Question

[Nguyễn Hàn Băng]

1) Tìm x , biết :

a) x² = 2x

b) x³ = x

c) 8^x : 2^x = 16²⁰¹⁷

^{2) Tính : A = ( \(\frac{1}{4}\)} - 1 ) ( \(\frac{1}{9}\) - 1 ) ( \(\frac{1}{16}\) - 1 ) ( \(\frac{1}{25}\) - 1 ) ..... ( \(\frac{1}{121}\) - 1 )

3) So sánh A và B , cho :

A=\(\frac{8^{2016+1}}{8^{2017+1}}\) và B = \(\frac{8^{2017+1}}{8^{2018+1}}\)

Answer 1

[Mysterious Person]

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 \(x=0;x=-1;x=1\)

Answer 2

[Mashiro Shiina]

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

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

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

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

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

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

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

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

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

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

3) Áp dụng tính chất:

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

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

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

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

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

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

\(B< A\)