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Bài 2:
\(A=x^2+4y^2-2x+10-4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay x + 2y = 5 vào biểu thức A ta được: \(A=5^2-2.5+10=25\)
\(B=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)\left(y-1\right)+y^2-2y+1\)
\(=x^2+4xy+4y^2-2xy+2x-4y^2+4y+y^2-2y+1\)
\(=x^2+2xy+y^2+2x+2y+1\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1\)
Thay x + y = 5 vào biểu thức B ta được: \(B=5^2+2.5+1=25+10+1=36\)
\(C=x^2-y^2-4x=\left(x^2-4x+4\right)-y^2-4\)
\(=\left(x-2\right)^2-y^2-4\) \(=\left(x-y-2\right)\left(x-2+y\right)-4\)
Thay x + y = 2 vào C ta được: \(C=\left(x-2-y\right)\left(2-2\right)-4=0-4=-4\)
\(D=x^2+y^2+2xy-4x-4y-3\)
\(=\left(x+y\right)^2-4\left(x+y\right)-3\) Thay x + y = 4 vào D ta được:
\(D=4^2-4.4-3=16-16-3=-3\)
Bài 3:
a) \(N=-9x^2+12x-5=-\left(9x^2-12x+4\right)-1\)
\(=-\left(3x-2\right)^2-1\)
Do \(\left(3x-2\right)^2\ge0\) nên \(-\left(3x-2\right)^2-1< 0\)
Vậy N < 0
b) ghi đề cẩn thận lại đi, mk k hiểu
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a) ( x + 2 )( x2 - 2x + 4 ) - ( 18 + x3 )
= x3 + 8 - 18 - x3 = -10
b) ( 2x - y )( 4x2 + 2xy + y2 ) - ( 2x + y )( 4x2 - 2xy + y2 )
= 8x3 - y3 - ( 8x3 + y3 )
= 8x3 - y3 - 8x3 - y3 = -2y3
c) ( x - 3 )( x + 3 ) - ( x + 5 )( x - 1 )
= x2 - 9 - ( x2 + 4x - 5 )
= x2 - 9 - x2 - 4x + 5 = -4x - 4
d) ( 3x - 2 )2 + ( x + 1 )2 + 2( 3x - 2 )( x + 1 )
= ( 3x - 2 + x + 1 )2
= ( 4x - 1 )2
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1) x2 + x - y2 + y = (x2 - y2) + (x + y) = (x - y)(x + y) + (x + y) = (x - y + 1)(x + y)
2) 4x2 - 9y2 + 4x - 6y = (4x2 - 9y2) + (4x - 6y) = (2x - 3y)(2x + 3y) + 2(2x - 3y) = (2x - 3y)(2x + 3y + 2)
3) x2 + x + y2 + y + 2xy = (x2 + 2xy + y2) + (x + y) = (x + y)2 + (x + y) = (x + y)(x + y + 1)
4) -x2 + 5x + 2xy - 5y - y2 = -(x2 - 2xy + y2) + (5x - 5y) = -(x - y)2 + 5(x - y) = (x - y)(y - x + 5)
5) x2 - y2 + 2x + 1 = (x2 + 2x + 1) - y2 = (x + 1)2 - y2 = (x + 1 + y)(x - y + 1)
6) x2 - 1 - y2 + 2y = x2 - (y2 - 2y + 1) = x2 - (y - 1)2 = (x - y + 1)(x + y - 1)
7) x2 + 2xz - y2 + 2uy + z2 - u2 =(x2 + 2xz + z2) - (y2 - 2uy + u2) = (x + z)2 - (y - u)2 = (x + z - y + u)(x + z + y - u)
8) x3 + 3x2y + x + 3xy2 + y + y3 = (x3 + 3x2y + 3xy2 + y3) + (x + y) = (x + y)3 + (x + y) = (x + y)(x2 + 2xy + y2 + 1)
9) x3 + y(1 - 3x2) + x(3y2 - 1) - y3 = x3 + y - 3x2y + 3xy2 - x - y3 = (x3 - 3x2y + 3xy2 - y3) - (x - y) = (x - y)3 - (x - y) = (x - y)(x2 - 2xy+y2-1)
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a: \(x^2+x-2x-2\)
\(=x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2\right)=\left(-1+1\right)\left(-1-2\right)=0\)
b: \(3x^2-2x+9x-6\)
\(=x\left(3x-2\right)+3\left(3x-2\right)\)
\(=\left(3x-2\right)\left(x+3\right)=\left(3\cdot7-2\right)\left(7+3\right)\)
\(=19\cdot10=190\)
c: \(2x^2-3xy-xy^2\)
\(=x\left(2x-3y-y^2\right)\)
\(=2\left(2\cdot2-3\cdot3-9\right)\)
\(=2\cdot\left(4-18\right)=-28\)
\(A=\left(x-y\right)^3=\left(\dfrac{1}{5}-\dfrac{1}{3}\right)^3=\left(-\dfrac{2}{15}\right)^3=-\dfrac{8}{1375}\)