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

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

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

\(2x-2y-x^2+2x-y^2\)

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

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

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

Có :\(\left(n+3\right)^3-\left(n-3\right)^3\)

\(=\left[\left(n+3\right)-\left(n-3\right)\right]\left[\left(n+3\right)^2+\left(n+3\right)\left(n-3\right)+\left(n-3\right)^2\right]\)

\(=6\left(3n^2+9\right)==18\left(n^2+3\right)\) chia hết 18 (DPCM)

Ta có:\(\left(n+3\right)^3-\left(n-3\right)^3\)

\(=\left[\left(n+3\right)-\left(n-3\right)\right]\left[\left(n+3\right)^2+\left(n+3\right)\left(n-3\right)+\left(n-3\right)^2\right]\)

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

\(=6.\left(3n^2+9\right)\)

\(=6.3.\left(n^2+3\right)\)

\(=18.\left(n^2+3\right)⋮18\)

Vậy \(\left(n+3\right)^3-\left(n-3\right)^3⋮18\)

2(x - y)(x + y) + (x - y)²

= 2(x - y)(x + y) + x² - 2xy + y²

= 2x² - 2y² + x² - 2xy + y²

= (2x² + x²) - 2xy + (-2y² + y³) 

= 3x² - 2xy - 3y³

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

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

\(=3x^2-2xy-y^2\)

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

(x + y)² + x - y²

= x² + 2xy + y² + x - y² 

= x² + 2xy + (y² - y²) + x

= x² + 2xy + x

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

\(=x^2+2xy+y^2+x-y^2\)

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

\(=x^2+2xy+x\)

\(f\left(x\right)\) chia \(x-2\) dư \(11\Leftrightarrow f\left(x\right)=\left(x-2\right)H\left(x\right)+11\Leftrightarrow f\left(x\right)-11=\left(x-2\right)H\left(x\right)\)

\(f\left(x\right)\) chia \(x-3\) dư \(23\Leftrightarrow f\left(x\right)=\left(x-3\right)G\left(x\right)+23\Leftrightarrow f\left(x\right)-23=\left(x-3\right)G\left(x\right)\)

Do vậy \(\left(f\left(x\right)-11\right)\left(f\left(x\right)-23\right)=\left(x-2\right)\left(x-3\right)H\left(x\right)G\left(x\right)\)

\(\Rightarrow f\left(x\right)^2-34f\left(x\right)+253⋮\left(x-2\right)\left(x-3\right)\)

Do vậy \(f\left(x\right)\) chia \(\left(x-2\right)\left(x-3\right)\) dư \(-253\)

a) Vì a chia 3 dư 1 nên a có dạng 3m+1 , vì b chia 3 dư 2 nên b có dạng 3n+2. \(\left(m,n\in N\right)\)

Ta có \(ab=\left(3m+1\right)\left(3n+2\right)=3mn+6m+3n+2\)

                \(=3\left(mn+2m+n\right)+2\)

Vậy ab chia 3 dư 2 .

b) Vì a chia 5 dư 4 nên a có dạng 5k-1 \(\left(k\in N\right)\)

Ta có \(a^2=\left(5k-1\right)^2=25k^2-10k+1=5\left(5k^2-2k\right)+1\)

Vậy \(a^2\) chia 5 dư 1 .

\(=>2^3-x^3+\left(x^2-3x\right)\left(x+4\right)-x^2+24=0\)

\(=>8-x^3+x^3+x^2-12x-x^2+24=0\)

\(=>-12x=-16=>x=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

\(\left(x^2+2x+4\right)\left(2-x\right)+x\left(x-3\right)\left(x+4\right)-x^2+24=0\)

\(\Leftrightarrow2x^2-x^3+4x-2x^2+8-4x+\left(x^2-3x\right)\left(x+4\right)-x^2+24=0\)

\(\Leftrightarrow2x^2-x^3+4x-2x^2+8-4x+x^3+4x^2-3x^2-12x-x^2+24=0\)

\(\Leftrightarrow-12x+8+24=0\)

\(\Leftrightarrow-12x+32=0\)

\(\Leftrightarrow-12x=-32\)

\(\Leftrightarrow x=\frac{8}{3}\)

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

<=> \(9x^2-9x+2=9x^2+6x+1\)

<=>  \(15x=1\) <=> \(x=\frac{1}{15}\)

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

<=> \(4x^2+3x-1=4x^2-12x+9\)

<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)

c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)

<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)

<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)

d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)

<=> 16 - 9x² = 12x - 9x2 - 3

<=> 12x = 19

<=> x = 19/12

e) x(x + 1)(x + 2)(x + 3) = 24

<=> (x² + 3x)(x² + 3x + 2) = 24

<=> (x² + 3x)²  + 2(x² + 3x) - 24 = 0

<=> (x² + 3x)² + 6(x² + 3x) - 4(x² + 3x) - 24 = 0

<=> (x² + 3x + 6)(x² + 3x - 4) = 0

<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

g) (7x - 2)² = (7x - 3)(7x + 2)

<=> 49x² - 28x + 4 = 49x² - 7x - 6

<=> 21x = 10 <=> x = 10/21