Bài 1.

Ta có : B = ( x + 2 )² + ( x - 2 )² - 2( x + 2 )( x - 2 )

= [ ( x + 2 ) - ( x - 2 ) ]²

= ( x + 2 - x + 2 )²

= 4² = 16

=> B không phụ thuộc vào x

Vậy với x = -4 thì B vẫn bằng 16

Bài 2.

4x² - 4x + 1 = ( 2x )² - 2.2x.1 + 1² = ( 2x - 1 )²

Bài 3.

Ta có : \(A=\frac{3}{2}x^2+2x+3\)

\(=\frac{3}{2}\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)+\frac{7}{3}\)

\(=\frac{3}{2}\left(x+\frac{2}{3}\right)^2+\frac{7}{3}\ge\frac{7}{3}\forall x\)

Dấu "=" xảy ra khi x = -2/3

=> MinA = 7/3 <=> x = -2/3

= 8x^3 -27 -8x^3 +8x

= -27+8x

1. y(y+1)-5y-5                    2. 4x³=x

=y(y+1)-(5y+5)                    <=>4x³-x=0

=y(y+1)-5(y+1)                    <=>x(4x²-1)=0

=(y+1)(y-5)                          <=>x(4x²-1)=0

                                            <=>\(\orbr{\begin{cases}x=0\\4x^2-1=0\end{cases}}\)=\(\orbr{\begin{cases}x=0\\4x^2=1\end{cases}}\)=\(\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\end{cases}}\)=\(\orbr{\begin{cases}x=0\\x=+_-\frac{1}{2}\end{cases}}\)

3. M= (x+3)² -(4x+1)-x(2x+1)

   M= (x²+6x+9)-4x-1-2x²-x

   M=x²+6x+9-4x-1-2x²-x

   M= -x²+x+8

\(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\\ ---\\ 4x^2-4x-3\\ =4x^2-4x+1-4\\ =\left(2x-1\right)^2-2^2=\left(2x-1-2\right)\left(2x-1+2\right)\\ =\left(2x-3\right)\left(2x+1\right)\)

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

=(2x-1)^2

2: =4x^2-6x+2x-3

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

=(2x-3)(2x+1)

Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

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

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

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

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

1 )

\(\left(2x-1\right)^2+\left(3x+1\right)^2+2\left(2x-1\right)\left(3x+1\right)=\left[\left(2x-1\right)+\left(3x+1\right)\right]^2=\left(5x\right)^2=25x^2\)

2 ) 

\(4x^4+1=\left(2x^2\right)^2+2.2x^2.1+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)

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

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

\(=\left(5x\right)^2=25x^2\)

1) (2x-1)²  + (3x+1)² +2(2x-1)(3x+1)

=(2x-1+3x+1)²

=(5x)²=25x²

2)4x⁴ +1 =(4x⁴ +4x² +1) -4x² =(2x² +1)² -4x² =(2x² +1 -2x)(2x²+1+2x)