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a) Gọi pt đường thẳng (d) đi qua 2 điểm A,B là : y= ax +b
Ta có A(-1,1), B(2,7) thuộc (d) ⇒ \(\left\{{}\begin{matrix}1=-1.a+b\\7=2.a+b\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
⇒ pt đi qua AB (d) là y=2x+3
b) Giả sử C(-2,-1) ∈ (d)
⇒ -1=-2.2 +3 ⇒ -1=-1( luôn đúng)
⇒ C(-2,-1) ∈(d) ⇒ A,B,C thẳng hàng
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có : \(a^2+b^2\ge2ab\) ( cauchuy )
\(\Rightarrow a^2+2ab+b^2=\left(a+b\right)^2\ge4ab\)
\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
b, Ta có : \(a^2+b^2\ge2ab\) ( cauchuy )
\(\Rightarrow ab\le\dfrac{a^2+b^2}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ giả thiết: \(\left\{{}\begin{matrix}a+b+c\ge3\sqrt[3]{abc}=3\\ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\end{matrix}\right.\) (1)
Ta có:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\ge3\left(a+b+c\right)-1\)
Nên ta chỉ cần chứng minh:
\(3\left(a+b+c\right)-1\ge2\left(1+a+b+c\right)\)
\(\Leftrightarrow a+b+c\ge3\) (hiển nhiên đúng theo (1))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
\(\Leftrightarrow a+b=a+c+b+c+2\sqrt{\left(a+c\right)\left(b+c\right)}\)
\(\Leftrightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\)
\(\Leftrightarrow c+\sqrt{\left(a+c\right)\left(b+c\right)}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\-c=\sqrt{\left(a+c\right)\left(b+c\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\c^2=\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\ab+bc+ac=0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\left(đúng\right)\)
Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0
Khi đó:
(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)
=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2
=a+b+2c+2|c|=a+b+2c+2|c|
Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c
Do đó:
(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b
⇒√a+c+√b+c=√a+b