a. 64x⁴+1

= (8x²)²+1²

= (8x²)²+16x²+1²-16x²

= (8x²+1)²-(4x)²

= (8x²+1-4x)(8x²+1+4x)

b. 81x⁴+4

= (9x²)²+2²

= (9x²)²+36x²+2²-36x²

= (9x²+2)²-(6x)²

= (9x²+2-6x)(9x²+2+6x)

a)x³+3x²+3x+1-27z³

=(x+1)³-(3z)³

=(x+1-3z)[(x+1)²+3z(x+1)+9z²

b)81x⁴+4

=(9x²)²+2²

=(9x²+2)²-(6x)²

=(9x²-6x+2)(9x²+6x+2)

81x²+4 = 4*(\(\frac{81}{4}x^2+1\))

81x²+4= (9x)² + 36x + 2² -36x= (9x+2)²-36x                            (à mà phân tích đa thức thành nhân tử là gì nhỉ)

c: \(x^4+x^3-4x^2+x+1\)

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

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

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

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

1. \(x^2-x+\frac{1}{4}-\frac{485}{4}=\left(x-\frac{1}{2}\right)^2-\frac{485}{4}=\left(x-\frac{1}{2}-\frac{\sqrt{485}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{485}}{2}\right)=\left(x-\frac{1+\sqrt{485}}{2}\right)\left(x+\frac{\sqrt{485}-1}{2}\right)\)

2) \(81x^2+4=4\left(\frac{81}{4}x^2+1\right)\)

3) \(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\)=> Min A =-3 <=> x=2

. Nhớ L I K E

1.

\(a,x^2-x-121\)\(=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{485}{4}\)\(=\left(x-\frac{1}{2}\right)^2-\frac{485}{4}\)\(=\left(x-\frac{1}{2}-\frac{\sqrt{485}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{485}}{2}\right)\)

\(b,81x^2+4\)\(=\left(9x^2\right)^2+2^2=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2\)

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

2.

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

Vì \(\left(x-2\right)^2\ge0\)\(\Rightarrow\left(x-2\right)^2-3\ge-3\)

Dấu ''='' xảy ra khi x-2=0    => x=2

Vậy GTNN của A là A=-3 khi x=2

a sai đề

b)\(x^2-x-12\)

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

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

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

c sai đề