Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1
a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)
y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)
y= \(\frac{-3}{14}\)
b.5x + 5x+2=650
5x . 1 + 5x + 52=650
5x(1+25)=650
5x.26=650
5x=25
x=2
Đặt \(B=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
...
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)
Vậy A<\(\frac{3}{4}\)
A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)=\(\frac{2013}{2014}\)<\(\frac{3}{4}\)
Ta có : A = \(\frac{\left(2^2.3\right)^6+8^4.3^5}{2^{12}.3^5-4^6.9^2}\)
= \(\frac{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}\)
= \(\frac{2^{12}.3^6+2^{12}.3^5}{2^{12}.3^5-2^{12}.3^4}\)
= \(\frac{2^{12}.\left(3^6+3^5\right)}{2^{12}.\left(3^5-3^4\right)}\)
= \(\frac{3^6+3^5}{3^5-3^4}\)
= \(6\)
\(A=\frac{\left(2^2.3\right)^6+8^4.3^5}{2^{12}.3^5-4^6.9^2}\)
\(A=\frac{2^{12}.3^6+\left(2^3\right)^4.3^5}{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}\)
\(A=\frac{2^{12}.3^6+2^{12}.3^5}{2^{12}.3^5-2^{12}.3^4}\)
\(A=\frac{2^{12}.\left(3^6+3^5\right)}{2^{12}.\left(3^5-3^4\right)}\)
\(\Rightarrow A=\frac{3^5.\left(3+1\right)}{3^4.\left(3-1\right)}=\frac{3^5.4}{3^4.2}=\frac{3.3^4.2.2}{3^4.2}=\frac{3.2}{1}=6\)