Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=\dfrac{2\cdot8^4\cdot27^2+44\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot11\cdot2^9\cdot3^9}{2^7\cdot3^7\cdot2^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9\cdot11}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\cdot11\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2\cdot301}{3\cdot31}=\dfrac{602}{93}\)
Bài 3:
a: a*S=a^2+a^3+...+a^2023
=>(a-1)*S=a^2023-a
=>\(S=\dfrac{a^{2023}-a}{a-1}\)
b: a*B=a^2-a^3+...-a^2023
=>(a+1)B=a-a^2023
=>\(B=\dfrac{a-a^{2023}}{a+1}\)
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2019}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}:\sqrt{\frac{9}{25}}\)\(=\frac{\frac{2^3}{3^3}.\frac{-3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{-5^3}{12^3}}:\frac{3}{5}\)
\(=\frac{\frac{2^3}{5^3}.\frac{-3^2}{2^4}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{-5^3}{2^6.3^3}}:\frac{3}{5}=\frac{\frac{-1}{3.2}}{\frac{-5}{2^4.3^3}}:\frac{3}{5}\)\(=\frac{-1}{3.2}.\frac{-2^4.3^3}{5}.\frac{5}{3}\)
\(=\frac{2^3.3^2}{5}.\frac{5}{3}=24\)
\(-\left|1,7-x\right|-\dfrac{5}{3}=\dfrac{2}{3}\\ \Rightarrow\left|1,7-x\right|=-\dfrac{5}{3}-\dfrac{2}{3}=-\dfrac{7}{3}\left(l\right)\)
Vậy không có giá trị x thoả mãn
a)
\(\left[\dfrac{3}{8}+\left(-\dfrac{3}{4}+\dfrac{7}{12}\right)\right].\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{25}{144}+\dfrac{1}{2}\)
\(=\dfrac{97}{144}\)
b)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}+\dfrac{1}{2}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}-\dfrac{1}{2}\)
\(=0\)
c)
\(\dfrac{6}{\dfrac{5}{12}}:\dfrac{2}{\dfrac{3}{4}}+\dfrac{11}{\dfrac{1}{4}}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(=\dfrac{27}{5}+\dfrac{11}{\dfrac{1}{4}}.\dfrac{2}{15}\)
\(=\dfrac{27}{5}+\dfrac{88}{15}\)
\(=\dfrac{169}{15}\)
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2019}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2018}{2019}\)
\(A=\frac{1.2.3.....2018}{2.3.4....2019}\)
\(A=\frac{1}{2019}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2018}{2019}\)
\(A=\frac{1}{2019}\)
Bn ơi tk cho mk nha!!!!!!!!!!!
Ý bạn đề bài thế này đúng ko: \(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2019}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\).
Nếu vậy mình làm như thế này nhé!
Ta có: B = \(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2019}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
= \(\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
= \(\frac{\frac{-2^3\cdot3^2}{3^3\cdot2^4}}{-\frac{5}{432}}\)
= \(\frac{\frac{1}{3\cdot2}}{\frac{5}{2^4\cdot3^3}}\)
= \(\frac{1}{3\cdot2}\div\left(\frac{5}{2^4\cdot3^3}\right)\)
= \(\left(\frac{1}{2\cdot3}\cdot\frac{2^4\cdot3^3}{5}\right)\)
= \(\left(\frac{2^3\cdot3^2}{5}\right)\)
= \(\frac{72}{5}.\)
Vậy B = \(\frac{72}{5}\).
đúng rồi
cảm ơn bạn nhiều nhé ^^