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Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)
Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì \(\left(a-b=b-c\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\dfrac{\sqrt{a}-\sqrt{c}}{b-c}\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)
Câu hỏi của hoàng thị huyền trang - Toán lớp 9 - Học toán với OnlineMath
Em tham khảo nhé!
Từ giả thiết: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}\)
Xét hạng tử: \(\frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}\)
Từ đó: \(N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}\)
\(=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)
Mặt khác: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13\)
Suy ra: \(N=\frac{4}{9-13}=-1\). Kết luận: N = -1.
Từ giả thiết: \sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}a+b+c=7⇔c=7−a−b
Xét hạng tử: \frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}ab+c−61=ab+7−a−b−61=(a−1)(b−1)1
Từ đó: N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}N=(a−1)(b−1)1+(b−1)(c−1)1+(c−1)(a−1)1
=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}=(a−1)(b−1)(c−1)a+b+c−3=abc−(ab+bc+ca)+(a+b+c)−1a+b+c−3
=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=3−(ab+bc+ca)+7−17−3=9−(ab+bc+ca)4
Mặt khác: \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13ab+bc+ca=2(a+b+c)2−(a+b+c)=13
Suy ra: N=\frac{4}{9-13}=-1N=9−134=−1. Kết luận: N = -1.
- Theo BĐT Cauchy ta có:
\(\sqrt{a.1}\le\dfrac{a+1}{2}\)
\(\sqrt{b.1}\le\dfrac{b+1}{2}\)
\(\sqrt{c.1}\le\dfrac{c+1}{2}\)
\(\sqrt{ab}\le\dfrac{a+b}{2}\)
\(\sqrt{bc}\le\dfrac{b+c}{2}\)
\(\sqrt{ca}\le\dfrac{c+a}{2}\)
\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3.3+3}{2}=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Mà ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)
\(\Rightarrow a=b=c=1\)
\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)