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Ta có : x4 - y4
= (x2)2 - (y2)2
= (x2 - y2)(x2 + y2)
= (x - y)(x + y)(x2 + y2)
b) 9(x - y)2 - 4(x + y)2
= [3(x - y) - 4(x + y)][3(x - y) + 4(x + y)]
= [3x - 3y - 4x - 4y][3x - 3y + 4x + 4y]
= (-x - 7y)(x + y)
e.\(x^4+2x^2+1=\left(x^2+1\right)^2\)
c.\(x^2-9y^2=\left(x-3y\right)\left(x+3y\right)\)
f.\(-x^2-2xy-y^2+1=-\left[\left(x+y\right)^2-1\right]=-\left(x+y-1\right)\left(x+y+1\right)=\left(x-y+1\right)\left(x+y+1\right)\)
g.\(x^3-x^2-x+1==x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)=\left(x-1\right)^2\left(x+1\right)\)
h.\(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
i.\(\left(x+y\right)^3-x^3-y^3=\left(x+y\right)^3-\left(x^3+y^3\right)=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-\left(x^2-xy+y^2\right)\right]=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
tíck mình nha bn thanks !!!!!
1,Ta có: \(A=2\left[\left(x^2\right)^3+\left(y^2\right)^3\right]-3x^4-3y^4\)
\(=2\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4\)
Thay \(x^2+y^2=1,\) ta có:
\(A=2.1\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4\)
\(=2x^4-2x^2y^2+2y^4-3x^4-3y^4\)
\(=-\left(x^4+2x^2y^2+y^4\right)=-\left(x^2+y^2\right)^2=-1\)
2,Ta có: \(B=\left(x^2-y^2\right)\left(x^2+y^2\right)+\left(x^4+x^2y^2\right)+3y^2\)
\(=\left(x^2-y^2\right).1+x^2\left(x^2+y^2\right)+3y^2\)
\(=x^2-y^2+x^2+3y^2=2\left(x^2+y^2\right)=2\)
\(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2\)
\(=3xy\left(x+y\right)\)
\(a,\left(2x+3y\right)^2-4\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x+3y-4\right)\)
\(b,\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x^3+y^3\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-\left(x^2-xy+y^2\right)\right]\)
\(=\left(x+y\right).3x\)
\(c,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
bạn chia hai đa thức cho nhau sẽ còn dư là (a-7)xy^3+(b+6)y^4.
Suy ra được: a=7 và b= -6
Các bạn giúp mk với tối nay mk đi học rùi!!!!!!!!!!!!!!!!!!!!!!!!!
b \(2x^4-y^4+x^2y^2+3y^2=\left(x^4-y^4\right)+\left(x^4+x^2y^2\right)+3y^2=\left(x^2-y^2\right)\left(x^2+y^2\right)+x^2\left(x^2+y^2\right)+3y^2\)
\(=\left(x^2-y^2\right)\cdot1+x^2\cdot1+3y^2=x^2-y^2+x^2+3y^2=2x^2+2y^2=2\left(x^2+y^2\right)=2\cdot1=2\)
a \(2\left(x^6+y^6\right)-3\left(x^4+y^4\right)=2\left(\left(x^2\right)^3+\left(y^2\right)^3\right)-3x^4-3y^4=2\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)
\(-3x^4-3y^4=2\cdot1\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4=2x^4-2x^2y^2+2y^4-3x^4-3y^4\)
\(=-x^4-2x^2y^2-y^4=-\left(x^4+2x^2y^2+y^4\right)=-\left(x^2+y^2\right)^2=-1^2=-1\)
\(B=2x^4-y^4+x^2y^2+3y^2\\ =\left(x^4-y^4\right)+\left(x^4+x^2y^2\right)+3y^2\\ =\left(x^2-y^2\right)\left(x^2+y^2\right)+x^2\left(x^2+y^2\right)+3y^2\\ =x^2-y^2+x^2+3y^2\\ =2x^2+2y^2\\ =2\left(x^2+y^2\right)\\ =2\)