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Lời giải:
Sử dụng công thức $(a-1)(a+1)=a^2-1$ ta có:
$8F=(9-1)(9+1)(9^2+1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^2-1)(9^2+1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^4-1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^8-1)(9^8+1)...(9^{32}+1)$
$=(9^{16}-1)...(9^{32}+1)=(9^{32}-1)(9^{32}+1)=9^{64}-1$
$\Rightarrow F=\frac{9^{64}-1}{8}$
\(B=10+9^2+9^3+...+9^{2005}\)
\(\Rightarrow B=1+9+9^2+...+9^{2005}\)
\(\Rightarrow9B=9+9^2+9^3+...+9^{2006}\)
\(\Rightarrow9B-B=\left(9+9^2+9^3+...+9^{2006}\right)-\left(1+9+9^2+...+9^{2005}\right)\)
\(\Rightarrow8B=9^{2006}-1\)
\(\Rightarrow B=\frac{9^{2006}-1}{8}\)
Vậy \(B=\frac{9^{2006}-1}{8}\)
_Chúc bạn học tốt_
Ta có \(\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\frac{1}{8}\left(9-1\right)\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\frac{1}{8}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
cứ như thế
\(=\frac{1}{8}\left(9^{64}-1\right)< 9^{64}-1\)=>đpcm
964 - 1 = (932 + 1)(932 - 1) = ... = (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(9 + 1)(9 - 1) > (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(9 + 1)
964=(932+1).(932-1)
=(932+1)(916+1)(916-1)
=(932+1)(916+1)(98+1)(98-1)
=(932+1)(916+1)(98+1)(94+1)(94-1)
=(932+1)(916+1)(98+1)(94+1)(92+1)(92-1)
=(932+1)(916+1)(98+1)(94+1)(92+1)(9+1)(9-1)
Vì (932+1)(916+1)(98+1)(94+1)(92+1)(9+1)(9-1)>(932+1)(916+1)(98+1)(94+1)(92+1)(9+1)
=>964-1>(932+1)(916+1)(98+1)(94+1)(92+1)(9+1)
đk : x khác -9 ; -10
\(\dfrac{x+9}{10}+1+\dfrac{x+10}{9}+1=\dfrac{9}{x+10}+1+\dfrac{10}{x+9}+1\)
\(\Leftrightarrow\dfrac{x+19}{10}+\dfrac{x+19}{9}=\dfrac{x+19}{x+10}+\dfrac{x+19}{x+9}\)
\(\Leftrightarrow\left(x+19\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{x+10}-\dfrac{1}{x+9}\right)=0\Leftrightarrow x=-19\)
\(19^{5^{1^{8^{9^0}}}}=19^5;2^{9^{1^{9^{6^9}}}}=2^9\)
195=194.19=...1.19=...9
29=24.24.2=16.16.2=...2
=>195+29 có tận cùng là 1
vậy chữ số tận cùng của \(19^{5^{1^{8^{9^0}}}}+2^{9^{1^{9^{6^9}}}}\)là 1
Những lần sau cố gắng đặt câu hỏi dưới dạng công thức nhé!
Lời giải:
ĐK: \(x\ne-10;x\ne-9\)
Phương trình tương đương:
\(9\left(x+10\right)\left(x+9\right)^2+10\left(x+9\right)\left(x+10\right)^2=810\left(x+9\right)+900\left(x+10\right)\\ \Leftrightarrow\left(x+10\right)\left(x+9\right)\left[9\left(x+9\right)+10\left(x+10\right)\right]=90\left[9\left(x+9\right)+10\left(x+10\right)\right]\\ \Leftrightarrow\left[\left(x+10\right)\left(x+9\right)-90\right].\left[9\left(x+9\right)+10\left(x+10\right)\right]=0\\ \Leftrightarrow\left(x^2+19x\right)\left(19x+181\right)=0\)
\( \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = - 19\\ x = - \dfrac{{181}}{{19}} \end{array} \right.\left( {tm} \right)\)
bằng báo cáo đó bn ơi
9 x ..... = ....... ok