anh yeu thow:đồ ngu

Trả lời:

7, 5( x + y )² + 15( x + y )

= 5( x + y )( x + y + 3 )

9, 7x( y - 4 )² - ( 4 - y )³ 

= 7x ( 4 - y )² - ( 4 - y )

= ( 4 - y )² ( 7x - 4 + y )

11, ( x + 1 )( y - 2 ) - ( 2 - y )²

= ( x + 1 )( y - 2 ) - ( y - 2 )²

= ( y - 2 )( x + 1 - y + 2 )

= ( y - 2 )( x - y + 3 )

8, 9x ( x - y ) - 10 ( y - x )² 

= 9x ( x - y ) - 10 ( x - y )²

= ( x - y )[ ( 9x - 10 ( x - y ) ]

= ( x - y )( 9x - 10x + 10y )

= ( x - y )( 10y - x )

10, ( a - b )² - ( a + b )( b - a ) 

= ( b - a )² - ( a + b )( b - a )

= ( b - a )( b - a - a - b )

= - 2a( b - a )

= 2a ( a - b )

12, 2x ( x - 3 ) + y ( x - 3 ) + ( 3 - x )

= 2x ( x - 3 ) + y ( x - 3 ) - ( x - 3 )

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

 \(x^2-\frac{y^2}{3}=x^2+\frac{y^2}{-5}\)nếu bạn chép sai đề => kq sài vô lý

sua de lam tiep

\(\left(xy\right)^{10}=1024=2^{10}=>xy=2=>\left(xy\right)^2=4\) 

\(\frac{x^2-y^2}{3}=\frac{x^2+y^2}{-5}=\frac{2x^2}{-2}=-x^2\)

\(\Leftrightarrow\frac{x^2-y^2}{3}=-x^2=>4x^2-y^2=0\)\(\Leftrightarrow4x^2=y^2\Leftrightarrow4x^2.y^2=y^2.y^2=>y^4=4.4=16=2^4=>y=!2!\)

KL:

y=!2!

x=!1!

(x,y)=(-1,-2); (1,2)

\(\dfrac{8x^3y^2-6x^2y^3}{-2xy}=\dfrac{8x^3y^2}{-2xy}+\dfrac{6x^2y^3}{2xy}=-4x^2y+3xy^2\)

⇒ Chọn A.

\(P=\left(x+2y\right)^2-2\left(x+2y\right)\left(y-1\right)+\left(y-1\right)^2\\ P=\left(x+2y-y+1\right)^2=\left(x+y+1\right)^2\\ Q.sai.đề\\ M=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\\ M=1^3-3xy\left(x+y-1\right)=1-3xy\left(1-1\right)=1-0=1\\ x+y=2\Leftrightarrow\left(x+y\right)^2=4\\ \Leftrightarrow x^2+y^2+2xy=4\\ \Leftrightarrow2xy=4-10=-6\\ \Leftrightarrow xy=-3\\ N=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\\ N=2\left(10+3\right)=2\cdot13=26\)

\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{y^2+x^2+y^2-x^2}{3+5}=\frac{2y^2}{8}=\frac{y^2}{4}\)

\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{\left(x^2+y^2\right)-\left(y^2-x^2\right)}{5-3}=\frac{2x^2}{2}=x^2\)

\(\frac{y^2}{4}=x^2\Rightarrow\frac{y^{10}}{1024}=\frac{x^{10}}{1}\Rightarrow x^{20}=\frac{x^{10}.y^{10}}{1024}=\frac{1024}{1024}=1\)

=>x=-1;1

xét x=-1=>y²=4=>y=-2;2

xét x=1=>y²=4=>y=-2;2

Vậy (x;y)=(-1;-2);(-1;2);(1;-2);(1;2)

\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{y^2+x^2+y^2-x^2}{3+5}=\frac{2y^2}{8}=\frac{y^2}{4}\)

\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{\left(x^2+y^2\right)-\left(y^2-x^2\right)}{5-3}=\frac{2x^2}{2}=x^2\)

\(\frac{y^2}{4}=x^2\Rightarrow\frac{y^{10}}{1024}=\frac{x^{10}}{1}\Rightarrow x^{20}=\frac{x^{10}.y^{10}}{1024}=\frac{1024}{1024}=1\)

=>x=-1;1

xét x=-1=>y²=4=>y=-2;2

xét x=1=>y²=4=>y=-2;2

Vậy (x;y)=(-1;-2);(-1;2);(1;-2);(1;2)

Áp dụng tính chất của dãy tỉ số bằng nhau có:  \(\frac{y^2-x^2}{3}=\frac{x^2+y^2}{5}=\frac{\left(y^2-x^2\right)+\left(x^2+y^2\right)}{3+5}=\frac{\left(y^2-x^2\right)-\left(x^2-y^2\right)}{3-5}\)

 => \(\frac{2y^2}{8}=\frac{-2x^2}{-2}\Rightarrow\frac{y^2}{4}=x^2\) => y² = 4x²

Ta có x¹⁰.y¹⁰ = x¹⁰. (4x²)⁵ = 1024.x²⁰ = 1024 => x²⁰ = 1 => x =1 hoặc x = -1

=> y² = 4 => y = 2 hoặc y = -2

Vậy ... 

\(\left(9+5\right)\)