a: Để phương trình có nghiệm duy nhất thì \(\left(m-3\right)\left(m+2\right)< >0\)

hay \(m\notin\left\{3;-2\right\}\)

Để phương trình vô nghiệm thì \(\left\{{}\begin{matrix}\left(m-3\right)\left(m+2\right)=0\\\left(m-3\right)\left(m-1\right)< >0\end{matrix}\right.\Leftrightarrow m=-2\)

Để phương trình có vô số nghiệm thì m=3

\(x^4+y^4+\left(x+y\right)^4=2\left(x^4+y^4+2x^3y+3x^2y^2+2xy^3\right)\)

\(=2\left(\left(x^4+y^4+2x^2y^2\right)+\left(2x^3y+2xy^3\right)+x^2y^2\right)\)

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

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

Đặt x² + xy + y² = a² ; x + y = b.Ta có :

a⁴ = (a²)² = (x² + xy + y²)² = x⁴ + y⁴ + x²y² + 2x³y + 2xy² + 2x²y² = x⁴ + y⁴ + x²y² + 2xy(x² + y² + xy) = x⁴ + y⁴ + x²y² + 2xya² (1)

mà b = x + y

=> b² = x² + y² + 2xy = a² + xy => b⁴ = a⁴ + x²y² + 2a²xy .Từ (1) và (2) ,ta có :

2a⁴ = x⁴ + y⁴ + a⁴ + x²y² + 2xya² = x⁴ + y⁴ + b⁴.Thay a² = x² + xy + y² ; b = x + y,ta có đpcm

<=> 

\(A=\dfrac{-1}{5}x^3\cdot\dfrac{1}{32}x^{20}y^5\cdot\dfrac{64}{27}x^3y^9\cdot z^{2022}=-\dfrac{2}{135}x^{26}y^{14}z^{2022}\)

`A=\frac{-1}{5}x^3 \times \frac{1}{32}x^{20}y^5 \times \frac{64}{27}x^3y^9 \times z^{2022}=-\frac{2}{135}x^{26}y^{14}z^{2022}`

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

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

\(=3x^2-3y^2-2x^2+4xy-2y^2\)

\(=x^2+4xy-5y^2\)

\(=x^2+4xy+4y^2-9y^2\)

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

\(=\left(x+2y-3y\right)\left(x+2y+3y\right)\)

\(=\left(x-y\right)\left(x+5y\right)\)

A/ \(16x-5x^2-3=\left(15x-3\right)-\left(5x^2-x\right)=3\left(5x-1\right)-x\left(5x-1\right)=\left(5x-1\right)\left(3-x\right)\)

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

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

C/ \(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

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

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\(\Delta'=\left(m-1\right)^2-2\left(m^2-1\right)=-m^2-2m+3>0\)

\(\Rightarrow-3< m< 1\)

Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\left(m-1\right)\\x_1x_2=\dfrac{m^2-1}{2}\end{matrix}\right.\)

\(P=\left(x_1-x_2\right)^2=x_1^2+x_2^2-2x_1x_2\)

\(P=x_1^2+x_2^2+2x_1x_2-4x_1x_2=\left(x_1+x_2\right)^2-4x_1x_2\)

\(P=\left(m-1\right)^2-4\left(\dfrac{m^2-1}{2}\right)\)

\(P=-m^2-2m+3=-\left(m^2+2m+1\right)+4\) 

\(P=-\left(m+1\right)^2+4\le4\)

\(P_{max}=4\) khi \(m+1=0\Leftrightarrow m=-1\) (thỏa mãn)