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\(y^2=a^2\left(1+b^2\right)+b^2\left(1+a^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(x^2=a^2b^2+\left(1+a^2\right)\left(1+b^2\right)+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)
\(=a^2+b^2+2a^2b^2+2ab\sqrt{\left(1+a^2\right)\left(1+b^2\right)}+1\)
\(\Rightarrow y^2+1=x^2\)
\(\Rightarrow y^2=x^2-1\)
\(\Rightarrow y=\sqrt{x^2-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)
\(\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2=4\)
\(\Leftrightarrow a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=4\)
\(\Leftrightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\frac{4-(a+b+c)}{2}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})\)
Tương tự:
$b+1=(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})$
$c+1=(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})$
Khi đó:
\(A=\left[\frac{\sqrt{a}}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})}+\frac{\sqrt{b}}{(\sqrt{b}+\sqrt{a})(\sqrt{b}+\sqrt{c})}+\frac{\sqrt{c}}{(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})}\right]\sqrt{(a+1)(b+1)(c+1)}\)
\(\frac{\sqrt{a}(\sqrt{b}+\sqrt{c})+\sqrt{b}(\sqrt{c}+\sqrt{a})+\sqrt{c}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.\sqrt{(\sqrt{a}+\sqrt{b})^2(\sqrt{b}+\sqrt{c})^2(\sqrt{c}+\sqrt{a})^2}\)
\(=\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})\)
\(=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(gt\Rightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\Rightarrow\frac{1}{a^2}+1=\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)\)
\(\frac{1}{ab}\sqrt{\frac{\left(a^2+1\right)\left(b^2+1\right)}{c^2+1}}=\sqrt{\frac{\left(1+\frac{1}{a^2}\right)\left(1+\frac{1}{b^2}\right)}{c^2\left(1+\frac{1}{c^2}\right)}}\)
\(=\frac{1}{c}.\sqrt{\frac{\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}+\frac{1}{c}\right)}{\left(\frac{1}{c}+\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{b}\right)}}=\frac{1}{c}\sqrt{\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
\(=\frac{1}{c}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{bc}+\frac{1}{ca}\)
Tương tự với các cụm còn lại, ta được
\(A=2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2\)
bài này khó thật, nhưng bạn đừng buồn, sẽ có nhiều bạn khác giúp bạn
nha Nguyễn Quang Linh à
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ ab + bc + ac =1
=> ab + bc + ac + a2 = 1 + a2
=> 1 + a2 = (a+b)(a+c) (1)
Tương tự: 1 + b2 = (a+b)(b+c) (2)
1 + c2 = (a+c)(b+c) (3)
Thay (1) (2) (3) vào P
P= a\(\sqrt{\left(b+c\right)^2}\)+ b\(\sqrt{\left(a+c\right)^2}\)+ c\(\sqrt{\left(a+b\right)^2}\)
= a|b+c| + b|a+c| + c|a+b|
= a(b+c) + b(a+c) + c(a+b) (do a,b,c >0)
= ab + ac +ab + bc +ac +bc
= 2(ab + ac + bc)
=2
![](https://rs.olm.vn/images/avt/0.png?1311)
ques này nhiều ng` hỏi r` thay ab+bc+ca=1 vào rồi phân tích rút gọn
![](https://rs.olm.vn/images/avt/0.png?1311)
Do ab + bc + ca = 1 nên ta có :
\(a\sqrt{\frac{\left(b^2+1\right)\left(c^2+1\right)}{a^2+1}}=a\sqrt{\frac{\left(b^2+ab+ac+bc\right)\left(c^2+ab+ac+bc\right)}{a^2+ab+ac+bc}}\)
\(=a\sqrt{\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}{\left(a+b\right)\left(a+c\right)}}=a\sqrt{\left(b+c\right)^2}=a\left(b+c\right)=ab+ac\text{ }\left(1\right)\)
Tương tự : \(b\sqrt{\frac{\left(a^2+1\right)\left(c^2+1\right)}{b^2+1}}=ab+bc\) (2)và \(c\sqrt{\frac{\left(b^2+1\right)\left(a^2+1\right)}{c^2+1}}=bc+ac\) (3)
Cộng vế với vế của (1) ; (2) ; (3) lại ta được :
\(a\sqrt{\frac{\left(b^2+1\right)\left(c^2+1\right)}{a^2+1}}+b\sqrt{\frac{\left(a^2+1\right)\left(c^2+1\right)}{b^2+1}}+c\sqrt{\frac{\left(b^2+1\right)\left(a^2+1\right)}{c^2+1}}=2\left(ab+bc+ac\right)=2\)
Hình như a+b=0
bai nay nhan lien hop voi tung thua so ben ve trai la ra
KQ: a+b=0
chuc ban thanh cong