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.

2:

a: \(x^2-12x+20\)

\(=x^2-2x-10x+20\)

=x(x-2)-10(x-2)

=(x-2)(x-10)

b: \(2x^2-x-15\)

=2x^2-6x+5x-15

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

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

c: \(x^3-x^2+x-1\)

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

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

d: \(2x^3-5x-6\)

\(=2x^3-4x^2+4x^2-8x+3x-6\)

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

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

e: \(4y^4+1\)

\(=4y^4+4y^2+1-4y^2\)

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

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

f; \(x^7+x^5+x^3\)

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

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

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

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

g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

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

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

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

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

h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)

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

\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)

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

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

\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)

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

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

i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)

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

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

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

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

j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)

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

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

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

a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

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

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

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

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

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

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

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

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

\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)

a) 2x²+6x=2x(x+3)
b) x⁴+3x³+x+3=(x⁴+x)+(3x³+3)=x(x³+1)+3(x³+1)=(x+3)(x³+1)
c) 64-x²-y²+2xy=-(x²-2xy+y²)+8²=8-(x+y)²=(8+x+y)(8-x-y)

A= (x+5)(x+1)+(x-2)(x²+2xx+4)-(x²+x-2)
A= x²+6x+5+x³-8-x²-x+2
A= x³+(x²-x²)+(6x-x)+(5-8+2)
A= x³+5x-1

Đề bạn có mấy chỗ thiếu mk bổ sung nha

\(a,2^3+4^2+6x=8+16+6x=6x+24=x\left(x+4\right)\\ b,x^2-4=\left(x-2\right)\left(x+2\right)\\ c,x^2-10x+25=\left(x-5\right)^2\\ d,x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\\ e,x^2+xy-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\\ g,x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)

Tick plzz

a: Ta có: \(2x^3+4x^2+6x\)

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

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

c: \(x^2-10x+25=\left(x-5\right)^2\)

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

e: \(x^2+xy-3x-3y\)

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

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

g: \(x^2-4x+4-y^2\)

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

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

a: 3x^2-9

=3*x^2-3*3

=3(x^2-3)

b: 1/2x^2-2y^2

=1/2(x^2-4y^2)

=1/2(x-2y)(x+2y)

c: 3x^2-12y^2

=3(x^2-4y^2)

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

d: 1/3x^2y^2-3x^2

=1/3x^2(y^2-9)

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

\(a,=7xy\left(x^2-2xy+y^2\right)=7xy\left(x-y\right)^2\\ b,=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\)

a) \(3x-1=\left(\sqrt{3x}\right)^2-1^2=\left(\sqrt{3x}-1\right)\left(\sqrt{3x}+1\right)\)

b) \(4x-25=\left(2\sqrt{x}\right)^2-5^2=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4\left(x\ge0\right)\Rightarrow x+\sqrt{x}-4\sqrt{x}-4\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)\)

b) \(4x-25=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4=\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)\)

a: \(9x^3y^2+3x^2y^2\)

\(=3x^2y^2\cdot3x+3x^2y^2\cdot1\)

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

b: \(x^2-2x+1-y^2\)

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

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

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