Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(C=B-\frac{1}{2}=\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2012}\)
\(\Rightarrow\frac{3}{2}\cdot C=\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+...+\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow\frac{3}{2}C-C=\frac{1}{2}C=\frac{3}{2}+\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow C=3+\left(\frac{3}{2}\right)^{2013}\cdot2\)
\(\Rightarrow B=\frac{1}{2}+3+\left(\frac{3}{2}\right)^{2013}\cdot2\)
do đó \(A-B=\left(\frac{3}{2}\right)^{2014}+\frac{7}{2}\)
Ta có: \(a^2+b+\frac{3}{4}=a^2+\frac{1}{4}+b+\frac{1}{2}\ge a+b+\frac{1}{2}\)
Và \(b^2+a+\frac{3}{4}\ge a+b+\frac{1}{2}\)
\(\Rightarrow(a^2+b+\frac{3}{4})(b^2+a+\frac{3}{4})\ge(a+b+\frac{1}{2})^2\)
Cần chứng minh \((a+b+\frac{1}{2})^2\ge\left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right)\)
\(\Leftrightarrow a^2+b^2+\frac{1}{4}+a+b+2ab\ge4ab+a+b+\frac{1}{4}\Leftrightarrow(a-b)^2\ge0\)
BDT cuối đúng hay \(VT\ge VP\)
Nên xảy ra khi \(a=b=\frac{1}{2}\)
2)
a) \(\frac{1}{3}x-\left(-\frac{2}{3}\right)^2=\sqrt{\frac{16}{81}}\)
\(\Rightarrow\frac{1}{3}x-\frac{4}{9}=\frac{4}{9}\)
\(\Rightarrow\frac{1}{3}x=\frac{4}{9}+\frac{4}{9}\)
\(\Rightarrow\frac{1}{3}x=\frac{8}{9}\)
\(\Rightarrow x=\frac{8}{9}:\frac{1}{3}\)
\(\Rightarrow x=\frac{8}{3}\)
Vậy \(x=\frac{8}{3}.\)
b) \(4x-\frac{2}{5}+\frac{3}{4}=\frac{11}{4}\)
\(\Rightarrow4x-\frac{2}{5}=\frac{11}{4}-\frac{3}{4}\)
\(\Rightarrow4x-\frac{2}{5}=2\)
\(\Rightarrow4x=2+\frac{2}{5}\)
\(\Rightarrow4x=\frac{12}{5}\)
\(\Rightarrow x=\frac{12}{5}:4\)
\(\Rightarrow x=\frac{3}{5}\)
Vậy \(x=\frac{3}{5}.\)
Chúc bạn học tốt!
