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\(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{a\left(bc+b+2018\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{ab+2018a+2018}+\frac{1}{ab+2018a+2018}\)
\(\Rightarrow M=\frac{2018a+ab+1}{2018a+ab+1}=1\)
Do : \(abc=2018\)nên : \(a,b,c\ne0\)
Ta có : \(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{abc+ab+2018a}+\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{2018+ab+2018a}+\frac{2018}{2018+ab+2018a}\)
\(=\frac{2018a+ab+2018}{ab+2018a+2018}=1\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có
\(VT:\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{b^{2018}\cdot k^{2018}+d^{2018}\cdot k^{2018}}{b^{2018}+d^{2018}}=\frac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\)
\(VP:\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{k^{2018}\cdot\left(b+d\right)^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\)
\(\Rightarrow VT=VP\)
Hay \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\left(đpcm\right)\)
1) ADTCDTSBN
có: \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-7}=\frac{x-y-z}{3-5+7}=\frac{20}{5}=4.\)
=> ...
Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\Rightarrow a=2016k;b=2017k;c=2018k\)
\(\frac{a}{24}+\frac{b}{4}=\frac{c}{2018}\)
\(\Rightarrow\frac{2016k}{24}+\frac{2017k}{4}=\frac{2018k}{2018}\)
\(\Rightarrow84k+504,25k=k\)
\(\Rightarrow k=0\)
\(\Rightarrow a,b,c=0\)
A=a/2018-c +b/2018-a +c/2018-b
A= a/a+b + b/b+c + c/c+a
Nhận thấy: a/a+b< a/a+b+c; b/b+c<b/a+b+c; c/c+a<c/a+b+c
Do đó A= a/a+b + b/b+c + c/c+a < a/a+b+c + b/a+b+c + c/a+b+c = a+b+c/a+b+c=1
=>A>1(1)
áp dụng t/c:a/b<1=>a/b<a+n/b+n(a,b,n khác 0), ta có:
a/a+b < a+c/a+b+c ; b/b+c < b+a/b+c+a ; c/c+a < c+b/c+a+b
Do đó :A= a/a+b + b/b+c + c/c+a < a+c/a+b+c + b+a/a+b+c + c+b/a+b+c= 2(a+b+c)/a+b+c=2
=>A<2(2)
từ (1);(2)=>1<A<2=> A không thuộc Z=>ĐPCM. chúc bạn học tốt