c) Ta có: \(C=4x^2+y^2-4xy+8x-4y+4\)

\(=\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot2+2^2\)

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

a) Thu gọn:

P(x) = x⁴+(-7x²+4x²)+(x+6x)-2x³-2

P(x) = x⁴-3x²+7x-2x³-2

Sắp xếp: P(x) = x⁴-2x³-3x²+7x-2

Thu gọn:

Q(x) = x⁴+(-3x+x)+(-5x³+6x³)+1

Q(x) = x⁴-2x+x³+1

Sắp xếp: Q(x)= x⁴+ x³-2x+1

b/ Nếu x=2, ta có:

P(2) = 2⁴-2.2³-3.2²+7.2-2

        = 16 - 2.8 - 3.4 + 14 -2

        = 16-16-12+14-2

        = -12+14-2 

        = 0

=> x=0 là nghiệm của P(x)

Q(2)= 2⁴+ 2³-2.2+1

= 16+8-4+1

= 24-4+1

=21

mà 21≠0

Vậy: x=2 không phải là nghiệm của Q(x)

=>

Lời giải:

$A=x^4-4x^3+7x^2-12x+75$

$=(x^2-2x)^2+3x^2-12x+75$

$=(x^2-2x)^2+3(x^2-4x+4)+63$

$=(x^2-2x)^2+3(x-2)^2+63\geq 63$

Vậy $A_{\min}=63$. Giá trị này đạt tại $x^2-2x=x-2=0$

$\Leftrightarrow x=2$

\(A=\left(x^4-4x^3+4x^2\right)+\left(3x^2-12x+12\right)+63\)

\(A=x^2\left(x^2-4x+4\right)+3\left(x^2-4x+4\right)+63\)

\(A=\left(x^2+3\right)\left(x-2\right)^2+63\ge63\)

\(A_{min}=63\) khi \(x=2\)

\(a,x^4-4x^3-19x^2+106x-120=0\\ \Rightarrow\left(x-4\right)\left(x^3-19x+30\right)=0\Rightarrow\left(x-4\right)\left(x+5\right)\left(x^2-5x+6\right)=0\\ \Rightarrow\left(x-4\right)\left(x+5\right)\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-5\\x=2\\x=3\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{-5;2;3;4\right\}\)

\(b,4x^4+12x^3+5x^2-6x-15=0\\ \Rightarrow\left(x-1\right)\left(4x^3+16x^2+21x+15\right)=0\\ \Rightarrow\left(x-1\right)\left[\left(4x^3+10x^2\right)+\left(6x^2+15x\right)+\left(6x+15\right)\right]=0\\ \Rightarrow\left(x-1\right)\left[2x^2\left(2x+5\right)+3x\left(2x+5\right)+3\left(2x+5\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(2x+5\right)\left(2x^2+3x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{2}\\2x^2+3x+3=0\left(vô.lí\right)\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{1;-\dfrac{5}{2}\right\}\)

Bài 3: 

\(B=x^4-4x^3-2x^2+12x+9\)

\(=x^4-3x^3-x^3+3x^2-5x^2+15x-3x+9\)

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

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

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

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

Bài 3: 

\(B=x^4-4x^3-2x^2+12x+9=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(3x^2+3x\right)+\left(9x+9\right)=\left(x^3-5x^2+3x+9\right)\left(x+1\right)=\left[\left(x^3+x^2\right)-\left(6x^2+6x\right)+\left(9x+9\right)\right]\left(x+1\right)=\left(x^2-6x+9\right)\left(x+1\right)^2=\left(x-3\right)^2\left(x+1\right)^2=\left[\left(x-3\right)\left(x+1\right)\right]^2\)

a) ( x 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).

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

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

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

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