Bài 5: Lũy thừa của một số hữu tỉ

Đặt \(A=2^{2009}+2^{2008}+...+2+2^0\)

\(=1+2+...+2^{2008}+2^{2009}\)

\(\Rightarrow2A=2+2^2+...+2^{2010}\)

\(\Rightarrow2A-A=\left(2+2^2+...+2^{2010}\right)-\left(1+2+...+2^{2009}\right)\)

\(\Rightarrow A=2^{2010}-1\)

\(\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)\)

\(=2^{2010}-2^{2010}+1=1\)

Vậy M = 1

Bài 1:

c) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.4^4.5^4}{5^{10}.4^5}=\dfrac{5^8.4^4}{5^8.5^2.4^4.4}=\dfrac{1}{25.4}=\dfrac{1}{100}\)

Bài 2: a) \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\\\left(y+0,4\right)^{100}\ge0\forall y\\\left(z-3\right)^{678}\ge0\forall z\end{matrix}\right.\) \(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)

Vậy ...

Bài 3: \(M=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256.\)

Vậy \(M=256.\)

Mấy bài kia dễ tự làm.

\(3)\)

\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256\)\(4)\)

\(2^{24}=\left(2^6\right)^4=64^4;3^{16}=\left(3^4\right)^4=81^4\)

\(\Leftrightarrow2^{24}< 3^{16}\)

Bài 4 :

\(2^{24}=\left(2^3\right)^8=8^8\) \(\left(1\right)\)

\(3^{16}=\left(3^2\right)^8=9^8\) \(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow8^8< 9^8\)

hay \(2^{24}< 3^{16}\) \((dpcm) \)

Ta có :

\(8^7-2^{18}=\left(2^3\right)^7-2^{18}\)

\(=2^{21}-2^{18}\)

\(=2^{18}\left(2^3-1\right)\)

\(=2^{18}.7\)

\(=2^{17}.2.7\)

\(=2^{17}.14⋮14\)

Vậy \(8^7-2^{18}⋮14\rightarrowđpcm\)

Ta có:

\(8^7-2^{18}=\left(2^3\right)^7-2^{18}\)

\(=2^{21}-2^{18}=2^{18}.\left(2^3-1\right)\)

\(=2^{17}.2.7=2^{17}.14\)

Vì \(14⋮14\) nên \(2^{17}.14⋮14\)

=> \(8^7-2^{18}⋮14\) (đpcm)

Chúc bạn học tốt!!!

Giải:

\(8^7-2^{18}\)

\(=\left(2^3\right)^7-2^{18}\)

\(=2^{21}-2^{18}\)

\(=2^{18}.\left(2^3-1\right)\)

\(=2^{18}.\left(8-1\right)\)

\(=2^{18}.7\)

\(=2^{17}.2.7\)

\(=2^{17}.14⋮14\) (đpcm)

Chúc bạn học tốt!

cho hỏi đề yêu cầu tính hay so sánh

C1: Ta có: \(2^{91}=\left(2^{13}\right)^7\)

\(5^{35}=\left(5^5\right)^7\)

Do \(2^{13}>5^5\Rightarrow2^{91}>5^{35}\)

cách trên thửu công quá nên ta có cách khác tìm số trung gian,

C2: \(2^{91}>2^{90}=\left(2^5\right)^{18}=32^{18}\left(1\right)\)

\(5^{35}< 5^{36}=\left(5^2\right)^{18}=25^{18}\left(2\right)\)

Do \(32>25\Rightarrow32^2>25^2\left(3\right)\)

Từ (1);(2);(3) Suy ra:

\(2^{91}>5^{35}\)

\(\Leftrightarrow6^x\cdot\dfrac{1}{6}+6^x\cdot36=6^7\cdot217\)

\(\Leftrightarrow6^x=1679616\)

hay x=8

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^{10}.10^2}\)

\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^{10}}-\dfrac{1}{10^2}\)

\(=1-\dfrac{1}{10^2}< 1\)

\(\Rightarrowđpcm\)

ta có

\(8^7-2^{18}=8^7-\left(2^3\right)^6\\ =8^7-8^6=8^6\left(8-1\right)\\ =8^6.7=8^5.8.7=8^5.\left(4.2\right).7=14.\left(4.8^5\right)\)

do 14 chia hết cho 14 nên \(14.\left(4.8^5\right)\)cũng chia hết cho 14

hay\(8^7-2^{18}\)chia hết cho 14

các câu b,c bạn làm tương tự. mình chỉ làm cho bạn câu a. chúc bạn học tốt. nếu có vấn đề j về bài toán câu a thì bạn cứ ib cho mình mình sẽ giải đáp.

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

Gọi \(N=2^{2009}+2^{2008}+...+2^1+2^0\)

\(2N=2^{2010}+2^{2009}+...+2^2+2^1\\ 2N-N=\left(2^{2010}+2^{2009}+...+2^2+2^1\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\\ N=2^{2010}-2^0\\ N=2^{2010}-1\)

Thay vào ta được

\(M=2^{2010}-\left(2^{2010}-1\right)\\ M=2^{2010}-2^{2010}+1\\ M=1\)

Vậy \(M=1\)

Ta có :

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^0\right)\)

Đặt A=2²⁰⁰⁹+2²⁰⁰⁸+..+2⁰

\(A=2^{2009}+2^{2008}+...+2^0\\ 2A=2^{2010}+2^{2009}+...+2^1\\ \Rightarrow2A-A=A=2^{2010}-2^0\\ \Rightarrow M=2^{2010}-\left(2^{2010}-2^0\right)\\ M=2^{2010}-2^{2010}+1\\ \Rightarrow M=1\)

Chúc bạn học tốt!