a) Giả sử Bm // Cn. Khi đó ta có:

\(\widehat{xBm}=\widehat{BCn}\) (hai góc đồng vị) và \(\widehat{BCn}+\widehat{CBm}=180^o\) (hai góc trong cùng phía)

\(\Rightarrow\widehat{xBm}+\widehat{mBC}=\widehat{xBC}=180^o\) (a)

Mà \(\widehat{ABC}\) và \(\widehat{xBC}\) là hai góc kề bù (vì \(\widehat{xBC}\) là góc ngoài đỉnh B)

\(\Rightarrow\widehat{ABC}+\widehat{mBC}=180^o\)

\(\Rightarrow\widehat{mBC}=180^o-\widehat{ABC}\)

\(\Rightarrow\widehat{mBC}< 180^o\) (b)

Từ (a) và (b) suy ra vô lí, suy ra Bm không song song với Cn

Vậy Bm cắt Cn

Vì At là tia phân giác của \(\widehat{xAy}\) nên \(\widehat{BAD}=\widehat{CAD}\)

Xét ΔBAD vuông tại B và ΔCAD vuông tại C có:

- AD là cạnh chung

- \(\widehat{BAD}=\widehat{CAD}\) (chứng minh trên)

Suy ra ΔBAD = ΔCAD (cạnh huyền - góc nhọn), từ đó AB = AC (vì hai cạnh tương ứng)

\(\dfrac{72^3\cdot54^2}{108^4}=\dfrac{\left(2^3\cdot3^2\right)^3\cdot\left(2\cdot3^3\right)^2}{\left(2^2\cdot3^3\right)^4}=\dfrac{2^9\cdot3^6\cdot2^2\cdot3^6}{2^8\cdot3^{12}}\)

\(=\dfrac{2^{11}\cdot3^{12}}{2^8\cdot3^{12}}=2^3=8\)

\(\dfrac{3}{5}+\dfrac{3}{11}-\left(-\dfrac{3}{7}\right)+\left(-\dfrac{2}{97}\right)-\dfrac{1}{35}-\dfrac{3}{4}+\left(-\dfrac{23}{44}\right)\)

\(=\left(\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{1}{35}\right)+\left(\dfrac{3}{11}-\dfrac{3}{4}-\dfrac{23}{44}\right)-\dfrac{2}{97}\)

\(=1-1-\dfrac{2}{97}=-\dfrac{2}{97}\)

\(x^2⋮6\) ⇒ \(x^2\in\left\{0;36;144;324;...\right\}\)

⇒ \(x\in\left\{0;6;12;18;...\right\}\) ⇒ \(x⋮6\)

Vậy mệnh đề trên là mệnh đề đúng.

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+...+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}+\dfrac{1}{3^{51}}\)

Và \(B=\dfrac{1}{3^2}+\dfrac{1}{3^4}+\dfrac{1}{3^6}+...+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}+\dfrac{1}{3^{50}}\)

Ta có:

\(9A=3+\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{45}}+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}\)

\(9A-A=\left(3+\dfrac{1}{3}+...+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{49}}+\dfrac{1}{3^{51}}\right)\)

\(8A=3-\dfrac{1}{3^{51}}\)

\(A=\dfrac{3-\dfrac{1}{3^{51}}}{8}\)

\(9B=1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{44}}+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}\)

\(9B-B=\left(1+\dfrac{1}{3^2}+...+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{48}}+\dfrac{1}{3^{50}}\right)\)

\(8B=1-\dfrac{1}{3^{50}}\)

\(B=\dfrac{1-\dfrac{1}{3^{50}}}{8}\)

Suy ra

\(-\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{50}}-\dfrac{1}{3^{51}}=B-A=\dfrac{1-\dfrac{1}{3^{50}}}{8}-\dfrac{3-\dfrac{1}{3^{51}}}{8}\)

\(=\dfrac{\left(1-\dfrac{1}{3^{50}}\right)-\left(3-\dfrac{1}{3^{51}}\right)}{8}=\dfrac{-2-\dfrac{1}{3^{50}}+\dfrac{1}{3^{51}}}{8}=\dfrac{-2+\dfrac{-3^{51}+3^{50}}{3^{101}}}{8}\)

\(=\dfrac{-2+\dfrac{3^{50}\left(-3+1\right)}{3^{101}}}{8}=\dfrac{-2-\dfrac{2}{3^{51}}}{8}=-\dfrac{2\left(1+\dfrac{1}{3^{51}}\right)}{8}=-\dfrac{1+\dfrac{1}{3^{51}}}{4}\)

b) Ta có:

 P(x) + H(x) = x⁴ - x³ + 2x² + x + 1

=> H(x) = x⁴ - x³ + 2x² + x + 1 - P(x)

=> H(x) = (x⁴ - x³ + 2x² + x + 1) - (2x⁴ - x² + x - 2)

=> H(x) = -x⁴ - x³ + 3x² + 3

Vậy H(x) = -x⁴ - x³ + 3x² + 3

\(\left(x^2-1\right)\left(x^2-5\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2-1< 0\\x^2-5>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x^2-1>0\\x^2-5< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2< 1\\x^2>5\end{matrix}\right.\)(vô lí) hoặc \(\left\{{}\begin{matrix}x^2>1\\x^2< 5\end{matrix}\right.\)

\(\Rightarrow1< x< \sqrt{5}\) hoặc \(-\sqrt{5}< x< -1\)

Vậy \(-\sqrt{5}< x< -1\) hoặc \(1< x< \sqrt{5}\)

Trước hết ta phải chứng minh \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).

Thật vậy, \(\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{a+ab}{b^2+b}\) và \(\dfrac{a+1}{b+1}=\dfrac{\left(a+1\right)b}{\left(b+1\right)b}=\dfrac{ab+b}{b^2+b}\).

Mà theo giả thuyết là a < b nên \(\dfrac{a+ab}{b^2+b}< \dfrac{ab+b}{b^2+b}\), suy ra \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).

Từ đây ta có:

\(B=\dfrac{2022^{2022}+1}{2022^{2023}+1}=\dfrac{2022^{2023}+2022}{2022^{2024}+2022}=\dfrac{2022^{2023}+2021+1}{2022^{2024}+2021+1}\)

Đặt \(A_1=\dfrac{2022^{2023}+2}{2022^{2024}+2}=\dfrac{2022^{2023}+1+1}{2022^{2024}+1+1}\), rõ ràng \(A_1>A\).

Đặt \(A_2=\dfrac{2022^{2023}+3}{2022^{2024}+3}=\dfrac{2022^{2023}+2+1}{2022^{2024}+2+1}\), rõ ràng \(A_2>A_1\).

...

Đặt \(A_{2020}=\dfrac{2022^{2023}+2021}{2022^{2024}+2021}=\dfrac{2022^{2023}+2020+1}{2022^{2024}+2020+1}\), rõ ràng \(A_{2020}>A_{2019}\) và \(B>A_{2020}\).

Suy ra \(B>A_{2020}>A_{2019}>...>A_2>A_1>A\). Vậy A < B.