x² - axy - bxy + aby² 

= ( x² - axy) - ( bxy - aby²)

= x( x-ay) - by( x - ay)

= ( x-ay)( x - by)

dạng tách hạng tử:

1// 3x²+5x-2

2// x²-x-30

3// -14x²+39x-10

4// 4x³+14x²+6x

dạng đặt ẩn phụ:

1// (x²+x+1)(x²+x+2)-12

2// (x+1)(x+2)(x+3)(x+4)-24

3// x⁴+6x³+7x²+6x+1

4// x⁴-x³-10x²+2x+4

dang thêm bớt:

1//x⁷+x²+1

2// 4x⁴+81

3//x⁷+x⁵+1

Bạn chuyển tất cả hạng tử từ vế phải sang vế trái ta được

\(^{x^2+5\text{x}^3+x^2y=5\text{x}^3+x^2y}\)

\(x^2+5\text{x}^3+x^2y-5\text{x}^3-x^2y=0\)

Rút gọn ta được

\(x^2=0\)

\(=>x=0\)

tick cho mình nha

\(3x^2-7x+4=\left(3x^2-3x\right)-\left(4x-4\right)=3x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(3x-4\right)\)

3x^2 -7x+4

= 3x^2 -3x-4x+4

= 3x ( x-1) -4(x-1)

= (3x-4)(x-1)

3x^2 - 7x  + 4 =3x ^2 - 4x - 3x  + 4 = 3x(x-1) - 4 ( x+ 1)= (3x-4)(x-1)

\(\left(x^2-3x\right)^2-14x^2+42x+40\\ =\left(x^2-3x-7\right)^2-9\\ =\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)

\(\left(x^2-3x\right)^2-14x^2+42x+40\)

\(=\left(x^2-3x-4\right)\left(x^2-3x-10\right)\)

\(=\left(x-4\right)\left(x+1\right)\left(x-5\right)\left(x+2\right)\)

K phải là đề sai mak là bạn làm sai.

x²-2x-15=(x²-5x)+(3x-15)=x(x-5)+3(x-5)=(x-5)(x+3)

\(x^2-3x-15=\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{69}{4}=\left(x-\dfrac{3}{2}\right)^2-\left(\dfrac{\sqrt{69}}{2}\right)^2\)

\(=\left(x-\dfrac{3}{2}-\dfrac{\sqrt{69}}{2}\right)\left(x-\dfrac{3}{2}+\dfrac{\sqrt{69}}{2}\right)\)

\(x^2-2x-15=\left(x-5\right)\left(x+3\right)\)

Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)

\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)

\(x^3-3x^2-4x+12\)

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

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

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

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

\(x^3-3x^2-4x+12\)

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

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

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