\(A=16x^2-y^2-16x^2+8x=8x-y^2\\ A=8\cdot3-\left(-1\right)^2=24-1=23\\ B=64x^3-80x-64x^3-1=-80x-1\\ B=-80\cdot\dfrac{1}{5}-1=-16-1=-17\)

2a) pt <=> (x + 6)^2 = 0

<=> x = -6

b) pt <=> (4x - 1)^2 = 0

<=> x = 1/4

c) pt<=> (x + 1)^3 = 0

<=> x = -1

Bài 1:

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)

Bài 2: 

a: Ta có: \(x^2+12x+36=0\)

\(\Leftrightarrow x+6=0\)

hay x=-6

b: Ta có: \(16x^2-8x+1=0\)

\(\Leftrightarrow4x-1=0\)

hay \(x=\dfrac{1}{4}\)

Bài 1: 

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)

\(=-12x^2-24\)

c: Ta có: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)

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

\(=4x^2\)

a, \(x^4-4x^3-6x^2-4x+1=0\)(*)

<=> \(x^4+4x^2+1-4x^3-4x+2x^2-12x^2=0\)

<=> \(\left(x^2-2x+1\right)^2=12x^2\)

<=>\(\left(x-1\right)^4=12x^2\) <=> \(\left[{}\begin{matrix}\left(x-1\right)^2=\sqrt{12}x\\\left(x-1\right)^2=-\sqrt{12}x\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x^2-2x+1-\sqrt{12}x=0\left(1\right)\\x^2-2x+1+\sqrt{12}x=0\left(2\right)\end{matrix}\right.\)

Giải (1) có: \(x^2-2x+1-\sqrt{12}x=0\)

<=> \(x^2-2x\left(1+\sqrt{3}\right)+\left(1+\sqrt{3}\right)^2-\left(1+\sqrt{3}\right)^2+1=0\)

<=> \(\left(x-1-\sqrt{3}\right)^2-3-2\sqrt{3}=0\)

<=> \(\left(x-1-\sqrt{3}\right)^2=3+2\sqrt{3}\) <=> \(\left[{}\begin{matrix}x-1-\sqrt{3}=\sqrt{3+2\sqrt{3}}\\x-1-\sqrt{3}=-\sqrt{3+2\sqrt{3}}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(ktm\right)\\x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(tm\right)\end{matrix}\right.\)

=> \(x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)

Giải (2) có: \(x^2-2x+1+\sqrt{12}x=0\)

<=> \(x^2-2x\left(1-\sqrt{3}\right)+\left(1-\sqrt{3}\right)^2-\left(1-\sqrt{3}\right)^2+1=0\)

<=> \(\left(x+\sqrt{3}-1\right)^2=3-2\sqrt{3}\) .Có VP<0 => PT (2) vô nghiệm

Vậy pt (*) có nghiệm x=\(-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)

\(16x^2-8x+1\\ =\left(16x^2-4x\right)-\left(4x-1\right)\\ =4x\left(4x-1\right)-\left(4x-1\right)\\ =\left(4x-1\right)\left(4x-1\right)\\ =\left(4x-1\right)^2\)

 (4x)^2 - 2.4x.1 + 1 = 0

 (4x - 1)^2 = 0

4x - 1 = 0

x = 1/4

\(16x^2-8x+1=\left(4x-1\right)^2\)

(4x - 1)²

\(16x^2-8x+1\\ =\left(16x^2-4x\right)-\left(4x-1\right)\\ =4x\left(4x-1\right)-\left(4x-1\right)\\ =\left(4x-1\right)^2\)

\(16x^2-8x+1=\left(4x-1\right)^2\)

Ta có

P   =   ( 4 x   +   1 ) 3   –   ( 4 x   +   3 ) ( 16 x 2   +   3 )     =   ( 4 x ) 3   +   3 . ( 4 x ) 2 . 1   +   3 . 4 x . 1 2   +   1 3   –   ( 64 x 3   +   12 x   +   48 x 2   +   9 )     =   64 x 3   +   48 x 2   +   12 x   +   1   –   64 x 3   –   12 x   –   48 x 2   –   9   =   - 8

Nên P = -8

Q   =   ( x   –   2 ) 3   –   x ( x   +   1 ) ( x   –   1 )   +   6 x ( x   –   3 )   +   5 x     =   x 3   –   3 . x 2 . 2   +   3 x . 2 2   –   2 3   –   x ( x 2   –   1 )   +   6 x 2   –   18 x   +   5 x     =   x 3   –   6 x 2   +   12 x   –   8   –   x 3   +   x   +   6 x 2   –   18 x   +   5 x   =   - 8

=> Q = -8

Vậy P = Q

Đáp án cần chọn là: A