Thay vì \(\alpha;\beta;\gamma\) khó gõ kí tự, mình chuyển thành \(a,b,c\) cho dễ, bạn tự thay lại.

Do ABCD là hbh \(\Rightarrow\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}\)

- Chứng minh chiều thuận: I, F, K thẳng hàng \(\Rightarrow\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\)

Do I, F, K thẳng hàng \(\Rightarrow\) tồn tại một số \(k\ne0\) để \(\overrightarrow{KF}=k.\overrightarrow{KI}\)

\(\Rightarrow\left(\overrightarrow{KA}+\overrightarrow{AF}\right)=k.\left(\overrightarrow{KA}+\overrightarrow{AI}\right)\Rightarrow\left(-c.\overrightarrow{AD}+b.\overrightarrow{AC}\right)=k\left(-c.\overrightarrow{AD}+a.\overrightarrow{AB}\right)\)

\(\Rightarrow\overrightarrow{AD}\left(ck-c\right)=k.a.\overrightarrow{AB}-b.\overrightarrow{AC}=ka.\overrightarrow{AB}-b.\overrightarrow{AB}-b.\overrightarrow{AD}\)

\(\Rightarrow\overrightarrow{AD}\left(ck-c+b\right)=\overrightarrow{AB}\left(ka-b\right)\) (1)

Do \(\overrightarrow{AD};\overrightarrow{AB}\) không cùng phương \(\Rightarrow\left(1\right)\) xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}ck-c+b=0\\ka-b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\dfrac{c-b}{c}\\k=\dfrac{b}{a}\end{matrix}\right.\)

\(\Rightarrow\dfrac{c-b}{c}=\dfrac{b}{a}\Rightarrow1=\dfrac{b}{a}+\dfrac{b}{c}\Rightarrow\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\) (đpcm)

- Chứng minh chiều nghịch: \(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\Rightarrow\) I, F, K thẳng hàng

\(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\Rightarrow b=\dfrac{ac}{a+c}\)

\(\overrightarrow{FI}=\overrightarrow{FA}+\overrightarrow{AI}=-b.\overrightarrow{AC}+a.\overrightarrow{AB}=-b\left(\overrightarrow{AB}+\overrightarrow{AD}\right)+a.\overrightarrow{AB}\)

\(\Rightarrow\overrightarrow{FI}=-\dfrac{ac}{a+c}\overrightarrow{AB}-\dfrac{ac}{a+c}\overrightarrow{AD}+a.\overrightarrow{AB}=\dfrac{a^2}{a+c}\overrightarrow{AB}-\dfrac{ac}{a+c}\overrightarrow{AD}\)

\(\Rightarrow\overrightarrow{FI}=\dfrac{a}{a+c}\left(a.\overrightarrow{AB}-c.\overrightarrow{AD}\right)\) (1)

Lại có \(\overrightarrow{KI}=\overrightarrow{KA}+\overrightarrow{AI}=-c.\overrightarrow{AD}+a.\overrightarrow{AB}=a.\overrightarrow{AB}-c.\overrightarrow{AD}\) (2)

Từ (1), (2) \(\Rightarrow\overrightarrow{FI}=\dfrac{a}{a+c}\overrightarrow{KI}\) ; mà \(\dfrac{a}{a+c}\) là hằng số \(\ne0\)

\(\Rightarrow F,I,K\) thẳng hàng (đpcm)

Vậy F, I, K thẳng hàng khi và chỉ khi \(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\)

2.

ĐK: \(2x-y\ge0;y\ge0;y-x-1\ge0;y-3x+5\ge0\)

\(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\left(1\right)\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left(1-y\right)\sqrt{2x-y}+y-1+2x-y-1-\left(2x-y-1\right)\sqrt{y}=0\)

\(\Leftrightarrow\left(1-y\right)\left(\sqrt{2x-y}-1\right)+\left(2x-y-1\right)\left(1-\sqrt{y}\right)=0\)

\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(1+\sqrt{y}\right)+\left(\sqrt{2x-y}-1\right)\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}+1\right)=0\)

\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(\sqrt{y}+\sqrt{2x-y}+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2x-1\end{matrix}\right.\) (Vì \(\sqrt{y}+\sqrt{2x-y}+2>0\))

Nếu \(y=1\), khi đó:

\(\left(1\right)\Leftrightarrow x-5=\sqrt{-x}+\sqrt{-3x+6}\)

Phương trình này vô nghiệm

Nếu \(y=2x-1\), khi đó:

\(\left(1\right)\Leftrightarrow2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\) (Điều kiện: \(2\le x\le4\))

\(\Leftrightarrow2x\left(x-3\right)+x-3+1-\sqrt{x-2}+1-\sqrt{4-x}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1\right)=0\)

Ta thấy: \(1+\sqrt{x-2}\ge1\Rightarrow-\dfrac{1}{1+\sqrt{x-2}}\ge-1\Rightarrow1-\dfrac{1}{1+\sqrt{x-2}}\ge0\)

Lại có: \(\dfrac{1}{1+\sqrt{4-x}}>0\); \(2x>0\)

\(\Rightarrow\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1>0\)

Nên phương trình \(\left(1\right)\) tương đương \(x-3=0\Leftrightarrow x=3\Rightarrow y=5\)

Ta thấy \(\left(x;y\right)=\left(3;5\right)\) thỏa mãn điều kiện ban đầu.

Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)=\left(3;5\right)\)

Ta có: 

A. \(\alpha< \beta\)

\(\Rightarrow\left(0,3\right)^{\alpha}>\left(0,3\right)^{\beta}\)

Sai 

B. \(\alpha< \beta\)

\(\Rightarrow\pi^{\alpha}< \pi^{\beta}\)

Sai

C. \(\alpha< \beta\)

\(\Rightarrow\left(\sqrt{2}\right)^{\alpha}< \left(\sqrt{2}\right)^{\beta}\)

Đúng

D. \(\alpha< \beta\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{\alpha}>\left(\dfrac{1}{2}\right)^{\beta}\)

Sai 

⇒ Chọn C