-4y+2\right)\)

\(a,xy-3x+y^2-3y=\left(xy-3x\right)+\left(y^2-3y\right)=x\left(y-3\right)+y\left(y-3\right)=\left(x+y\right)\left(y-3\right)\\ b,x^2-16y^2+4x+4=\left(x^2+4x+4\right)-16y^2=\left(x+2\right)^2-\left(4y\right)^2=\left(x-4y+2\right)\left(x+4y+2\right)\)

\(1,x^2-16y^2=\left(x-4y\right)\left(x+4y\right)\)

\(2,21x-21y+ax-ay=21\left(x-y\right)+a\left(x-y\right)=\left(21+a\right)\left(x-y\right)\)

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

Đặt x+5 = a ; x-4 = b

=> 2x+1 = a+b

pt <=> a^4+b^4=(a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^+b^4

<=> 4a^3b+6a^2b^2+4ab^3 = 0

<=> 2a^3b+3a^2b^2+3ab^3 = 0

<=> ab.(2a^2+3ab+2b^2) = 0

<=> ab=0 ( vì 2a^2+3ab+b^2 > 0 )

<=> a=0 hoặc b=0

<=> x+5=0 hoặc x-4=0

<=> x=-5 hoặc x=4

Vậ y ............

Tk mk nha

Bạn xem lại đề đi , đề này sai rùi ko phân tích đa thức thành nhân tử được đâu

cảm ơn bạn nha

\(=14x^2-2x+35x-5\)

\(=2x\left(7x-1\right)+5\left(7x-1\right)\)

\(\left(7x-1\right)\left(2x+5\right)\)

Phân tích đa thức thành nhân tử: 14x^2 + 33x - 5

=14x2−2x+35x−5

=2x(7x−1)+5(7x−1)

(7x−1)(2x+5)

Phân tích đa thức thành nhân tử: 14x^2 + 33x - 5 là :

         \(14x^2-2x+35x-5\)

          \(2x\left(7x-1\right)+5\left(7x-1\right)\)

           \(\Rightarrow\left(7x-1\right)\left(2x+5\right).\)

\(3x^3+5x^2-14x+4\\ =\left(3x^3-x^2\right)+\left(6x^2-2x\right)-\left(12x-4\right)\\ =x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\\ =\left(x^2+2-4\right)\left(3x-1\right)\)

16y² - 4x² - 12x - 9 = 16y² - (4x² + 12x + 9) = 16y² - (2x + 3)² = (4y - 2x - 3)(4y + 2x + 3)

\(16y^2-4x^2-12x-9=16y^2-\left(4x^2+12x+9\right)=\left(4y\right)^2-\left(2x+3\right)^2=\left[4y-\left(2x+3\right)\right]\left[4y+\left(2x+3\right)\right]\)

=16y²  - (4y² + 12x + 9)

=(4y² )-(2y + 3)²

=(4y-2y-3)(4y+2y+3)

mik nha

\(x^3+4x^2+4x-16y^2\)

\(=\left(x^3+2x^2\right)+\left(2x^2+4x\right)-16y^2\)

\(=x^2.\left(x+2\right)+2x.\left(x+2\right)-16y^2\)

\(=\left(x+2\right).\left(x^2+2x\right)-16y^2\)

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

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

\(=\left[\sqrt{x}.\left(x+2\right)\right]^2-4y^2\)

\(=\left[\sqrt{x}.\left(x+2\right)-4y\right].\left[\sqrt{x}.\left(x+2\right)+4y\right]\)

Tham khảo nhé~

nếu đưa vô căn phải có điều kiện là x > 0

\(x^3+4x^2+4x-16y^2=x\left(x+2\right)^2-\left(4y\right)^2\)

\(=\left(x\sqrt{x}+2\sqrt{x}\right)^2-\left(4y\right)^2=\left(x\sqrt{x}+2\sqrt{x}-4y\right)\left(x\sqrt{x}+2\sqrt{x}+4y\right)\)

       \(x^3-9x^2+14x\)

= \(x^3-7x^2-2x^2+14x\)

= \(x^2.\left(x-7\right)-2x.\left(x-7\right)\)

= \(\left(x-7\right).\left(x^2-2x\right)\)

= \(\left(x-7\right).\left(x-2\right).x\)

\(x^3-9x^2+14x\)

\(=x\left(x^2-9x+14\right)\)

\(=x\left(x^2-7x-2x+14\right)\)

\(=x\left[x\left(x-7\right)-2\left(x-7\right)\right]\)

\(=x\left(x-2\right)\left(x-7\right)\)

khó quá tui ko biết làm..

k cho tui nha

thanks

Bài làm

   3x² + 14x - 15

= 3x² + 9x + 5x - 15

= -( 9x - 3x² ) - ( 15 - 5x )

= -3x( 3 - x ) - 5( 3 - x )

= ( 3 - x )( -5 - 3x )

# Hokc tốt #