a) Ta có:

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

\(\dfrac{1}{x+y}=\dfrac{1\cdot x}{x\cdot\left(x+y\right)}=\dfrac{x}{x\left(x+y\right)}\)

b) Ta có:

\(\dfrac{1}{x^2+xy}=\dfrac{1}{x\left(x+y\right)}=\dfrac{1\cdot y}{xy\left(x+y\right)}=\dfrac{y}{xy\left(x+y\right)}\)

\(\dfrac{1}{xy+y^2}=\dfrac{1}{y\left(x+y\right)}=\dfrac{1\cdot x}{xy\left(x+y\right)}=\dfrac{x}{xy\left(x+y\right)}\)

c) Ta có: 

\(\dfrac{2x}{x^2+4x+4}=\dfrac{2x}{x^2+2\cdot2\cdot x+2^2}=\dfrac{2x}{\left(x+2\right)^2}\)

\(\dfrac{x+1}{x+2}=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x+2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)^2}\)

d: \(\dfrac{3x-24}{x+y}=\dfrac{x\left(3x-24\right)}{x\left(x+y\right)}=\dfrac{3x^2-24x}{x^2+xy}\)

\(\dfrac{xy+5}{x^2+xy}=\dfrac{xy+5}{x^2+xy}\)

e: \(\dfrac{2xy-3y^2}{x^2-3xy}=\dfrac{y\left(2x-3y\right)}{x\left(x-3y\right)}=\dfrac{3y\left(2x-3y\right)}{3x\left(x-3y\right)}\)

\(\dfrac{x}{3x-9y}=\dfrac{x}{3\left(x-3y\right)}=\dfrac{x^2}{3x\left(x-3y\right)}\)

f: \(\dfrac{1}{x-2}=\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}\)

\(\dfrac{3}{x+1}=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{3x-6}{\left(x-2\right)\left(x+1\right)}\)

Thay x=0 và y=3 vào (d), ta được:

\(0\left(\sqrt{3}-2\right)+3=3\)

=>3=3(đúng)

=>A(0;3) thuộc (d)

\(C=1-6y-5y^2-12xy-9x^2\)

\(\Rightarrow C=-4y^2-12xy-9x^2-y^2-6y+1\)

\(\Rightarrow C=-\left(4y^2+12xy+9x^2\right)-\left(y^2+6y+9\right)+1+9\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\)

mà \(\left\{{}\begin{matrix}-\left(2y-3x\right)^2\le0,\forall x;y\\-\left(y+3\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\le10\)

\(\Rightarrow GTLN\left(C\right)=10\left(tạix=-2;y=-3\right)\)

Đặt \(A=-x^2-y^2+xy+2x+2y\)

\(\Rightarrow2A=-2x^2-2y^2+2xy+4x+4y\)

\(=-\left(x^2-4x+4\right)-\left(y^2-y+4\right)-\left(x^2-2xy+y^2\right)+8\)

\(=8-\left(x-2\right)^2-\left(y-2\right)^2-\left(x-y\right)^2\)

anh lm chi tiết hộ e ạ 
 

a) \(A=\left(15x^5y^3-10x^3y^2+20x^4y^4\right):5x^2y\)

\(A=5x^2y\cdot\left(3x^3y^2-2xy+4x^2y^3\right):5x^2y\)

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

Thay x=1; y=-2 vào A ta có:

\(A=3\cdot1^3\cdot\left(-2\right)^2-2\cdot1\cdot-2+4\cdot1^2\cdot\left(-2\right)^3=-16\)

Vậy: ...

b) \(B=\left(4x^4y^2+3x^4y^3-6x^3y^2\right):\left(-x^2y^2\right)\)

\(B=\left(-x^2y^2\right)\cdot\left(-4x^2-3x^2y+6x\right):\left(-x^2y^2\right)\)

\(B=-4x^2-3x^2y+6x\)

Thay x=y=-2 vào B ta có:

\(B=-4\cdot\left(-2\right)^2-3\cdot\left(-2\right)^2\cdot-2+5\cdot-2=-4\)

Vậy: ...

\(a.\\ A=\left(15x^5y^3-10x^3y^2+20x^4y^4\right):5x^2y\\ A=\left(15x^5y^3:5x^2y\right)+\left(-10x^3y^2:5x^2y\right)+\left(20x^4y^4:5x^2y\right)\\ A=3x^3y^2-2xy+4x^2y^3\left(1\right)\)

Thay x = 1 , y = -2 vào (1)

\(A=3.1^3.\left(-2\right)^2-2.1.-2+4.1^2.\left(-2\right)^3=-16\)

Tương tự như vậy với câu b nha , nếu ko hiểu hỏi mình

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

Vậy biểu thức trên không phụ thuộc vào giá trị của x

`(x-1)(x^2 + x + 1) - (x+1)(x^2 - x + 1)`

`=x^3 - 1 - (x^3 + 1)`

`=x^3 - 1 - x^3 -1`

`=-2`
Vậy biểu thức trên không phụ thuộc vào giá trị của `x`

-2x(x + 3) + x(2x - 1) = 10

-2x² - 6x + 2x² - x = 10

-7x = 10

x = -10/7