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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
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
\(\left(a^2+b^2+c^2\right)+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
a,(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)+(a^2+2ac+c^2)
=(a+b)^2+(b+c)^2+(a+c)^2
b,(2a-b)(c-b)+2(b-a)(c-a)+2(b-c)(a-c)
=2(a-b)(c-b-c+a)+2(b-c)(c-a)
=2(a-b)(a-b)+2(b-c)(c-a)
=2(a-b)^2+2(b-c)(c-a)
=2(a^2-2ab+b^2)+(ab-bc-ca+c^2)
=2(a^2+b^2+c^2-ab-bc-ca)
=(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)
=(a-b)^2+(b-c)^2+(c-a)^2
chúc bạn học tốt!!!
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= \(\left(a-c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2+\left(b-c\right)^2-\left(b-a\right)^2-\left(c-a\right)^2+\left(b-a\right)^2-\left(b-c\right)^2-\left(a-c\right)^2\)
\(=\left(a-c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2+\left(b-c\right)^2-\left(a-b\right)^2-\left(a-c\right)^2+\left(a-b\right)^2-\left(b-c\right)^2-\left(a-c\right)^2\)
\(=-\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\)
=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
a) (a + b + c)2 + a2 + b2 + c2
= a2 + b2 + c2 + 2ab + 2bc + 2ac +a2 + b2 + c2
= (a2 + 2ab + b2 ) + (b2 + 2bc + c2 ) + (a2 + 2ac + c2)
= (a + b)2 + (b + c)2 + (c+a)2
a,\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\)
Chúc bạn học tốt!!!
\(\)a) \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(=2a^2+2b^2+2c^2+2ab+2ac+2bc\)
b) \(2\left(a-b\right)\left(c-b\right)+2\left(b-a\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(a-b\right)\cdot\left(c-b\right)+2\left(b-a\right)\cdot\left(c-a\right)+2\left(b-c\right)\cdot\left(-\left(c-a\right)\right)\)
\(=2\left(a+b\right)\cdot\left(c-b\right)+2\left(b-a\right)\cdot\left(c-a\right)-2\left(b-c\right)\cdot\left(c-a\right)\)
\(=2\left(\left(a-b\right)\cdot\left(c-b\right)+\left(b-a\right)\cdot\left(c-a\right)-\left(b-c\right)\cdot\left(c-a\right)\right)\)
\(=2\left(ac-ab-bc+b^2+bc-ab-ac+a^2-\left(bc-ab-c^2+ac\right)\right)\)
\(=2\left(-ab+b^2-ab+a^2-bc+ab+c^2-ac\right)\)
\(=2\left(b^2-ab+a^2-bc+c^2-ac\right)\)
chết rồi mik tưởng chỉ cần có 3 cái mũ 2 là đc :))
a)(a+b+c)2+a2+b2+c2=a2+b2+c2+2ab+2ac+2bc+a2+b2+c2
=(a2+2ab+b2)+(a2+2ac+c2)+(b2+2bc+c2)=(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)
= 2(ac - ab - bc + b2) + 2(bc - ab - ac + a2) + 2(ab - bc - ac + c2)
= 2ac - 2ab - 2bc + 2b2 + 2bc - 2ab - 2ac + 2a2 + 2ab - 2bc - 2ac + 2c2
= 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc
= a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2ac - 2bc
= (a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 - 2bc + c2)
= (a - b)2 + (a - c)2 + (b - c)2
b) \(2\left(a-b\right)\left(c-b\right)+2\left(b-a\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)
= \(2\left(ac-ab-bc+b^2+bc-ab-ac+a^2+ab-bc-ac+c^2\right)\)
= \(2a^2+2b^2+2c^2-2ab-2bc-2ac\)
= \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)
= \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)
bn lm j v?
làm kỉu j vậy má :v