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\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
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)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
=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+c^2)
=(a+b)^2+(b+c)^2+(c+b)^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
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)\(\left[\left(a-b\right)^2-2\left(a-b\right)\left(c-b\right)+\left(c-b\right)^2\right]-\left(a-b\right)^2-\left(b-c\right)^2=\left(a-b-c+b\right)^2-\left(a-b\right)^2-\left(b-c\right)^2\)
\(=\left(a-c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2\) tương tự thì
A= \(\left(a-c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2+\left(b-c\right)^2-\left(b-a\right)^2-\left(c-a\right)^2+\left(b-a\right)^2-\left(b-c\right)^2-\left(a-c\right)^2\)
\(=\left(a-c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2+\left(b-c\right)^2-\left(a-b\right)^2-\left(a-c\right)^2+\left(a-b\right)^2-\left(b-c\right)^2-\left(a-c\right)^2\)
\(=-\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\)
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
\(\left(a^2+b^2+c^2\right)+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)