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1) ADTCDTSBN
có: \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-7}=\frac{x-y-z}{3-5+7}=\frac{20}{5}=4.\)
=> ...
\(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{a\left(bc+b+2018\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{ab+2018a+2018}+\frac{1}{ab+2018a+2018}\)
\(\Rightarrow M=\frac{2018a+ab+1}{2018a+ab+1}=1\)
Do : \(abc=2018\)nên : \(a,b,c\ne0\)
Ta có : \(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{abc+ab+2018a}+\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{2018+ab+2018a}+\frac{2018}{2018+ab+2018a}\)
\(=\frac{2018a+ab+2018}{ab+2018a+2018}=1\)
Câu 2: A = \(^{1+2+2^2+2^{ }^3+...+2^{2017}}\)
2A = \(2+2^2+2^3+...+2^{2018}\)
Suy ra 2A - A =\(2^{2018}-1\) Do đó A < B
1. Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=t\Rightarrow a=2016t,b=2017t,c=2018t\)
\(\left(a-c\right)^3=\left(2016t-2018t\right)^3=\left(-2t\right)^3=-8t^3\)
\(8\left(a-b\right)^2\left(b-c\right)=8\left(2016t-2017t\right)^2\left(2017t-2018t\right)=8.\left(-t\right)^2.\left(-t\right)=-8t^3\)
Vậy \(\left(a-c\right)^3=8\left(a-b\right)^2\left(b-c\right)\)
Sửa đề : Cần chứng minh \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Đặt :\(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}=k\)
\(\Rightarrow\hept{\begin{cases}a=2017k\\b=2018k\\c=2019k\end{cases}}\)
Khi đó :
\(4\left(a-b\right)\left(b-c\right)=4\left(2017k-2018k\right)\left(208k-2019k\right)\)
\(=4\cdot\left(-k\right)\cdot\left(-k\right)=4k^2\)
\(\left(c-a\right)^2=\left(2019k-2017k\right)^2=\left(2k\right)^2=4k^2\)
Do đó : \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\) (đpcm)
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
Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\Rightarrow a=2016k;b=2017k;c=2018k\)
\(\frac{a}{24}+\frac{b}{4}=\frac{c}{2018}\)
\(\Rightarrow\frac{2016k}{24}+\frac{2017k}{4}=\frac{2018k}{2018}\)
\(\Rightarrow84k+504,25k=k\)
\(\Rightarrow k=0\)
\(\Rightarrow a,b,c=0\)
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