A = \(4x^2-3x+7x^2+2x-5\)

\(11x^2-3x+2x-5\)

\(11x^2-x-5\)

B = \(3x+7y-6x-8+y-2\)

\(3x+7y-6x-10+y\)

\(- 3x+7y-10+y\)

\(3x+8y-10\)

C =  chịu

D= \(6x^4-3x^2+x^2-4x+3.4-x+2\)

\(6x^4-3x^2+x^2-4x;12-x+2\\ \)

\(6x^4-3x^2+x^2-4x+14-x\)

\(6x^4-2x^2-4x+14-x\)

\(6x^4-2x^2-5x+14\)

Bài 1:

\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)

\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)

\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)

Bài 2:

\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)

Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:

\(\left(y-1\right)\left(y+1\right)=120\)

\(\Leftrightarrow y^2-1=120\)

\(\Leftrightarrow y^2=121\)

\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)

+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)

\(\Leftrightarrow x^2+5x-6=0\)

\(\Leftrightarrow x^2-x+6x-6=0\)

\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)

+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)

\(\Leftrightarrow x^2+5x+16=0\)

\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)

\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)

Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

\(\Rightarrow\) loại

Vậy \(x\in\left\{1;-6\right\}\).

\(b,\) Đề thiếu vế phải rồi bạn.

a: =3x^2-3x-8x+8=(x-1)(3x-8)

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

c: =x^2-6x+2x-12=(x-6)(x+2)

a) \(\left(2x^4-3x^3-3x^2-2+6x\right):\left(x^2-2\right)=2\left(x^2-\dfrac{3}{2}x+\dfrac{1}{2}\right)\left(x^2-2\right):\left(x^2-2\right)=2x^2-3x+1\)

a: \(\dfrac{2x^4-3x^3-3x^2+6x-2}{x^2-2}\)

\(=\dfrac{2x^4-4x^2-3x^3+6x+x^2-2}{x^2-2}\)

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

b: \(\dfrac{5x^3-3x^2+7}{x^2+1}\)

\(=\dfrac{5x^3+5x-3x^2-3-5x+10}{x^2+1}\)

\(=5x-3+\dfrac{-5x+10}{x^2+1}\)

a) \(x^5-3x^2+x^4-4x-x^5+5x^4+x^2-1\)

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

\(=-2x^2+6x^4-4x-1\)

\(=6x^4-2x^2-4x-1\)

- Hệ số tự do: \(-1\)

- Hệ số cao nhất:  \(6\)

b) \(x-x^9+x^2-5x^3+x^6-x+3x^9+2x^6-x^3+7\)

\(=\) \((x-x)+(x^9+3x^9)+x^2+(-5x^3-x^3)+(x^6+2x^6)+7\)

\(=4x^9+x^2-6x^3+3x^6+7\)

\(=4x^9+3x^6-6x^3+x^2+7\)

- Hệ số tự do: \(7\)

- Hệ số cao nhất: \(4\)

mong trả lời bài dễ này

1+1=2

1+2=3

1+3=4

1+4=5

được chưa vậy??????????????

kết bạn nha!!!!!!!!!!!

8:

a: M(x)=x^4+2x^2+1

N(x)=x^4+2x^2-3x-14

P(x)=M(x)-N(x)=3x+15

P(x)=0

=>3x+15=0

=>x=-5

b: M(x)=x^2(x^2+1)+1>0

=>M(x) vô nghiệm