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(a - b - 2)2 - (2a - 2b)(a - b - 2) + a2 + b2 - 2ab
= (a - b - c)(a - b - c) - (2a - 2b)(a - b - 2) + a2 + b2 - 2ab
= -2ab + a2 - 4a + b2 + 4b + 4 + 4ab - 2a2 + 4a - 2b2 - 4b + a2 + b2 - 2ab
= 4
(x + y + 3)2 - (x + y + 3)(2x + y) + (x + y)2
= (x + y + 3)2 - (x + y + 3).2(x + y) + (x + y)2
= (x + y + 3)2 - 2(x + y + 3)(x + y) + (x + y)2
= [x + y + 3 - (x + y)]2
= (x + y + 3 - x - y)2
= [(x - x) + (y - y) + 3]2
= 32
= 9
\(a^6-1=\left(a^3-1\right)\left(a^3+1\right)=\left(a-1\right)\left(a+1\right)\left(a^2+a+1\right)\left(a^2-a+1\right)\)
\(a^6-1=\left(a^3-1\right).\left(a^3+1\right)=\left(a-1\right).\left(a^2+a+1\right).\left(a-1\right).\left(a^2-a+1\right)\)
\(=\left(a-1\right).\left(a+1\right).\left(a^4+a^2+1\right)=\left(a-1\right).\left(a+1\right).\left(a^4-13a^2+14a^2+1\right)\)
\(=\left(a-1\right).\left(a+1\right).\left(a^2-4\right).\left(a^2-9\right)+14a^2.\left(a-1\right).\left(a+1\right)\)
đến đây dễ rồi, b tự làm tiếp :))
Ta có:
\(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^5=-c^5\)
\(\Rightarrow a^5+5a^4b+10a^3b+10a^2b^3+5ab^4+b^5=-c^5\)
\(\Rightarrow a^5+b^5+c^5=5ab\left(a^3+b^3+2a^2b+2ab^2\right)\)
\(\Rightarrow a^5+b^5+c^5=5ab\left[\left(a^3+b^3\right)+2ab\left(a+b\right)\right]\)
\(\Rightarrow a^5+b^5+c^5=5ab\left[\left(a+b\right)\left(a^2-ab+b^2\right)+2ab\left(a+b\right)\right]\)
\(\Rightarrow a^5+b^5+c^5=5ab\left(a+b\right)\left(a^2+ab+b^2\right)\)
\(\Rightarrow a^5+b^5+c^5=-5abc\left(a^2+ab+b^2\right)\)
\(\Rightarrowđpcm\)
Vì :
a^2; b^2 là số chính phương
a,b không chia hết cho 3
Nên a^2; b^2 chia 3 dư 1
=> a^2 - b^2 chia hết cho 3 (1)
Ta có :
(a^2 - 1) - (b^2 - 1) = (a - 1)(a + 1) - (b - 1)(b + 1) chia hết cho 8 (2)
Vì :
(a - 1); (a + 1)(a - 1); (a + 1) là 2 số chẵn liên tiếp
(b - 1); (b + 1)(b - 1), (b + 1) là 2 số chẵn liên tiếp
Từ (1), (2)
=> a^2 - b^2 chia hết cho 3.8
=> a^2 - b^2 chia hết cho 24
\(a,\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)
\(=\left(2a+3\right)\left(x-y+1\right)\)
\(b,\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)
\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)
\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)
\(=\left(4x-y\right)\left(a+b-c-1\right)\)
\(c,x^k+1-x^k-1\)
\(=0?!?!\)
\(d,x^m+3-x^m+1\)
\(=4\)
\(e,3\left(x-y\right)^3-2\left(x-y\right)^2\)
\(=\left(x-y\right)^2\left(3\left(x-y\right)-2\right)\)
\(=\left(x-y\right)^2\left(3x-3y-2\right)\)
\(f,81a^2+18a+1\)
\(=\left(9a\right)^2+2.9a+1\)
\(=\left(9a+1\right)^2\)
\(g,25a^2.b^2-16c^2\)
\(=\left(5ab\right)^2-\left(4c\right)^2\)
\(=\left(5ab+4c\right)\left(5ab-4c\right)\)
\(h,\left(a-b\right)^2-2\left(a-b\right)c+c^2\)
\(=\left(a-b-c\right)^2\)
\(i,\left(ax+by\right)^2-\left(ax-by\right)^2\)
\(=\left(ax+by-ax+by\right)\left(ax+by+ax-by\right)\)
\(=2by.2ax\)
\(=4axby\)