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Lời giải:
Ta có: \(\sqrt{a-c}+\sqrt{b-c}=\sqrt{a+b}\)
\(\Rightarrow (\sqrt{a-c}+\sqrt{b-c})^2=a+b\)
\(\Leftrightarrow a-c+b-c+2\sqrt{(a-c)(b-c)}=a+b\)
\(\Leftrightarrow \sqrt{(a-c)(b-c)}=c\)
Bình phương hai vế: \(c^2=(a-c)(b-c)\)
\(\Leftrightarrow ab=ac+bc(*)\)
----------------------------
Ta có: \(P=\frac{bc}{a^2}+\frac{ac}{b^2}-\frac{ab}{c^2}\)
\(P=\frac{(bc)^3+(ac)^3-(ab)^3}{(abc)^2}\)
Xét tử số kết hợp với $(*)$
\((bc)^3+(ac)^3-(ab)^3=(bc+ac)^3-3bc.ac(bc+ac)-(ab)^3\)
\(=(ab)^3-3bc.ac.ab-(ab)^3=-3(abc)^2\)
Do đó: \(P=\frac{-3(abc)^2}{(abc)^2}=-3\)
ta có \(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4a+4\sqrt{abc}\)
=> \(4a+4\sqrt{abc}=16-4b-4c\Leftrightarrow4a+4\sqrt{abc}+bc=16-4b-4c+bc\)
=> \(\left(2\sqrt{a}+\sqrt{bc}\right)^2=\left(4-b\right)\left(4-c\right)\Rightarrow a\left(4-b\right)\left(4-c\right)=a\left(2\sqrt{a}+\sqrt{bc}\right)^2\)
=> \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a}\left(2\sqrt{a}+\sqrt{bc}\right)=2a+\sqrt{abc}\)
tương tự như thế thay vào , thì A=8
Ta có:
\(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4c+4\sqrt{abc}\)
\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\Leftrightarrow4a+4\sqrt{abc}+bc=16-4b-4c+bc\)
\(\Rightarrow\left(2\sqrt{a}+\sqrt{bc}\right)^2=\left(4-b\right)\left(4-c\right)\Rightarrow a\left(4-b\right)\left(4-c\right)=a\left(2\sqrt{a}+\sqrt{bc}\right)^2\)
\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a}\left(2\sqrt{a}+\sqrt{bc}\right)=2a+\sqrt{abc}\)
Tương tự như thế thay vào, thì A = 8
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.
Áp dụng bất đẳng thức Cô-si, ta được: \(P=\frac{bc\sqrt{a-1}+ca\sqrt{b-4}+ab\sqrt{c-9}}{abc}\)\(=\frac{bc\sqrt{\left(a-1\right).1}+\frac{1}{2}ca\sqrt{4.\left(b-4\right)}+\frac{1}{3}ab\sqrt{9.\left(c-9\right)}}{abc}\)\(\le\frac{bc.\frac{\left(a-1\right)+1}{2}+\frac{1}{2}ca.\frac{4+\left(b-4\right)}{2}+\frac{1}{3}ab.\frac{9+\left(c-9\right)}{2}}{abc}\)\(=\frac{\frac{1}{2}abc+\frac{1}{4}abc+\frac{1}{6}abc}{abc}=\frac{\frac{11}{12}abc}{abc}=\frac{11}{12}\)
Đẳng thức xảy ra khi a = 2; b = 8; c = 18
- 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: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)
\(\Rightarrow a=b=c=1\)
\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)
chờ bạn trả lời xong thì tui nghĩ ra hết chục bài thế rùi