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$\textbf{Ta có:}$
$A=\dfrac{-5^2-5\cdot3^2}{5^3+5^2\cdot3^2}$
$=\dfrac{-25-45}{125+225}$
$=\dfrac{-70}{350}$
$=-\dfrac15.$
Và
$B=\dfrac{2^{12}\cdot3^{10}+6^9\cdot120}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}$
$=\dfrac{2^{12}3^{10}+2^{12}3^{10}\cdot5}{2^{11}3^{11}(2\cdot3-1)}$
$=\dfrac{2^{12}3^{10}(1+5)}{2^{11}3^{11}\cdot5}$
$=\dfrac{2^{12}3^{10}\cdot6}{2^{11}3^{11}\cdot5}$
$=\dfrac{2^2}{5}$
$=\dfrac45.$
=> $M=B-A$$=\dfrac45-\left(-\dfrac15\right)$
$=\dfrac55$$=1.$
Ta có; \(\frac{\left(2^3.3^4\right)}{\left(2^2.3^3.5\right)}\Leftrightarrow\frac{\left(2^2.2^1.3^3.3^1\right)}{2^2.3^3.5}\Leftrightarrow\frac{\left(2.3\right)}{5}=\frac{6}{5}\)
Đs: \(\frac{6}{5}\)
Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)
=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)
Mình làm ngắn gọn nhé.
\(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}\)
\(\Rightarrow A=2^{51}-1\)
\(B=1+3+...+3^{66}\)
\(3B=3+3^2+...+3^{67}\)
\(2B=3+3^2+...+3^{67}-1-3-...-3^{66}\)
\(2B=3^{67}-1\)
\(B=\frac{3^{67}-1}{2}\)
\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{2^5.3^3.5^3.7.\left(-2\right)}{3.2^5.5^3.7}=\frac{3^2.\left(-2\right)}{1}=-18\)
dễ, nhưng phai giai dc câu nay 60% nhan x cong 2 phan 3 = 1 phan 3 nhan 6va 1 phan 3
a/ 26176
b/ 8
c/472392
d/ Ko tính dc
Thấy mình đẳng cấp ko
Cho mình 10 k nhé
$\textbf{A)}$
$A=\dfrac{4^2\cdot36+64^4}{16^4\cdot100}$
$=\dfrac{576+2^{24}}{2^{16}\cdot2^2\cdot5^2}$
$=\dfrac{576+16777216}{1638400}$
$=\dfrac{16777792}{1638400}$
$=\dfrac{262153}{25600}.$