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.

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

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

\(\Rightarrow x^2+2xy+y^2=4\)

\(\Rightarrow2xy=4-\left(x^2+y^2\right)=4-10=-6\Rightarrow xy=-3\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\left(10+6\right)=32\)

Ta có HPT:

\(\left\{{}\begin{matrix}x+y=2\\x^2+y^2=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-y\\x=\sqrt{10-y^2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2-y=\sqrt{10-y^2}\\x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=3\end{matrix}\right.\)

Thay x = 3, y = -1 vào x³ + y³, ta được:

3³ + (-1)³ = 27 + (-1) = 26

a) Ta có:x+y = 2 <=> (x+y)^2 = 4 <=> x^2 + y^2+ 2xy = 4 (1)
mà x^2 +y^2=10. Thay vào (1) => xy= - 3
=> x^3 + y^3 = (x+y)(x^2+y^2-xy) = 1(10+3) =13

a) Ta có:x+y = 2 <=> (x+y)^2 = 4 <=> x^2 + y^2+ 2xy = 4 (1)
mà x^2 +y^2=10. Thay vào (1) => xy= - 3
=> x^3 + y^3 = (x+y)(x^2+y^2-xy) = 1(10+3) =13

chcú bn hok tốt @_@