a) A = x2( x + y ) - y( x2 + y2 )
= x3 + x2y - x2y - y3
= x3 - y3
Với x = 1 ; y = -1
A = 13 - (-1)3 = 1 + 1 = 2
b) B = 5x( x - 4y ) - 4y( y - 5x )
= 5x2 - 20xy - 4y2 + 20xy
= 5x2 - 4y2
Với x = -0, 6 ; y = -0, 75
B = 5.(-0, 6)2 - 4.(-0, 75)2 = 5.9/25 - 4.9/16 = 9/5 - 9/4 = -9/20
C = x( x - y + 1 ) - y( y + 1 - x )
= x2 - xy + x - y2 - y + xy
= x2 + x - y2 - y
= ( x2 - y2 ) + ( x - y )
= ( x - y )( x + y ) + ( x - y )
= ( x - y )( x + y + 1 )
Thế x = -2/3 ; y = -1/3 ta được
C = [ -2/3 - (-1/3 ) ][ -2/3 - 1/3 + 1 ]
= ( -2/3 + 1/3 ).0
= 0
a, \(A=x^2\left(x+y\right)-y\left(x^2+y^2\right)+2002=x^3-y^3+2002\)
Thay x = 1; y = -1 ta có : \(1^3-\left(-1\right)^3+2002=1-1+2002=2002\)
b, \(5x\left(x-4y\right)-4y\left(y-5x\right)-\frac{11}{20}=5x^2-4y^2-\frac{11}{20}\)
Thay x = -0,6 ; y = -0,75 ta có : \(5.\left(-0,6\right)^2-4\left(-0,75\right)^2-\frac{11}{20}=-1\)
c, \(x\left(x-y+1\right)-y\left(y+1-x\right)=x^2+x-y^2-y\)
Thay x = -2/3 ; y = -1/3 ta có : \(\left(-\frac{2}{3}\right)^2-\frac{2}{3}-\left(-\frac{1}{3}\right)^2+\frac{1}{3}=0\)
a) \(A=x^2\left(x+y\right)-y\left(x^2+y^2+2002\right)\)với \(x=1;y=-1\)
\(A=x^3-y^3-2002y\)
Thay: x=1;y=-1, ta có:
\(1^3-\left(-1\right)^3-2002\left(-1\right)\Rightarrow1+1+2002=2004\)
b) \(B=4x\left(x-4y\right)-4y\left(y-5x\right)-\frac{11}{20}\)với \(x=-0,6;y=-0,75\)
\(B=4x^2-4y^2+4xy-\frac{11}{20}\)
\(B=4\left(x^2-y^2+xy\right)-\frac{11}{20}\)
Thay: x= -0,6; y=-0.75, ta có:
\(4\left(-0.6^2--0.75^2+\left(-0,6.-0,75\right)\right)\Rightarrow4\left(0.35.8875\right)=0,143.55\)
c) \(C=x\left(x-y+1\right)-y\left(y+1-x\right)\)với x=-2/3; y=-1/3
\(C=x^2+x-y^2-y\)
Thay: x=-2/3; y=-1/3, ta có:
\(C=\left(-\frac{2}{3}\right)^2-\frac{2}{3}-\left(-\frac{1}{3}\right)^2+\frac{1}{3}=\frac{4}{9}-\frac{6}{9}-\frac{1}{9}=-\frac{3}{9}=-\frac{1}{3}\)
