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Mạn phép bỏ câu a :))
b) a2(b2 - a2) + b2(b2 + a2)
= a2.b2 + a2.(-a2) + b2.b2 + b2.a2
= a2.b2 - a4 + b4 + a2.b2
= a4 + 2a2b2 + b2 (hđt)
c) x2(x3 + 2y - x2y) - y(x2 - x4 + y)
= x2.x3 + x2.2y + x2.(-x2y) + (-y).x2 + (-y).(-x)4 + (-y).y
= x5 + 2x2y - x4y - x2y + x4y - y2
= x5 + (2xy2 - xy2) + (-x4y + x4y) - y2
= x5 + xy2 - y2
a, Xét tử thức \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left[\left(x-z\right)-\left(y-z\right)\right]\)
\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-z\right)-z^2\left(y-z\right)\)
\(=\left(x^2-z^2\right)\left(y-z\right)-\left(y^2-z^2\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(x+z\right)\left(y-z\right)-\left(y-z\right)\left(y+z\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(xy-xz+yz-z^2-y^2-yz+yz+z^2\right)\)
\(=\left(x-z\right)\left(xy-xz+yz-y^2\right)=\left(x-z\right)\left[x\left(y-z\right)-y\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)
Mẫu thức \(x^2y-x^2z+y^2z-y^3=x^2\left(y-z\right)-y^2\left(y-z\right)=\left(x-y\right)\left(x+y\right)\left(y-z\right)\)
Vậy \(\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}=\frac{x-z}{x+y}\)
b, \(\frac{x^5+x+1}{x^3+x^2+x}=\frac{x^5-x^2+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1}{x\left(x^2+x+1\right)}=\frac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}=\frac{x^3-x^2+1}{x}\)
Bài làm
a) 2(x + y)3 + 2(x - y)3
= 2[(x + y)3 + (x - y)3]
= 2[x3 + 3x2y + 3xy2 + y3 + x3 - 3x2y + 3xy2 - y3]
= 2[(x3 + x3) + (3x2y - 3x2y) + (3xy2 + 3xy2) + (y3 - y3)]
= 2[2x3 + 6xy2]
= 4x3 + 12xy2
b)uhm... Mình sửa đề chút, thay vì là -3(x + y)2(x - y) thì mình sẽ thành +3(x + y)2(x - y)
(x - y)3 - (x + y)3 + 3(x + y)2(x - y) - 3(x + y)(x - y)2
= -[(x + y)3 - 3(x + y)2(x - y) + 3(x + y)(x - y)2 - (x - y)3]
= -[(x + y) - (x - y)]3
= -[x + y - x + y ]3
= -[y]3
= -y
a)(x+y)3-3xy(x+y)
\(=\left(x+y\right)\left(x^2+xy+y^2\right)-3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+xy+y^2-3xy\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2-4ab\)
\(=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)+\left(a-b\right)\right]-4ab\)
\(=\left(a+b-a+b\right)\left(a+b+a-b\right)-4ab\)
\(=2b.2a-4ab\)
\(=4ab-4ab=0\)
giải
Áp dụng hàng đẳng thức đáng nhớ :
a ) \(5.\left(x+2\right)\left(x-2\right)-\left(3-4x\right)^2\)
\(=5\left(x^2-2^2\right)-\left(9-24x+16x^2\right)\)
\(=5x^2-20-9+24x-16x^2\)
\(=-11x^2+24x-29\)
b ) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
\(=2x^2-2y^2+2x ^2+2y^2=4x^2\)
Chúc bạn học tốt !!!
Bài làm
a) 5 . ( x + 2 )( x - 2 ) - ( 3 - 4x )2
= 5 . ( x2 - 22 ) - [ 32 - 2.3.4x + ( 4x )2 ]
= 5x2 - 20 - ( 9 - 24x + 16x2 )
= 5x2 - 20 - 9 + 24x - 16x2
= ( 5x2 - 16x2 ) - ( 20 + 9 ) + 24x
= -11x2 - 29 + 24x
b) 2( x - y )( x + y ) + ( x + y )2 + ( x - y )2
= 2( x2 - y2 ) + x2 + 2xy + y2 + x2 - 2xy + y2
= 2x2 - 2y2 + x2 + 2xy + y2 + x2 - 2xy + y2
= ( 2x2 + x2 + x2 ) + ( -2y2 + y2 + y2 ) + ( 2xy - 2xy )
= 4x2
# Học tốt #
\(\frac{x^2-3x+2}{x^3-1}=\frac{x^2-2x-x+2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\frac{x.\left(x-2\right)-\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{\left(x-1\right).\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(=\frac{x-2}{x^2+x+1}\)
a)Ta có: x(x-y) + y(x+y)
= x2-xy+xy+y2
=x2+y2
Thay x=-6 và y=8 vào biểu thức ta được:
(-6)2+82=36+64=100
Vậy tại x=-6 và y=8 thì giá trị biểu thức là 100
