1.
a.\(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=199+195+...+3\)
\(=\dfrac{\left(\dfrac{199-3}{4}+1\right).\left(199+3\right)}{2}=5050\)
b.\(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1^2=2^{128}-1^2+1^2=2^{128}\)
c.\(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca-2a^2-4ab-2b^2=2c^2\)
2.
a. \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3=a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2\)
\(\Leftrightarrow a^3+b^3=a^3+b^3\left(đúng\right)\)
b. \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\left(đpcm\right)\)
Bài 2:
i. \(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow a+b+c=0\) hay \(\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
*\(\left(a^2+b^2+c^2-ab-bc-ca\right)=0\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left(a-b\right)^2=0;\left(b-c\right)^2=0;\left(c-a\right)^2=0\Rightarrow a=b=c\)
ii. \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{bc+ca+ab}{abc}=0\Rightarrow bc+ca+ab=0\)
\(A=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{b^3c^3+c^3a^3+a^3b^3}{a^2b^2c^2}=\dfrac{\left(bc+ca+ab\right)\left(b^2c^2+c^2a^2+a^2b^2-bc^2a-b^2ca-bca^2\right)}{a^2b^2c^2}=\dfrac{0.\left(b^2c^2+c^2a^2+a^2b^2-bc^2a-b^2ca-bca^2\right)}{a^2b^2c^2}=0\)
c. \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a+b+c=0\) hay \(a=b=c\)
*\(a+b+c=0\Rightarrow a+b=-c;b+c=-a;c+a=-b\)
\(B=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)*\(a=b=c\)
\(B=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)
Bài 3:
a. \(A=4x^2+4x+11=4x^2+4x+1+10=\left(2x+1\right)^2+10\ge10\)
-Dấu bằng xảy ra \(\Leftrightarrow x=\dfrac{-1}{2}\).
b. \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+6x-x-6\right)\left(x^2+3x+2x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x-6+12\right)=\left(x^2+5x-6\right)^2+12\left(x^2+5x-6\right)=\left(x^2+5x-6\right)^2+12\left(x^2+5x-6\right)+36-36=\left(x^2+5x-6+6\right)^2-36=\left(x^2+5x\right)^2-36\ge-36\)
-Dấu bằng xảy ra \(\Leftrightarrow x=0\) hoặc \(x=-5\)
