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\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
Lời giải:
\(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2019.2020}\)
\(\Rightarrow 2B=\frac{2}{1.2}+\frac{2}{3.4}+\frac{2}{5.6}+....+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\( 2B< 1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
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Đặt \(2^{2018}=a; 3^{2019}=b; 5^{2020}=c(a,b,c>0)\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}> \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(\Rightarrow A>1> \frac{3}{4}> B\)
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
+\(A>1\)
Ta có :\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2018^2}+\frac{1}{2019^2}>1\) 1
+\(A< 2\)
Ta có:\(1=1\)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...........................
\(\frac{1}{2019^2}< \frac{1}{2018.2019}\)
\(\Rightarrow A< 1+1-\frac{1}{2019}=2-\frac{1}{2019}< 2\)2
Từ 1 và 2 => A không là số tự nhiên
\(a,\frac{21}{36}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(2018-2019\right)^0\)
=\(\frac{7}{12}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(-1\right)\)
= \(\frac{7}{12}.\left(\frac{5}{2}+\frac{2}{7}\right)+\left(-1\right)\)
=\(\frac{7}{12}.\frac{39}{14}+\left(-1\right)\)
=\(\frac{13}{8}+\left(-1\right)\)
= \(\frac{5}{8}\)
\(b,-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)
=\(-\frac{37}{3}+\frac{-5}{7}+\frac{22}{3}+\frac{12}{7}+1\)
=\(\left(\frac{-37+22}{3}\right)+\left(\frac{-5+12}{7}\right)+=1\)
= \(-5+1+1\)
=\(-3\)
a)
⇒ \(\frac{11x-1}{4}=\frac{10}{4}\)
⇒ 11x - 1 = 10
11x = 10 + 1 = 11
x = 11 : 11 = 1
b)
\(\left[{}\begin{matrix}3x-6=0\\\frac{x}{9}-\frac{1}{3}=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3x=0+6\\\frac{x}{9}=0+\frac{1}{3}\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}3x=6\\\frac{x}{9}=\frac{1}{3}\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}x=6:3\\\frac{x}{9}=\frac{3}{9}\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy x = 2 hoặc x = 3
c)
\(M=c\left(\frac{5}{7}+\frac{7}{14}-\frac{17}{14}\right)\)
\(M=c\left(\frac{10}{14}+\frac{7}{14}-\frac{17}{14}\right)\)
\(M=\left(\frac{2018}{2019}-\frac{2019}{2020}\right).0\)
M = 0
d)
\(N=\frac{-7}{13}+2-\frac{19}{13}+\frac{2020}{2018}.\frac{2018}{202}\)
\(N=\left(\frac{-7}{13}-\frac{19}{13}\right)+2+10\)
N = \(-2+2+10\)
N = 10
1 < S < 2
=> S ko phải là số tự nhiên
1< S< 2
=> S không phải số tự nhiên