Cho A= 1/2 + 3/2 +(3/2)^2+...+(3/2)^2012
B=(3/2)^2013 : 2
tinh B-A
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\(\frac{3}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow\frac{3}{2}.A-A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}-\left(\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\right)\)
\(\Rightarrow\frac{1}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}-\frac{3}{2}=\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}\Rightarrow A=2.\left(\frac{3}{2}\right)^{2013}-\frac{5}{2}\)
\(B-A=\frac{1}{2}.\left(\frac{3}{2}\right)^{2013}-2.\left(\frac{3}{2}\right)^{2013}+\frac{5}{2}=-\left(\frac{3}{2}\right)^{2014}+\frac{5}{2}\)
gt\(\Rightarrow1\ge a^2\Rightarrow-1\le a\le1\).Tương tự:\(-1\le b\le1;-1\le c\le1\)
\(\Rightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)\le0\)
\(\Leftrightarrow a^2-a^3+b^2-b^3+c^2-c^3\le0\)
\(\Leftrightarrow a^2+b^2+c^2\le a^3+b^3+c^3\)
Dấu "=" xảy ra khi \(\left(a,b,c\right)=\left(1,0,0\right)\) và các hoán vị
\(\Rightarrow S=1\)
\(\Rightarrow a,b,c\in\left\{-1;1\right\}\\ \Rightarrow a^3+b^3+c^3-\left(a^2+b^2+c^2\right)\\ =a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)\le0\\ \Rightarrow a^3+b^3+c^3\le1\\ \Rightarrow a,b,c.nhận.2.Giá.trị.là.0.hay.1\\ \Rightarrow b^{2012}=b^2;c^{2013}=c^2\\ \Rightarrow S=a^2+b^{2012}+c^{2013}=1\)
Lời giải:
$A-\frac{1}{2}=\frac{3}{2}+(\frac{3}{2})^2+....+(\frac{3}{2})^{2012}$
$\frac{3}{2}(A-\frac{1}{2})=(\frac{3}{2})^2+(\frac{3}{2})^3+....+(\frac{3}{2})^{2013}$
$\Rightarrow \frac{3}{2}(A-\frac{1}{2}) - (A-\frac{1}{2})=(\frac{3}{2})^{2013}-\frac{3}{2}$
$\Rightarrow \frac{1}{2}(A-\frac{1}{2})=(\frac{3}{2})^{2013}-\frac{3}{2}$
$\Rightarrow A=2(\frac{3}{2})^{2013}-\frac{5}{2}$
$\Rightarrow A-B=2(\frac{3}{2})^{2013}-\frac{5}{2}- \frac{1}{2}.(\frac{3}{2})^{2013}$
$\Rightarrow A-B=\frac{3}{2}(\frac{3}{2})^{2013}-\frac{5}{2}=(\frac{3}{2})^{2014}-\frac{5}{2}$
so sánh A và B hả bạn