Tìm tất cả các số nguyên dương a,b,c đôi một khác nhau sao cho biểu thức :\(A=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)nhận giá trị nguyên dương
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Ez còn nhờ :
Để \(A=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\) nghuyên \(\Leftrightarrow\hept{\begin{cases}1⋮a\text{ };\text{ }1⋮b\text{ };\text{ }1⋮c\\1⋮ab\text{ };\text{ }1⋮bc\text{ };\text{ }1⋮ac\end{cases}}\)
\(\Rightarrow\left(a;b;c\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)Mà a;b;c nguyên dương \(\Rightarrow\left(a;b;c\right)=\left(1;1;1\right)\)
Vậy \(\left(a;b;c\right)=\left(1;1;1\right)\)
1) Ta có : \(\frac{2016a+b+c+d}{a}=\frac{a+2016b+c+d}{b}=\frac{a+b+2016c+d}{c}=\frac{a+b+c+2016d}{d}\)
Trừ 4 vế với 2015 ta được : \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu a + b + c + d = 0
=> a + b = -(c + d)
=> b + c = (-a + d)
=> c + d = -(a + b)
=> d + a = (-b + c)
Khi đó M = (-1) + (-1) + (-1) + (-1) = - 4
Nếu a + b + c + d\(\ne0\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)
Khi đó M = 1 + 1 + 1 + 1 = 4
2) a) Ta có : \(\hept{\begin{cases}\left|x+2013\right|\ge0\forall x\\\left(3x-7\right)^{2004}\ge0\forall y\end{cases}\Rightarrow\left|x+2013\right|+\left(3x-7\right)^{2014}\ge0}\)
Dấu "=" xảy ra \(\hept{\begin{cases}x+2013=0\\3y-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2013\\y=\frac{7}{3}\end{cases}}}\)
b) 72x + 72x + 3 = 344
=> 72x + 72x.73 = 344
=> 72x.(1 + 73) = 344
=> 72x = 1
=> 72x = 70
=> 2x = 0 => x = 0
c) Ta có :
\(\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{5}{x+4}\Leftrightarrow\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{10}{2x+8}=\frac{7-10}{2x+2-2x-8}=\frac{1}{2}\)(dãy tỉ số bằng nhau)
=> 2x + 2 = 14 => x = 6 ;
2y - 4 = 6 => y = 5 ;
6 + 5 + z = 17 => z = 6
Vậy x = 6 ; y = 5 ; z = 6
3) a) Ta có : \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)(dãy ti số bằng nhau)
=> a + b + c = a + b - c => a + b + c - a - b + c = 0 => 2c = 0 => c = 0;
Lại có : \(\frac{a+b+c}{a+b-c}-1=\frac{a-b+c}{a-b-c}-1\Leftrightarrow\frac{2c}{a+b-c}=\frac{2c}{a-b-c}\Rightarrow a+b-c=a-b-c\) => b = 0
Vậy c = 0 hoặc b = 0
c) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b+b+c+a+c}{c+a+b}=2\)(dãy tỉ số bằng nhau)
=> \(\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)
Khi đó P = \(\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{b}{a}\right)=\frac{b+c}{b}.\frac{c+a}{c}=\frac{a+b}{a}=\frac{2a.2b.2c}{abc}=8\)
Vậy P = 8
2. b) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
\(A=\frac{ab}{a+c+b+c}+\frac{bc}{a+b+a+c}+\frac{ca}{a+b+b+c}\)
\(\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{a+b}+\frac{ca}{b+c}\right)\)
\(=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)
Nên max A là \(\frac{1}{4}\) khi \(a=b=c=\frac{1}{3}\)
Dễ dàng chứng minh được:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với \(x,y>0\)(1)
Dấu bằng xảy ra \(\Leftrightarrow x=y>0\)
Ta có:
\(\frac{a}{bc\left(a+1\right)}=\frac{a}{abc+bc}=\frac{a}{ab+bc+ca+bc}=\frac{a}{\left(ab+bc\right)+\left(bc+ca\right)}\)
Áp dụng (1), ta được:
\(\frac{1}{ab+bc}+\frac{1}{bc+ca}\ge\frac{4}{\left(ab+bc\right)+\left(bc+ca\right)}\)
\(\Leftrightarrow\frac{1}{4\left(ab+bc\right)}+\frac{1}{4\left(bc+ca\right)}\ge\frac{1}{ab+bc+bc+ca}\)
\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{ab+bc+bc+ca}\)
\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{bc\left(a+1\right)}\left(2\right)\)
Dấu bằng xảy ra \(\Leftrightarrow b=c>0\)
Chúng minh tương tự, ta được:
\(\frac{b}{4}\left(\frac{1}{ab+ca}+\frac{1}{bc+ca}\right)\ge\frac{b}{ca\left(b+1\right)}\left(3\right)\)
Dấu bằng xảu ra \(\Leftrightarrow a=c>0\).
\(\frac{c}{4}\left(\frac{1}{ac+ab}+\frac{1}{ab+bc}\right)\ge\frac{c}{ab\left(c+1\right)}\left(4\right)\)
Từ (2), (3) và (4), ta được:
\(\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\le\)\(\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ac}\right)+\frac{b}{4}\left(\frac{1}{ac+bc}+\frac{1}{ac+ab}\right)\)\(+\frac{c}{4}\left(\frac{1}{ab+bc}+\frac{1}{ab+ac}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}.\left(\frac{a}{ab+bc}+\frac{c}{ab+bc}\right)+\frac{1}{4}\left(\frac{a}{bc+ac}+\frac{b}{bc+ac}\right)\)\(+\frac{1}{4}\left(\frac{b}{ab+ac}+\frac{c}{ab+ac}\right)\)
\(\Leftrightarrow P\le\frac{a+c}{4\left(ab+bc\right)}+\frac{a+b}{4\left(bc+ac\right)}+\frac{b+c}{4\left(ab+ac\right)}\)
\(\Leftrightarrow P\le\frac{a+c}{4b\left(a+c\right)}+\frac{a+b}{4c\left(a+b\right)}+\frac{b+c}{4a\left(b+c\right)}\)
\(\Leftrightarrow P\le\frac{1}{4b}+\frac{1}{4c}+\frac{1}{4a}\)
\(\Leftrightarrow P\le\frac{1}{4}\left(\frac{ab+bc+ca}{abc}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}.\frac{abc}{abc}=\frac{1}{4}.1=\frac{1}{4}\)( vì \(ab+bc+ca=abc\))
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\ab+bc+ca=abc\end{cases}}\Leftrightarrow a=b=c=3\)
Vậy \(minP=\frac{1}{4}\Leftrightarrow a=b=c=3\)
Áp dụng giả thiết từ đề bài :
\(M=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(\Leftrightarrow M=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\)
\(\Leftrightarrow M=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)
\(\Leftrightarrow M=\frac{1+b+bc}{b+1+bc}=1\)
Vậy M = 1