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(a + b + c)(ab + bc + ac) - abc
= a2b + abc +a2c + ab2 + b2c + abc + abc + bc2 + ac2
= (a2b + 2abc + bc2) + (ac2 + a2c) + (ab2 + b2c)
= b(a2 + 2ac + c2) + ac(c + a) + b2(a + c)
= b(a + c) + ac(a + c) + b2(a + c)
= (a + c)[b(a + c) + ac + b2]
= (a + c)(ab + bc + ac + b2)
= (a + c)[b(a + b) + c(a + b)]
= (a + c)(b + c)(a + b)
\(\left(a+b+c\right)\left(ab+bc+ac\right)-abc\)
\(=a^2b+abc+a^2c+b^2a+b^2c+abc+abc+c^2b+c^2a-abc\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a^2+b^2+2ab-2ab\right)+2abc\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2-2abc+2abc\)
\(=\left(a+b\right)\left(ab+c^2+ca+cb\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Ta có:
\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)
\(A=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)
\(A=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)
\(A=\left(2c+1\right)\left(4ab+2a+2b+1\right)\)
\(A=\left(2c+1\right)\left[2a\left(2b+1\right)+\left(2b+1\right)\right]\)
\(A=\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)\)
Ta có:\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)
\(=8abc+4ab+4bc+4ca+2a+2b+2c+1\)
\(=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)
\(=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)
\(=\left(2c+1\right)\left(4ab+2b+2a+1\right)\)
\(=\left(2c+1\right)\left[2b\left(2a+1\right)+\left(2a+1\right)\right]\)
\(=\left(2c+1\right)\left(2b+1\right)\left(2a+1\right)\)
= (abc - ab) + (a - ca) + (b - bc) + (c -1) = ab.(c -1) - a.(c - 1) - b(c -1) + (c -1) = (c -1).(ab - a - b + 1)
abc-(ab+bc+ca)+(a+b+c)-1
=abc-ab-bc-ca+a+b+c-1
=(abc-ab)+(-bc+b)+(-ca+a)+(c-1)
=ab.(c-1)-b.(c-1)-a.(c-1)+(c-1)
=(c-1)(ab-b-a+1)
=(c-1)[b.(a-1)-(a-1)]
=(c-1)(a-1)(b-1)
Mình tính thử a ,b ,c bằng nhau đó
Mình nghĩ là 0,037037037037037037
abc - (ab + ac + bc) + (a + b + c) - 1
= abc - bc - ab + b - ac + c + a - 1
= bc(a - 1) - b(a - 1) - c(a - 1) + (a - 1)
= (a - 1)(bc - b - c + 1)
= (a - 1)(b - 1)(c - 1)