a: \(P=-3x^3+5x\)

\(=x\cdot\left(-3x^2\right)+x\cdot5\)

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

b: \(Q=\left(2x-1\right)+\left(x-2\right)\left(2x-1\right)\)

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

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

c: \(R=4-16x^2\)

\(=4\cdot1-4\cdot4x^2\)

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

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

d: \(S=36-4x^2\)

\(=4\cdot9-4\cdot x^2\)

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

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

e: \(T=8x^3-1\)

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

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

f: \(Q=8-x^3\)

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

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

g: \(N=64-x^3\)

\(=4^3-x^3\)

\(=\left(4-x\right)\left(16+4x+x^2\right)\)

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

Đặt \(t=x^2+3x\) thì biểu thức có dạng \(t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)

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

Đặt \(k=x^2-x+2\) thì biểu thức có dạng

k²+4(k-3)=k²+4k-12=k²-2k+6k-12=k(k-2)+6(k-2)=(k-2)(k+6)=(x²-x)(x²-x+8)=(x-1)x(x²-x+8)

c)làm tương tự câu a

\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

Sao đề là phân tích mà lại "= 0" vậy bạn?

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10+16+8\right)+16\)

\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)

\(=\left(x^2+10x+16+4\right)^2\)

\(=\left(x^2+10+20\right)^2\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right) \left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\)
\(\left(1\right)\Rightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\)
\(=\left(x^2+10x+20\right)^2\)
 

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\)

\(=\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+16\)

\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\)(1) 

Đặt \(x^2-10x+20=t\)thay vào (1) ta được : 

\(\left(t-4\right)\left(t+4\right)+16\)

\(=t^2-16+16\)

\(=t^2\)Thay \(t=x^2-10x+20\)ta được :

\(\left(x^2-10x+20\right)^2\)

\(=\left(x^2-2.5.x+25-25+20\right)^2\)

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

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

      \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+18\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)-16\)

\(=\left(x^2+10x+20\right)^2-16+16=\left(x^2+10x+20\right)^2\)

Chúc bạn học tốt.

      \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(\Rightarrow\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+6\right)\left(x+8\right)\right]+16\)

\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(\Rightarrow\left(x^2+10x+16\right)\left[\left(x^2+10x+16\right)+8\right]+16\)

\(\Rightarrow\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+4^2\)

\(\Rightarrow\left(x^2+10x+20\right)^2\)

B

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

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

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