(x+8)(2x+15)(2x^2+35x+120

T ko biết làm, chỉ hỏi liên thiên thôi :)))

Hủ phải không???? OvO Dưa Trong Cúc

a: \(A=5x^2-3x-x^3+x^2+x^3-6x^2-10x+3x+x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)

\(=-10x+x^3+x^2+x-x^3-x^2-x+5\)

=-11x+5

b: \(=6x-3-5x+15+18x-24-19x+3x^2+12x-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)

\(=3x^2+12x-12-7x+20+2x^3-3x^2-2x^3-5x\)

\(=8\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

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

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

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

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

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

8: \(x^2y-xy^2-2x+2y\)

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

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

9: \(x^2-y^2-2x+2y\)

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

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

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

b)(x²+x+1)(x²+x+2)-12

Đặt t=x²+x+1

t(t+1)-12=t²+t-12

=(t-3)(t+4)=(x²+x+1-3)(x²+x+1+4)

=(x²+x-2)(x²+x+5)

=(x-1)(x+2)(x²+x+5)

c)(x²+8x+7)(x²+8x+15)+15

Đặt t=x²+8x+7 

t(t+8)+15=t²+8t+15

=(t+3)(t+5)

=(x²+8x+7+3)(x²+8x+7+15)

=(x²+8x+10)(x²+8x+22)

d)(x+2)(x+3)(x+4)(x+5)-24

=(x²+7x+10)(x²+7x+12)-24

Đặt t=x²+7x+10

t(t+2)-24=(t-4)(t+6)

=(x²+7x+10-4)(x²+7x+10+6)

=(x²+7x+6)(x²+7x+16)

=(x+1)(x+6)(x²+7x+16)

a/ Đặt x² + 4x + 8 = a

Thì đa thức ban đầu thành

a² + 3ax + 2x² = (a² + 2ax + x²) + (ax + x²)

= (a + x)² + x(a + x) = (a + x)(a + 2x)

b/ Đặt x² + x + 1 = a thì đa thức ban đầu thành

a(a + 1) - 12 = a² + a - 12 = (a² - 3a) + (4a - 12) 

= (a - 3)(a + 4)

B

\(5x(2x+3)+6x+9\\=5x(2x+3)+3(2x+3)\\=(2x+3)(5x+3)\)

a: \(5x\left(2x+3\right)+6x+9\)

\(=5x\left(2x+3\right)+\left(6x+9\right)\)

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

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

b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+4\right)\left(3x+48+5\right)\)

=(x+4)(3x+53)

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