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a) \(2\left(c\right)^2+2ac+b^2-2b+a^2\)
b)\(\left(-\left(b+a\right)\right)\left(c+a\right)\left(c-b\right)\)
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\Rightarrowđcpm\)
a²+b²+c²+3=2(a+b+c)
=>a²-2a+1+b²-2b+1+c²-2c+1=1
=>(a-1) ² +(b-1) ² +(c-1) ²=1
=>a=b=c=1 dpcm
A=( a2 + b2 + c2 )2 - ( a2 - b2 + c2 )2=(a^2+b^2+c^2-a^2+b^2-c^2)(a^2+b^2+c^2+a^2-b^2+c^2)=2(b^2+c^2).2.(a^2+c^2)=4(b^2+c^2)(a^2+c^2)
B=( a + b + c )2 + ( a + b - c)2 - 2 ( a + b )2=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2-2ab-2bc+2ac-2a^2-2ab-2b^2=b^2+2bc+c^2=(b+c)^2
câu C làm tương tự
a)\(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a-c\right)\)
b)\((a+b)(a^2-b^2)+(b+c)(b^2-c^2)+(c+a)(c^2-a^2)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
c)\(a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+bc+ca\right)\)
d)\(a^4(b-c)+b^4(c-a)+c^4(a-b)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2+ab+bc+ca\right)\)
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
a: \(=a^2+2a\left(b-c\right)+\left(b-c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2-2\left(b-c\right)^2\)
\(=2a^2+2\left(b-c\right)^2-2\left(b-c\right)^2=2a^2\)
b: \(=a^2+2a\left(b+c\right)+\left(b+c\right)^2+a^2-2a\left(b+c\right)+\left(b+c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
\(=2a^2+2\left(b+c\right)^2+\left(a-b+c\right)^2+\left(a+b-c\right)^2\)
\(=2a^2+2\left(b+c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2+a^2+2a\left(b-c\right)+\left(b-c\right)^2\)
\(=2a^2+2\left(b+c\right)^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2\left(b^2+2bc+c^2+b^2-2bc+c^2\right)\)
\(=4a^2+4b^2+4c^2\)