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a) (a + b + c)2 + a2 + b2 + c2
= (a2 + b2 + c2 + 2ab + 2bc + 2ac) + a2 + b2 + c2
= (a2 + b2 + 2ab) + (a2 + c2 + 2ac) + (b2 + c2 + 2bc)
= (a + b)2 + (a + c)2 + (b + c)2
b) 2(a - b)(c - b) + 2(b - a)(c - a) + 2(b - c)(a - c)
= 2ac - 2ab - 2bc + 2b2 + 2bc - 2ab - 2ac + 2a2 + 2ab - 2bc - 2ac + 2c2
= 2b2 - 2ab + 2a2 - 2bc - 2ac + 2c2
= (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ac + a2)
= (a - b)2 + (b - c)2 + (c - a)2
a) (a+b+c)2 +a2 +b2 +c2 = a2 +b2 +c2 +2ab+2bc +2ca + a2 +b2 +c2 = 2a2 +2b2 +2c2 +2ab+2bc+2ac
=(a2 +2ab+b2 ) +(c2 +2bc+b2) +(c2 +2ca +a2 ) =(a+b)2 +(b+c)2 +(c+a)2
a) \(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(\Leftrightarrow a^2+2ab+b^2+b^2+2bc+c^2+a^2+2ac+c^2\)
\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
nha
1 : \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(c^2+2ac+a^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
2 : \(2\left(a-b\right)\left(c-b\right)+2\left(b-a\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(ac-bc-ab+b^2\right)+2\left(bc-ac-ab+a^2\right)+2\left(ab-ac-bc+c^2\right)\)
\(=2ac-2bc-2ab+2b^2+2bc-2ac-2ab+2a^2+2ab-2ac-2bc+2c^2\)
\(=2a^2+2b^2+2c^2-2ac-2ab-2bc\)
\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(\)\(1.\) \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(\Leftrightarrow\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)
\(\Leftrightarrow\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)
\(2.\) \(2\left(a-b\right)\left(c-b\right)+2\left(b-a\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)
\(\Leftrightarrow2ac-2ab-2bc+2b^2+2bc-2ab-2ac+2a^2+2ab-2bc-2ac+2c^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\)
Tới đây dùng HĐT
a) \(9x^2-6x+1\)
\(=\left(3x\right)^2-2\cdot3\cdot x+1^2\)
\(=\left(3x-1\right)^2\)
b) \(\left(2x+3y\right)^2+2\left(2x+3y\right)+1\)
\(=\left(2x+3y+1\right)^2\)
Lời giải:
$(a-b+5)(a+b-5)=[a-(b-5)][a+(b-5)]=a^2-(b-5)^2$
\(\left(a-b+5\right)\left(a+b-5\right)=a^2-\left(b-5\right)^2\)
a: \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=2a^2+2b^2+2c^2+2ab+2bc+2ac\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\)
b: \(=2\left(a-b\right)\left(c-b\right)-2\left(a-b\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)
\(=\left(2a-2b\right)\left(c-b-c+a\right)+2\left(b-c\right)\left(a-c\right)\)
\(=\left(2a-2b\right)\left(a-b\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(a^2-2ab+b^2+ab-bc-ac+c^2\right)\)
\(=2\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=2a^2+2b^2+2c^2-2ab-2bc-2ac\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)