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Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
Lời giải:
Do \(ab+bc+ac=1\) nên:
\(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)
\(b^2+1=b^2+ab+bc+ac=(b+a)(b+c)\)
\(c^2+1=c^2+ab+bc+ac=(c+a)(c+b)\)
Do đó:
\(A=a\sqrt{\frac{(b^2+1)(c^2+1)}{a^2+1}}+b\sqrt{\frac{(a^2+1)(c^2+1)}{b^2+1}}+c\sqrt{\frac{(b^2+1)(a^2+1)}{c^2+1}}\)
\(=a\sqrt{\frac{(b+c)(b+a)(c+a)(c+b)}{(a+b)(a+c)}}+b\sqrt{\frac{(a+b)(a+c)(c+a)(c+b)}{(b+a)(b+c)}}+c\sqrt{\frac{(b+a)(b+c)(a+b)(a+c)}{(c+a)(c+b)}}\)
\(=a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ac)=2\)
Vậy \(A=2\)
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
Lời giải:
\(a+b+c=abc\)
\(\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=a^2bc+bc\)
\(\Rightarrow (a+b)(a+c)=bc(a^2+1)\)
\(\Rightarrow \frac{a}{\sqrt{bc(a^2+1)}}=\frac{a}{\sqrt{(a+b)(a+c)}}\leq \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\) (theo BĐT AM-GM ngược dấu)
Hoàn toàn tương tự:
\(\frac{b}{\sqrt{ca(b^2+1)}}\leq \frac{1}{2}\left(\frac{b}{b+a}+\frac{b}{b+c}\right)\)
\(\frac{c}{\sqrt{ab(c^2+1)}}\leq \frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)
Cộng theo vế những BĐT thu được ở trên ta có:
\(S\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
Vậy \(S_{\max}=\frac{3}{2}\Leftrightarrow a=b=c=\sqrt{3}\)
• Vì a, b, c đều dương và a + b + c = 2
nên \(0< a,b,c< 2\)
• Theo gt, ta có:
\(\Leftrightarrow\left\{{}\begin{matrix}b+c=2-a\\\left(b+c\right)^2-2bc=2-a^2\end{matrix}\right.\)
\(\Rightarrow\left(2-a\right)^2-2+a^2=2bc\)
\(\Rightarrow bc=\dfrac{\left(4-4a+a^2\right)-2+a^2}{2}=\dfrac{2a^2-4a+2}{2}=\left(a-1\right)^2\)
\(\Rightarrow b^2c^2=\left(a-1\right)^4\)
• Ta lại có: \(a\sqrt{\dfrac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}=a\sqrt{\dfrac{1+b^2+c^2+b^2c^2}{1+a^2}}\)
\(=a\sqrt{\dfrac{3-a^2+\left(a-1\right)^4}{1+a^2}}=a\sqrt{\dfrac{a^4-4a^3+5a^2-4a-4}{1+a^2}}\)
\(=a\sqrt{\dfrac{\left(1+a^2\right)\left(a-2\right)^2}{1+a^2}}=a\left(2-a\right)\)
• Tương tự, ta cũng có: \(b\sqrt{\dfrac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}=b\left(2-b\right)\)
\(c\sqrt{\dfrac{\left(1+b^2\right)\left(1+a^2\right)}{1+c^2}}=c\left(2-c\right)\)
• Suy ra \(a\sqrt{\dfrac{\left(1+a^2\right)\left(a-2\right)^2}{1+a^2}}+b\sqrt{\dfrac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}+c\sqrt{\dfrac{\left(1+b^2\right)\left(1+a^2\right)}{1+c^2}}\)
\(=2\left(a+b+c\right)-\left(a^2+b^2+c^2\right)=2\left(đpcm\right)\)
Xét \(n^2+1=n^2+mn+np+pm=n\left(m+n\right)+p\left(m+n\right)=\left(m+n\right)\left(n+p\right)\)
Tương tự: \(m^2+1=\left(m+n\right)\left(m+p\right)\)
\(p^2+1=\left(p+m\right)\left(p+n\right)\)
\(\Rightarrow\dfrac{\left(n^2+1\right)\left(p^2+1\right)}{m^2+1}=\dfrac{\left(n+p\right)^2\left(m+n\right)\left(m+p\right)}{\left(m+n\right)\left(m+p\right)}\)
\(=\left(n+p\right)^2\)
\(\Rightarrow\sqrt{\dfrac{\left(n^2+1\right)\left(p^2+1\right)}{m^2+1}}=n+p\)
Tương tự: \(\sqrt{\dfrac{\left(p^2+1\right)\left(m^2+1\right)}{n^2+1}}=m+p\)
\(\sqrt{\dfrac{\left(m^2+1\right)\left(n^2+1\right)}{p^2+1}}=m+n\)
\(\Rightarrow B=m\left(n+p\right)+n\left(m+p\right)+p\left(m+n\right)\)
\(=2\left(mn+np+pm\right)=2\)
Vậy B=2