a=\(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{79}{80}\)

a<\(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{80}{81}\)

\(\text{a}^2< \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot\dfrac{7}{8}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{79}{80}\cdot\dfrac{80}{81}\)

\(\Rightarrow\text{a}^2< \dfrac{1}{81}=\left(\dfrac{1}{9}\right)^2\)

\(\Rightarrow\text{a}< \dfrac{1}{9}\)(dpcm)

Nho tich cho mk nhe

có bị sai đề không đấy bạn

CMR A> 1/9 thì mới làm được chứ

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2\cdot2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3\cdot3}< \dfrac{1}{2\cdot3}\)

\(\dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}\)

...

\(\dfrac{1}{9^2}=\dfrac{1}{9\cdot9}< \dfrac{1}{8\cdot9}\)

\(\dfrac{1}{10^2}=\dfrac{1}{10\cdot10}< \dfrac{1}{9\cdot10}\)

\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A< 1-\dfrac{1}{10}\)

\(\Rightarrow A< \dfrac{9}{10}\)

\(\Rightarrow A< 1\) (vì: \(\dfrac{9}{10}< 1\))

221​=2⋅21​<1⋅21​

132=13⋅3<12⋅3321​=3⋅31​<2⋅31​

142=14⋅4<13⋅4421​=4⋅41​<3⋅41​

...

192=19⋅9<18⋅9921​=9⋅91​<8⋅91​

1102=110⋅10<19⋅101021​=10⋅101​<9⋅101​

⇒�=122+132+142+...+1102<11⋅2+12⋅3+13⋅4+...+19⋅10⇒A=221​+321​+421​+...+1021​<1⋅21​+2⋅31​+3⋅41​+...+9⋅101​

⇒�<1−12+12−13+...+19−110⇒A<1−21​+21​−31​+...+91​−101​

⇒�<1−110⇒A<1−101​

⇒�<910⇒A<109​

$\Rightarrow A<1$ (vì: 910<1109​<1)

đơn giản quá!

Bạn có bt làm bài 5 ko?

Lời giải:

Gọi tổng trên là $A$. Ta có:

\(2A>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(2A>\frac{\sqrt{2}-1}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3})}+...+\frac{\sqrt{81}-\sqrt{80}}{(\sqrt{80}+\sqrt{81})(\sqrt{81}-\sqrt{80})}\)

\(2A>(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{81}-\sqrt{80})\)

\(2A>\sqrt{81}-1=8\Rightarrow A>4\)

Ta có đpcm.

Ta có:

A=999993¹⁹⁹⁹−555557¹⁹⁹⁷

A=999993¹⁹⁹⁸.999993−555557¹⁹⁹⁶.555557

A=(999993²)⁹⁹⁹.999993 − (555557²)⁹⁹⁸.555557

A=\(\overline{\left(....9\right)}^{999}\) . 999993 - \(\overline{\left(...1\right)}.\text{555557}\)

A=\(\overline{\left(...7\right)}-\overline{\left(...7\right)}\)

A= \(\overline{\left(...0\right)}\)

Vì A có tận cùng là 0 nên \(A⋮5\)