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Giải trâu:
Xét \(A-B=\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}-\dfrac{a^{2019}-b^{2019}}{a^{2019}+b^{2019}}\)
\(=\dfrac{\left(a^{2018}-b^{2018}\right)\left(a^{2019}+b^{2019}\right)-\left(a^{2018}+b^{2018}\right)\left(a^{2019}-b^{2019}\right)}{\left(a^{2018}+b^{2018}\right)\left(a^{2019}+b^{2019}\right)}\)
\(=\dfrac{a^{4037}+a^{2018}b^{2019}-a^{2019}b^{2018}-b^{4037}-a^{4037}+a^{2018}b^{2019}-a^{2019}b^{2018}+b^{4037}}{\left(a^{2018}+b^{2018}\right)\left(a^{2019}+b^{2019}\right)}\)
\(=\dfrac{2a^{2018}b^{2019}-2a^{2019}b^{2018}}{\left(a^{2018}+b^{2018}\right)\left(a^{2019}+b^{2019}\right)}=\dfrac{2a^{2018}b^{2018}\left(b-a\right)}{\left(a^{2018}+b^{2018}\right)\left(a^{2019}+b^{2019}\right)}\)
\(\Rightarrow\)Nếu \(a>b\Rightarrow b-a< 0\Rightarrow A-B< 0\Rightarrow A< B\)
Nếu \(a< b\Rightarrow b-a>0\Rightarrow A-B>0\Rightarrow A>B\)
Sửa đề cmr a=2018 hoặc b=2018 hoặc c=2018, đây là toán 8
\(a+b+c=2018\Rightarrow\frac{1}{a+b+c}=\frac{1}{2018}\)
=>\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
<=>\(\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\Leftrightarrow\left(a+b\right)c\left(a+b+c\right)=-ab\left(a+b\right)\)
<=>\(\left(a+b\right)\left(ca+bc+c^2\right)+ab\left(a+b\right)=0\)
<=>\(\left(a+b\right)\left(ca+bc+c^2+ab\right)=0\)
<=>\(\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)
<=>\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
<=>a+b=0 hoặc b+c=0 hoặc c+a=0
Mà a+b+c=2018
=>c=2018 hoặc a=2018 hoặc b=2018 (đpcm)
\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)
\(\Rightarrow A=(1-\frac{1}{2017})+(1-\frac{1}{2018})+(1-\frac{1}{2019})\)
\(\Rightarrow A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)<\(\frac{3}{2017}\)<\(1\)
\(\Rightarrow A\)>\(3-1=2\)
\(B=\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow B=1-\frac{3}{6054}\)
\(\Rightarrow B=1-\frac{1}{2018}\)
\(B\)<\(1\);\(A\)>\(2\)
\(\Rightarrow A\)>\(B\)
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: \(=\dfrac{3}{2}\left(-21-\dfrac{1}{3}+1+\dfrac{1}{3}\right)=\dfrac{3}{2}\cdot\left(-20\right)=-30\)
b: \(=\dfrac{2018}{2019}\left(13-13-\dfrac{2018}{2019}-\dfrac{1}{2019}\right)=-\dfrac{2018}{2019}\)