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Ta có: \(3^{2x-1}=27^{x+2}\)

\(\Leftrightarrow3x+6=2x-1\)

hay x=-7

3^{2x - 1} = 27^{x + 2}

<=> 3^{2x - 1} = 3^{3(x + 2)}

<=> 3^{2x - 1} = 3^{3x + 6}

<=> 2x - 1 = 3x + 6

<=> -1 - 6 = 3x - 2x

<=> x = -7

\(3^{2x-1}=3^{3+x+2}\)
2x-1=3+x+2
2x-1=5+x
x=6

      \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

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

       \(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

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

\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)=\left(x+y+3\right)\left(x+y-4\right)\)  \(P=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

   \(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (nhóm 2 cái đầu với cuối lại với nhau, 2 cái giữa vào 1 nhóm)

Đặt \(x^2+7x+11=a\)

Ta có: \(P=\left(a-1\right)\left(a+1\right)-24\)

\(=a^2-25=\left(a-5\right)\left(a+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

d,   \(4x^4-32x^2+1\)

\(=4x^4+4x^2+1-36x^2\)

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

1. a) \(8x^3-32x=8x\left(x^2-4\right)=8x\left(x-4\right)\left(x+4\right)\)

b) \(y^3+64+\left(y+4\right)\left(y-16\right)=\left(y^3+4^3\right)+\left(y+4\right)\left(y-16\right)\)

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

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

2) a)

\(4x^3-9x=0\)

\(\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow x\left(2x+3\right)\left(2x-3\right)=0\)

<=> x=0 hoặc 2x+3=0 hoặc 2x-3=0

<=> x=0 hoặc x=-3/2 hoặc x=3/2

b) \(A=x^3-9x^2+27x-27=x^3-3.x^2.3+3.x.3^2-3^3=\left(x-3\right)^3\)

Tại x=203

A=(203-3)³=200³

Bài 1 :

a) \(8x^3-32x\)

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

\(=8x\left(x-2\right)\left(x+2\right)\)

b) \(y^3+64+\left(y+4\right)\left(y-16\right)\)

\(=\left(y^3+4^3\right)+\left(y+4\right)\left(y-16\right)\)

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

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

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

Bài 2 :

a) \(4x^3-9x=0\)

\(x\left(4x^2-9\right)=0\)

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

\(x\left(2x-3\right)\left(2x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\2x-3=0\\2x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\\x=\frac{-3}{2}\end{cases}}}\)

P.s: ở trên dùng ngoặc vuông nhé

b) \(A=x^3-9x^2+27x-27\)

\(A=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)

\(A=\left(x-3\right)^3\)

Thay x = 203 vào biểu thức ta có :

\(A=\left(203-3\right)^3\)

\(A=200^3\)

\(A=8000000\)

i. (x²+y²+z²).(x+y+z)²+(xy+yz+zx)²

\(x^8+x+1\)

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

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

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

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

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

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

b)\(27x^3-27x^2+18x-4=27x^3-18x^2-9x^2+12x+6x-4\)

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

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

c)\(4x^4-32x^2+1=4x^4+4x^2-36x^2+1\)

\(=\left(4x^4+4x^2+1\right)-36x^2\)

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

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