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\(a,\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)\(b,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3x^2\)\(c,\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2=\left(x-2y\right)^2\)
a) \(\left(x+y\right)^2+\left(x-y\right)^2\)
=\(\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
=\(x^2+2xy+y^2+x^2-2xy+y^2\)
\(2x^2+2y^2=2\left(x^2+y^2\right)\)
b) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
=\(\left[\left(x-y\right)+\left(x+y\right)\right]^2\)
= \(\left(x-y+x+y\right)^2\)
\(=2x^2\)
c) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(z-y\right)+\left(z-y\right)^2\)
\(=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2\)
= \(\left(x-y+z-z+y\right)^2=x^2\)
Bài giải:
a) (a + b)2 – (a – b)2 = (a2 + 2ab + b2) – (a2 – 2ab + b2)
= a2 + 2ab + b2 – a2 + 2ab - b2 = 4ab
Hoặc (a + b)2 – (a – b)2 = [(a + b) + (a – b)][(a + b) – (a – b)]
= (a + b + a – b)(a + b – a + b)
= 2a . 2b = 4ab
b) (a + b)3 – (a – b)3 – 2b3
= (a3 + 3a2b + 3ab2 + b3) – (a3 – 3a2b + 3ab2 – b3) – 2b3
= a3 + 3a2b + 3ab2 + b3 – a3 + 3a2b - 3ab2 + b3 – 2b3
= 6a2b
Hoặc (a + b)3 – (a – b)3 – 2b3 = [(a + b)3 – (a – b)3] – 2b3
= [(a + b) – (a – b)][(a + b)2 + (a + b)(a – b) + (a – b)2] – 2b3
= (a + b – a + b)(a2 + 2ab + b2 + a2 – b2 + a2 – 2ab + b2) – 2b3
= 2b . (3a2 + b2) – 2b3 = 6a2b + 2b3 – 2b3 = 6a2b
c) (x + y + z)2 – 2(x + y + z)(x + y) + (x + y)2
= x2 + y2 + z2+ 2xy + 2yz + 2xz – 2(x2 + xy + yx + y2 + zx + zy) + x2 + 2xy + y2
= 2x2 + 2y2 + z2 + 4xy + 2yz + 2xz – 2x2 – 4xy – 2y2 – 2xz – 2yz = z2
Bài làm:
Ta có: \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)(hằng đẳng thức đầu)
\(=\left(x-y+z+y-z\right)^2=x^2\)
\(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left[\left(x-y+z\right)+\left(y-z\right)\right]^2=\left(x-y+z+y-z\right)^2=x^2\)
b) Ta có nhận xét này nếu a+b+c=0 thì\(a^3+b^3+c^3=3abc\) (nếu cần chứng minh thì hỏi sau nhé)
Khi đó: tử=(x-y)(y-z)(z-x)
Mẫu nó cứ thế nào ấy. Rút gọn cũng chỉ được một chút thôi, chẳng gọn lắm
a) chịu chưa nghĩ ra
\(\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(\left[\left(x+y-z\right)-\left(x+y\right)\right]^2=z^2\)
\(\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-z-x+y\right)^2\)
\(=-z^2\)