=\(\frac{2^{12}.3^5+2^{12}.3^4}{2^{12}.3^6+2^{12}.3^3}\)

=\(\frac{2^{12}\left(3^5+3^4\right)}{2^{12}\left(3^6+3^3\right)}\)

\(=\frac{324}{756}\)

=\(\frac{3}{7}\)

1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)

\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)

\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)

\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)

\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)

2: \(\left(x^2-y^2\right)\cdot C=-8\)

=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)

=>\(\left(x-y\right)^3=-8\)

=>x-y=-2

=>x=y-2

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

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

\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)

\(=\left(y-1\right)\left(-4y+4\right)+4xy\)

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

\(=-4y^2+8y-4+4y^2-8y\)
=-4

A = |x|² + 4xy - 3y. |y|² = 25 + 4xy - 3y 

+ Nếu x = 5; y = 1 => A = 25 + 4.5.1 - 3.1 = 42

+ Nếu x = 5; y = -1 => A = 25 + 4.5. (-1) - 3.(-1) = 8

+ Nếu x = -5 ; y = 1 => A = 25 + 4.(-5).1 - 3.1 = 2

+ Nếu x = -5; y = -1 => A = 25 + 4.(-5). (-1) - 3.(-1) = 48

Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1

(x + 20)⁴ + (2y - 1)²⁰²⁴ ≤ 0

⇒ (x + 20)⁴ = 0 và (2y - 1)²⁰²⁴ = 0

*) (x + 20)⁴ = 0

x + 20 = 0

x = 0 - 20

x = -20

*) (2y - 1)²⁰²⁴ = 0

2y - 1 = 0

2y = 1

y = 1/2

M = 5.(-20)².1/2 - 4.(-2).(1/2)²

= 1000 + 2

= 1002

Trường hợp 1: x=1,5 và y=-0,75

\(P=\dfrac{3}{2}-4\cdot\dfrac{3}{2}\cdot\dfrac{-3}{4}+\dfrac{-3}{4}=\dfrac{3}{4}+3\cdot\dfrac{3}{2}=\dfrac{3}{4}+\dfrac{9}{2}=\dfrac{3}{4}+\dfrac{18}{4}=\dfrac{21}{4}\)

Trường hợp 2: x=-1,5 và y=-0,75

\(P=\dfrac{-3}{2}-4\cdot\dfrac{-3}{2}\cdot\dfrac{-3}{4}+\dfrac{-3}{4}\)

\(=\dfrac{-9}{4}-4\cdot\dfrac{9}{8}=\dfrac{-9}{4}-\dfrac{9}{2}=\dfrac{-27}{4}\)

Lời giải:

\(|x|=1,5\Rightarrow \left[\begin{matrix} x=1,5\\ x=-1,5\end{matrix}\right.\)

Nếu $x=1,5$ và $y=-0,75$:

\(P=x-4xy+y=1,5-4.1,5.(-0,75)+(-0,75)=\frac{21}{4}\)

Nếu \(x=-1,5; y=-0,75\):

\(P=x-4xy+y=(-1,5)-4(-1,5)(-0,75)+(-0,75)=\frac{-27}{4}\)