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Sửa đề:
\(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(=2001.\dfrac{1}{10}-3\)
\(=200,1-3=197,1\)
Vậy S = 197,1
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{10}\)
\(\Rightarrow2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)=\dfrac{2017}{10}\)
\(\Rightarrow\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{a+c}=201,7\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}=201,7\)
\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{a+c}{a+c}+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow3+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=198,7\)
Ta có: \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{10}\)
\(=>2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2017}{10}\)
\(=>\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{c+a}=201,7\)
Mà 2017 = a+b+c nên ta có:
\(=>\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=201,7\)
\(=>1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)
\(=>\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=201,7-3=198,7\)
CHÚC BẠN HỌC TỐT....
1. Câu hỏi của Cuber Việt ( Câu b í -.- )
2. Quy đồng mẫu số:
\(\dfrac{a}{b}=\dfrac{a.\left(b+2018\right)}{b.\left(b+2018\right)}=\dfrac{ab+2018a}{b.\left(b+2018\right)}\)
\(\dfrac{a+2018}{b+2018}=\dfrac{\left(a+2018\right).b}{\left(b+2018\right).b}=\dfrac{ab+2018b}{b.\left(b+2018\right)}\)
Vì \(b>0\) \(\Rightarrow\) Mẫu 2 phân số ở trên dương.
So sánh \(ab+2018a\) và \(ab+2018b\):
. Nếu \(a< b\Rightarrow\) Tử số phân số thứ 1 < Tử số phân số thứ 2.
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)
. Nếu \(a=b\) \(\Rightarrow\) Hai phân số bằng 1.
. Nếu \(a>b\Rightarrow\) Tử số phân số thứ 1 > Tử số phân số thứ 2.
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)
3. \(\dfrac{x}{6}-\dfrac{1}{y}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{y}=\dfrac{x}{6}-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{y}=\dfrac{x-3}{6}\)
\(\Rightarrow y.\left(x-3\right)=6\)
Ta có: \(6=1.6=2.3=(-1).(-6)=(-2).(-3)\)
Tự lập bảng ...
Vậy ta có những cặp x,y thỏa mãn là:
\(\left(1,7\right);\left(6,2\right);\left(2,4\right);\left(3,3\right);\left(-1,-5\right);\left(-6,0\right);\left(-2,-2\right);\left(-3,-1\right)\)
\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{a\left(b+2018\right)}{b\left(b+2018\right)}\\\dfrac{a+2018}{b+2018}=\dfrac{b\left(a+2018\right)}{b\left(b+2018\right)}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{ab+2018a}{b^2+2018b}\\\dfrac{a+2018}{b+2018}=\dfrac{ab+2018b}{b^2+2018b}\end{matrix}\right.\)
Cần so sánh:
\(ab+2018a\) với \(ab+2018b\)
Cần so sánh \(2018a\) với \(2018b\)
Cần so sánh \(a\) với \(b\)
\(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2018}{b+2018}\)
\(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)
\(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2018}{b+2018}\)
\(a+b+c=2016\Rightarrow\left\{{}\begin{matrix}a=2016-\left(b+c\right)\\b=2016-\left(c+a\right)\\c=2016-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow S=\dfrac{2016-\left(b+c\right)}{b+c}+\dfrac{2016-\left(c+a\right)}{c+a}+\dfrac{2016-\left(a+b\right)}{a+b}\)\(\Rightarrow S=2016\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(\Rightarrow S=2016.\dfrac{1}{90}-3\)
\(\Rightarrow S=\dfrac{97}{2}\)
Nhân cả hai vế của đẳng thức cho a+b+c ta được
\(\dfrac{a+b+c}{a+b}\)+\(\dfrac{a+b+c}{a+b}\)=\(\dfrac{a+b+c}{c+a}\)=\(\dfrac{a+b+c}{90}\)
=> a+ \(\dfrac{c}{a+b}\)+1+\(\dfrac{a}{b+c}\)+1+\(\dfrac{b}{c+a}\)=\(\dfrac{2007}{90}\)
=>\(\dfrac{a}{b+c}\)+\(\dfrac{b}{c+a}\)+\(\dfrac{c}{a+b}\)=\(\dfrac{2007}{90}\)-3= 22,3-3=19,3