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\(x^2-\frac{y^2}{3}=x^2+\frac{y^2}{-5}\)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)
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 )
\(\dfrac{8x^3y^2-6x^2y^3}{-2xy}=\dfrac{8x^3y^2}{-2xy}+\dfrac{6x^2y^3}{2xy}=-4x^2y+3xy^2\)
⇒ Chọn A.
\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{y^2+x^2+y^2-x^2}{3+5}=\frac{2y^2}{8}=\frac{y^2}{4}\)
\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{\left(x^2+y^2\right)-\left(y^2-x^2\right)}{5-3}=\frac{2x^2}{2}=x^2\)
\(\frac{y^2}{4}=x^2\Rightarrow\frac{y^{10}}{1024}=\frac{x^{10}}{1}\Rightarrow x^{20}=\frac{x^{10}.y^{10}}{1024}=\frac{1024}{1024}=1\)
=>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)