bạn kiểm tra lại B xem sao

A=\(\frac{8^{18^{ }}+1}{8^{19}+1}\)=> 8A=8.

đợi mk tí nha, nãy mk làm rồi nhma máy cập nhật lại nên mất bài =)

ta có: \(8^{10}-8^9-8^8=8^8\left(8^2-8-1\right)\)
\(=8^8\cdot55\)
vậy \(\left(8^{10}-8^9-8^8\right)\) chia hết cho 55

1) a) \(x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)

b) \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\) \(\Leftrightarrow x\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\) vậy \(x=0;x=-1;x=1\)

\(x^2=2x\Rightarrow x^2-2x=0\Rightarrow x\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\end{matrix}\right.\)

\(x^3=x\Rightarrow x^3-x=0\Rightarrow x\left(x^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\end{matrix}\right.\)

\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\left(\dfrac{1}{25}-1\right)...\left(\dfrac{1}{121}-1\right)\)

\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}.\dfrac{-24}{25}...\dfrac{-120}{121}\)

\(A=\dfrac{3.8.15.24....120}{4.9.16.25...121}\)

\(A=\dfrac{1.3.2.4.3.5.4.6....10.12}{2.2.3.3.4.4.5.5....11.11}\)

\(A=\dfrac{1.2.4....10}{2.3.4.5...11}.\dfrac{3.4.5....12}{2.3.4.5....11}\)

\(A=\dfrac{1}{11}.6=\dfrac{6}{11}\)

3) Áp dụng tính chất:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(B=\dfrac{8^{2017}+1}{8^{2018}+1}< 1\)

\(B< \dfrac{8^{2017}+1+8}{8^{2018}+1+8}\)

\(B< \dfrac{8^{2017}+8}{8^{2018}+8}\)

\(B< \dfrac{8\left(8^{2016}+1\right)}{8\left(8^{2017}+1\right)}\)

\(B< \dfrac{8^{2016}+1}{8^{2017}+1}=A\)

\(B< A\)

ta có A= \(\frac{8^{18}+1}{8^{19} +1}\)=> 8A=\(\frac{8^{19}+8}{8^{19}+1}\)= \(\frac{\left(8^{19}+1\right)+7}{8^{19}+1}\)=\(\frac{8^{19}+1}{8^{19} +1}\)+\(\frac{7}{8^{19}+1}\) =1+\(\frac{7}{8^{19}+1}\) =\(\frac{7}{8^{19}+1}\) 

         B= \(\frac{8^{23}+1}{8^{24}+1}\)=> 8B=\(\frac{8^{24}+8}{8^{24}+1}\)= \(\frac{\left(8^{24}+1\right)+7}{8^{24}+1}\)=\(\frac{8^{24}+1}{8^{24}+1}\)+\(\frac{7}{8^{24}+1}\) =1+\(\frac{7}{8^{24} +1}\)=\(\frac{7}{8^{24}+1}\)

       vì  \(8^{19}\)<\(8^{24}\)=> \(8^{19}\)+1 >\(8^{24}\)+1 => \(\frac{7}{8^{19}+1}\)<\(\frac{7}{8^{24}+1}\)=> A<B

a) ta có \(8A=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\\ 8B=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)

Vì \(8^{24}+1>8^{19}+1\\\frac{7}{8^{24}+1}< \frac{7}{8^{19}+1} \)

vậy 8A>8B nên A>B

a) \(\frac{5}{6}\)x - \(\frac{3}{8}\)x - 10 = 12

=> \(\left(\frac{5}{6}-\frac{3}{8}\right)\)x = 12 + 10

=> \(\frac{11}{24}\)x = 22

=> x = 22 : \(\frac{11}{24}\)

=> x = 48

Vậy x = 48.

b) (\(\left|x\right|\) - \(\frac{1}{8}\)) . \(\left(-\frac{1}{8}\right)^5\) = \(\left(-\frac{1}{8}\right)^7\)

=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^7\) : \(\left(-\frac{1}{8}\right)^5\)

=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^{7-5}\)

=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\frac{1}{64}\)

=> \(\left|x\right|\) = \(\frac{1}{64}\) + \(\frac{1}{8}\)

=> \(\left|x\right|\) = \(\frac{9}{64}\)

=> x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)

Vậy x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)

hên quá làm đúng hì hì, cảm ơn nhen, hết sợ bị sai ồi