*Đường tròn ngoại tiếp tam giác AEJ cắt JF tại K (K khác J).

\(\Rightarrow AJKE\) nội tiếp nên \(\widehat{EKF}=\widehat{JAF}\) (vì \(\widehat{EKF}\) là góc ngoài đỉnh K của tg AJKE).

Xét △EKF và △JAF có: \(\widehat{JFA}\) là góc chung, \(\widehat{EKF}=\widehat{JAF}\).

\(\Rightarrow\)△EKF∼△JAF (g-g).

\(\Rightarrow\dfrac{FE}{JF}=\dfrac{FK}{FA}\Rightarrow FE.FA=FK.FJ\left(1\right)\)

Ta có: A,C,B,E cùng thuộc (O) \(\Rightarrow AEBC\) nội tiếp.

Nên \(\widehat{JAE}=\widehat{JBC}\) (vì \(\widehat{JAE}\) là góc ngoài đỉnh A của tg AEBC).

Mà \(\widehat{JBC}+\widehat{EBF}=180^0\Rightarrow\widehat{JAE}+\widehat{EBF}=180^0\)

\(\Rightarrow\widehat{EKF}+\widehat{EBF}=180^0\) mà \(\widehat{EKF}+\widehat{EKJ}=180^0\)

\(\Rightarrow\widehat{JKE}=\widehat{JBF}\)

Xét △JEK và △JFB có: \(\widehat{JKE}=\widehat{JFB}\), \(\widehat{BJF}\) là góc chung.

\(\Rightarrow\)△JEK∼△JFB (g-g).

\(\Rightarrow\dfrac{JK}{JB}=\dfrac{JE}{JF}\Rightarrow JE.JB=JK.JF\left(2\right)\)

\(\left(1\right)+\left(2\right)\Rightarrow FE.FA+JE.JB=JF\left(JK+JK\right)=JK^2\left(đpcm\right)\)

`x xx 2/3 xx 3/4 xx 4/5 xx ... xx 2010/2011 = 2/2012`

`<=> x/2011 = 1/1006`

`=> x = 2011/1006`

Cách 1 : \(100=49+51\)

Cách 2 : \(100=1+3+5+7+9+11+13+15+17+19\)

2²⁰²² = (2⁴)⁵⁰⁵.2² = \(\overline{...6}\)⁵⁰⁵. 4 = \(\overline{...4}\) =  \(\overline{...0}\) + 4 ⇒ 2²⁰²² : 5 dư 4 

Ta có: `2^2022 = (2^4)^505 . 2^2`.

`= 16^505 . 4 equiv 1^505 . 4 equiv 1 . 4 equiv 4 (mod 5)`.

c) P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

Dễ thấy \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}.50=\dfrac{1}{3}\)(1)

Tương tự

 \(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>50.\dfrac{1}{200}=\dfrac{1}{4}\)(2) 

Từ (1) và (2) ta được

\(P>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) 

P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

         \(\overline{50\text{ hạng tử }}\)                            \(\overline{50\text{ hạng tử }}\)

\(< \left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)+\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)\) 

\(=\dfrac{1}{100}.50+\dfrac{1}{150}.50=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\Rightarrow P< \dfrac{5}{6}< 1\)

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