Viết các biểu thức sau dưới dạng tổng:
(a+b-c)2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
Ta có: (a+b+c)^2 + a^2 + b^2 + c^2
= a^2 +b^2 +c^2 + 2ab + 2ac + 2bc + a^2 + b^2 + c^2
= (a^2 +2ab+ b^2) + (b^2 +2bc+ c^2) +(c^2 +2ac+ a^2 )
= (a+b)^2 +(b+c)^2 +(c+a)^2
\(\left(a^2+b^2+c^2\right)+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
2(a-b)(c-b)+2(b-a)(c-a)+2(b-c)(a-c)
=2a^2+2b^2+2c^2-2bc-2ab-2ac
=a^2-2ac+c^2+a^2-2ab+b^2+b^2-2bc+c^2
=(a-c)^2+(a-b)^2+(b-c)^2
\(b,=\left(x^4y^8\right)^2+2\cdot2x^4y^8+2^2=\left(x^4y^8+2\right)^2\)
a. (x + y)2 = x2 + 2xy + y2
b. (x - 2y)2 = x2 - 4xy - 4x2
c. (xy2 + 1)(xy2 - 1) = x2y4 - 1
d. (x + y)2(x - y)2 = (x2 + 2xy + y2)(x2 - 2xy + y2) = x4 - (2xy + y2)2 = x4 - (4x2y2 + y4) = x4 - 4x2y2 - y4
Chucs hocj toots
Câu 2:
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(x^2+10x+25=\left(x+5\right)^2\)
d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)
e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)
a)x2-6x+9
=x2-2.x.3+32
=(x-3)2
b)4x2+4x+1
=(2x)2+2.2x.1+12
=(2x+1)2
c)4x2+12xy+9y2
=(2x)2+2.2x.3y+(3y)2
=(2x+3y)2
d)4x4-4x2+4
=(2x2)2-2.2x2.2+22
=(2x2-2)2
\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2ac-2bc\\ =a^2+b^2+c^2+2ab+\left(-2ac\right)+\left(-2bc\right)\)