Tìm x,y :
A. x^3 - y^3 = 91
B. 9x + 2 = y^2 + y
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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 )
Bài 1:
\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)