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=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)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
\(A=\frac{100.101}{2}=5050\)
b)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^8-1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(B=2^{128}-1+1=2^{128}\)
c)
\(C=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\)
\(C=2c^2\)
2) b)
Do \(a+b+c=9\Rightarrow\left(a+b+c\right)^2=81\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=81\)
\(\Rightarrow2\left(ab+bc+ac\right)=81-141=-60\)
\(ab+bc+ac=-60:2=-30\)
a, B=x^3 + 3xy +y^3 = x^3 +3xy(x+y)+y^3 (vì x+y=1)
= (x+y)^3
= 1^3 =1
b, (a+b+c)^2 =a^2 +b^2 +c^2 +2ab +2bc +2ac
9^2 = 141 +2(ab+bc+ac)
-60 = 2(ab+bc+ac)
ab+ac+bc=-30
Vậy M=-30
c, N =(x+y)^3 -3(x+y)(x^2+y^2) +2(x^3+y^3)
= x^3 + 3x^2 .y + 3xy^2 + -3(x^3+xy^2 +x^2 .y+y^3)+ 2x^3 +2y^3
= x^3 +3x^2 .y + 3xy^2 - 3x^3 -3xy^2 -3x^2 .y -3y^3 +2x^3 +2y^3
= 0
Vậy N=0 .Chúc bạn học tốt.
\(2\left(ab+bc+ca\right)=\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=2^2-2\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=2\Leftrightarrow ab+bc+ca=1\)
\(M=\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[c\left(b+c\right)+a\left(b+c\right)\right]\)
\(=\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2=\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2\)
Ta có : theo điều kiện cho trước:
a + b + c =2
<=> \(\left(a+b+c\right)^2=4\)
<=> \(a^2+b^2+c^2+2ab+2ac+2bc=4\)
<=> \(2+2\left(ab+ac+bc\right)=4\)
<=> \(2\left(ab+ac+bc\right)=2\)
<=> \(ab+ac+bc=1\)
<=> \(\left(ab+ac+bc\right)^2=1\)
<=> \(a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)=1\)
<=> \(a^2b^2+b^2c^2+a^2c^2=1-2\left(ab^2c+a^2bc+abc^2\right)\)
Theo đề bài ta có :
M = \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)
<=> \(\left(a^2b^2+a^2+b^2+1\right)\left(c^2+1\right)\)
<=> \(a^2b^2c^2+a^2b^2+a^2c^2+a^2+b^2c^2+b^2+c^2+1\)
<=> \(a^2b^2c^2+1-2ab^2c-2a^2bc-2abc^2+3\)
<=> \(a^2b^2c^2-2ab^2c-2a^2bc-2abc^2+4\)
<=> \(abc\left(abc-2b-2a-2c\right)+4\)
<=> \(abc\left\{abc-2\left(a+b+c\right)\right\}+4\)
<=> \(abc\left(abc-4\right)+4\)
<=> \(a^2b^2c^2-4abc+4\)
<=> \(\left(abc\right)^2-4abc+4\)
<=> \(\left(abc-2\right)^2\left(đpcm\right)\)
a) \(x^2-6x+9=x^2-2\cdot x\cdot3+3^2=\left(x-3\right)^2\)
b) \(4x^2-12xy+9y^2=\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2=\left(2x-3y\right)^2\)
c) \(4x^2-2x+1=\left(2x-1\right)^2\)
d) \(x^2+8xy+16y^2=\left(x+4y\right)^2\)
1) a) a^2+b^2=ab+ba
<=> a^2+b^2-2ab=0
<=> (a-b)^2=0
<=> a-b=0 <=> a=b (đpcm)
b) a^2+b^2+c^2=ab+bc+ca
<=> 2a^2+2b^2+2c^2=2ab+2bc+2ca
<=> (a^2-2ab+b^2)+(a^2-2ca+c^2)+(b^2-2bc+c^2)=0
<=> (a-b)^2+(a-c)^2+(b-c)^2=0
<=> a-b=0 và a-c=0 và b-c=0
<=> a=b và a=c và b=c
<=> a=b=c (đpcm)
a) \(\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)
b) \(\left(z^2-6z+9\right)+\left(t^2+4t+4\right)=\left(z-3\right)^2+\left(t+2\right)^2\)
c) \(\left(4x^2-4xz+z^2\right)+\left(z^2-2z+1\right)=\left(4x-z\right)^2+\left(z-1\right)^2\)
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= \(\left(a-c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2+\left(b-c\right)^2-\left(b-a\right)^2-\left(c-a\right)^2+\left(b-a\right)^2-\left(b-c\right)^2-\left(a-c\right)^2\)
\(=\left(a-c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2+\left(b-c\right)^2-\left(a-b\right)^2-\left(a-c\right)^2+\left(a-b\right)^2-\left(b-c\right)^2-\left(a-c\right)^2\)
\(=-\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\)