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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 :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
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) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=676\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=676\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=673\)
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= \(\left(\dfrac{a}{b+c}+1\right)\)+\(\left(\dfrac{b}{a+c}+1\right)\)+\(\left(\dfrac{c}{a+b}+1\right)\)-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
Ta có :
\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}:\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}\cdot\dfrac{2}{1}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{2}{c}\)
\(\Rightarrow\dfrac{b}{ab}+\dfrac{a}{ab}=\dfrac{2}{c}\)
\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{2}{c}\)
\(\Rightarrow2ab=\left(a+b\right)c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
Vậy \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)
VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)
Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)
\(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\\ \Rightarrow P+3=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\\ \Rightarrow P+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\\ =\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=2018.\dfrac{2021}{4034}=1011.000992\\ \Rightarrow P=1008.000992\)