a) \(x^2-5x+6\)

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

b) \(x^2-9x+18=x^2-3x-6x+18\)

\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)

c) \(x^2-6x+5=x^2-x-5x+5\)

\(=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)

d) \(3x^2+5x-30=3\left(x^2+\dfrac{5x}{3}-10\right)=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{5347}{500}\right)\)

Câu này bạn xem lại đề nha

e) \(3x^2-5x-2=3x^2-6x+x-2\)

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

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

=x(x²+2x+1)-3²

=x(x+1)²-3²

=x(x+1-3)(x+1+3)

a) x² + 5x + 6

= x² + 2x + 3x + 6

= (x² + 2x) + (3x + 6)

= x(x + 2) + 3 (x + 2)

= (x + 2) (x + 3)

b) x² + 6x + 8

= x² + 2x + 4x + 8

= (x² + 2x) + (4x + 8)

= x(x + 2) + 4(x + 2)

= (x + 2)(x + 4)

c) x² - 5x - 14

= x² + 2x - 7x - 14

= (x² + 2x) - (7x + 14)

= x(x + 2) - 7(x + 2)

= (x + 2)(x - 7)

d) x² - 9x + 18

= x² - 3x - 6x + 18

= (x² - 3x) - (6x + 18)

= x(x - 3) - 6 (x - 3)

= (x - 3)(x - 6)

e) x² - 7x + 12

= x² -3x - 4x + 12

= (x² - 3x) - (4x + 12)

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

= (x - 3)(x - 4)

f) 3x² + 9x - 30

= 3(x² + 3x - 10)

= 3\(\left[\left(x^2+5x-2x-10\right)\right]\)

= 3\(\left[\left(x^2+5x\right)-\left(2x-10\right)\right]\)

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

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

 Chuc ban hoc tot

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

b) \(x^2+6x+8=\left(x+2\right)\left(x+4\right)\)

c) \(x^2-5x-14=\left(x-7\right)\left(x+2\right)\)

d) \(x^2-9x+18=\left(x-3\right)\left(x-6\right)\)

e) \(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)

f) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x+5\right)\left(x-2\right)\)

\(x^3-5x^2+3x+9\)

Gỉa sử \(x^3-5x^2+3x+9=\left(x+a\right)\left(x^2+bx+c\right)\)

\(=x^3+bx^2+cx+ax^2+abx+ac\)

\(=x^3+\left(a+b\right)x^2+\left(c+ab\right)x+ac\)

Từ \(x^3-5x^2+3x+9=x^3+\left(a+b\right)x^2+\left(c+ab\right)x+ac\) Hợp nhất ta được :

\(\hept{\begin{cases}a+b=-5\\c+ab=3\\ac=9\end{cases}}\)

Dự đoán \(a=-3\Rightarrow\hept{\begin{cases}b=-2\\c=-3\end{cases}}\) 

\(\Rightarrow\left(x+a\right)\left(x^2+bx+c\right)=\left(x-3\right)\left(x^2-2x-3\right)=\left(x-3\right)^2\left(x+1\right)\)

Vậy \(x^3-5x^2+3x+9=\left(x-3\right)^2\left(x+1\right)\)

DỄ ẸT MÀK KO BT LÀM CÒN LÊN ĐÂY HỎI ...LP 8 ĐỂ LÀM J DANH HIỆU HỞ~~~~~CÁI NÀY THÌ LP 5 CX LÀM ĐC NHÉ!

vậy bạn làm ngay cho tôi

x³ ​- 3x² - 9x - 5 = (x³ - 5x²) + (2x² -10x) + (x - 5) = x² (x - 5) + 2x(x - 5) + (x - 5) = (x - 5)(x²  + 2x + 1) = (x - 5)(x + 1)²

x^3-3x^2-9x-5

=x^3-5x^2+2x^2-10x+x-5

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

=(x-5)(x^2+2x+1)

=(x-5)(x+1)^2

=x^3-3x+3-1

=(x^3-1)-(3x-3)

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

=(x-1)(X^2+x+1-3)

=(x-1)(x^2+x+2)

A=x⁴+3x³-9x-9

=x⁴-9+3x³-9x

=(x²-3)(x²+3)+3x.(x²-3)

=(x²-3)(x²+3+3x)

bạn phân tích chưa hết kìa

a: \(x^2-y^2+3x+3y\)

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

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

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

b: Sửa đề: \(x^2-4y^2+4x+4\)

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

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

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

\(\left(x^2-3x+2\right)\left(x^2-9x+20\right)-40=\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-40\)

\(=\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40\)
Đặt \(t=x^2-6x+5\) thì ta có \(t\left(t+3\right)-40=t^2+3t-40=\left(t+8\right)\left(t-5\right)\)

Suy ra \(\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40=\left(x^2-6x+13\right)\left(x^2-6x\right)=x\left(x-6\right)\left(x^2-6x+13\right)\)