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\(a+b+c=2016\Rightarrow\left\{{}\begin{matrix}a=2016-\left(b+c\right)\\b=2016-\left(c+a\right)\\c=2016-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow S=\dfrac{2016-\left(b+c\right)}{b+c}+\dfrac{2016-\left(c+a\right)}{c+a}+\dfrac{2016-\left(a+b\right)}{a+b}\)\(\Rightarrow S=2016\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(\Rightarrow S=2016.\dfrac{1}{90}-3\)
\(\Rightarrow S=\dfrac{97}{2}\)
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
A= \(\left(\dfrac{a}{b+c}+1\right)\)+\(\left(\dfrac{b}{a+c}+1\right)\)+\(\left(\dfrac{c}{a+b}+1\right)\)-3
= \(\dfrac{a+b+c}{b+c}\)+\(\dfrac{a+b+c}{a+c}\)+ \(\dfrac{c+a+b}{a+b}\) -3
= (a+b+c). (\(\dfrac{1}{b+c}\) + \(\dfrac{1}{a+c}\) + \(\dfrac{1}{a+b}\)) -3
= 2016. 1-3=2013
b) Tìm min
\(SV=\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|\)
\(SV=\left|x-2016\right|+\left|2018-x\right|+\left|x-2017\right|\)
\(SV\ge\left|x-2016+2018-x\right|+\left|x-2017\right|=2+\left|x-2017\right|\ge2\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}2016\le x\le2018\\x=2017\end{matrix}\right.\Leftrightarrow x=2017\)
3) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=676\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=676\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=673\)
Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0
2/ Ta có :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
Làm lại cho you đây -_- vừa nãy bấm mt nhầm,đời t nhọ vãi
1)\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+....+16\right)\)
\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+\dfrac{1+2+3+4}{4}+...+\dfrac{1+2+3+...+16}{16}\)
Xét thừa số tổng quát: \(\dfrac{1+2+3+...+t}{t}=\dfrac{\left[\left(t-1\right):1+1\right]:2.\left(t+1\right)}{t}=\dfrac{\dfrac{t}{2}\left(t+1\right)}{t}=\dfrac{\dfrac{t^2}{2}+\dfrac{t}{2}}{t}=\dfrac{t\left(\dfrac{t}{2}+\dfrac{1}{2}\right)}{t}=\dfrac{t}{2}+\dfrac{1}{2}\)
Như vậy: \(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)
\(P=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+....+\dfrac{17}{2}\)
\(P=\dfrac{2+3+4+5+...+17}{2}\)
\(P=\dfrac{152}{2}=76\)
2) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)
\(\Rightarrow2016\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{2016}{a+b}+\dfrac{2016}{b+c}+\dfrac{2016}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2016}{3}-1-1-1=\dfrac{2007}{3}\)
Ta có: a+b-c/c = b+c-a/a = c+a-b/b = a+b-c+b+c-a+c+a-b/c+a+b
= a+b+c/a+b+c = 1 (Áp dụng tính chất dãy tỉ số bằng nhau)
Trường hợp 1 : Nếu a+b+c = 0 => a=0; b=0 ; c=0 => P =1
Trường hợp 2: Nếu a+b+c khác 0 => a+b+c = 1
=> a+b = 1-c => b+c = 1-a
=> a+c = 1-b
Ta lại có:
1-c-c/c =1 => 1- 2c/c =1 => 1-2c = c => 1 = 3c=> c= 1/3
1-a-c/a = 1 => 1- 2a/a=1 => 1-2a =a => 1 = 3a => a= 1/3
1-b-b/b = 1 => 1-2b/b = 1 => 1-2b = b => 1= 3b => b= 1/3
=> P= (1+ 1/3 : 1/3). (1+ 1/3 : 1/3). ( 1+ 1/3 :1/3)
= 2 . 2. 2 =8
Vậy P = 1 hoặc = 8
Câu 1:
\(\frac{a^{2016}+b^{2016}}{c^{2016}+d^{2016}}=\frac{a^{2016}-b^{2016}}{c^{2016}-d^{2016}}\)
\(\Rightarrow (a^{2016}+b^{2016})(c^{2016}-d^{2016})=(a^{2016}-b^{2016})(c^{2016}+d^{2016})\)
\(\Leftrightarrow 2(bc)^{2016}=2(ad)^{2016}\Rightarrow (bc)^{2016}=(ad)^{2016}\)
\(\Rightarrow (\frac{a}{b})^{2016}=(\frac{c}{d})^{2016}\)
\(\Rightarrow \frac{a}{b}=\pm \frac{c}{d}\) (đpcm)
Câu 2:
Nếu $a+b+c+d=0$ thì: \(\left\{\begin{matrix} a+b=-(c+d)\\ b+c=-(d+a)\\ c+d=-(a+b)\\ d+a=-(b+c)\end{matrix}\right.\)
\(\Rightarrow M=(-1)+(-1)+(-1)+(-1)=-4\)
Nếu $a+b+c+d\neq 0$
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{5(a+b+c+d)}{a+b+c+d}=5\)
\(\Rightarrow \left\{\begin{matrix} 2a+b+c+d=5a\\ a+2b+c+d=5b\\ a+b+2c+d=5c\\ a+b+c+2d=5d\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} b+c+d=3a(1)\\ a+c+d=3b(2)\\ a+b+d=3c(3)\\ a+b+c=3d(4)\end{matrix}\right.\)
Từ \((1);(2)\Rightarrow b+a+2(c+d)=3(a+b)\Rightarrow c+d=a+b\)
\(\Rightarrow \frac{a+b}{c+d}=1\)
Tương tự: \(\frac{b+c}{d+a}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\Rightarrow M=1+1+1+1=4\)
theo bài ra ta có:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{`1}{4}\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{1}{4}\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{a+b+c}{4}\)
\(\Rightarrow1+\dfrac{c}{a+b}+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1=\dfrac{2016}{4}\)
\(\Rightarrow\left(1+1+1\right)+\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=504\)
\(\Rightarrow3+\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=504\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=504-3\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=501\)
vậy \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=501\)
(a+b+c)(1/a+b+1/b+c+1/c+a)=(a+b+c)/4
(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)=(a+b+c)/4
=> 1+c/(a+b)+1+a/(b+c)+1+b/(c+a)=2016/4
<=>c/(a+b)+a/(b+c)+b/(c+a)+3=504
=> A=a/(b+c)+b/(c+a)+c/(a+b)=504-3=501