tính giá trị các biểu thức sau:
a, A= ( 2019 - \(\frac{1}{2019}\)) - ( 2018 - \(\frac{1}{2019}\))
b,B= ( 1 - \(\frac{1}{5}\)) * ( 1 - \(\frac{1}{6}\)) *.......* ( 1 - \(\frac{1}{100}\))
các bạn gipus mình với ạ! cảm ở nhiều!!!
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Những câu hỏi liên quan
18 tháng 1 2019
\(\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}\)
\(=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}\)
\(=2+\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\left|2018x-2019\right|+1\)có giá trị nhỏ nhất
Mà \(\left|2018x-2019\right|\ge0\)
\(\Rightarrow\left|2018x-2019\right|+1\ge1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left|2018x-2019\right|=0\)
\(\Leftrightarrow x=\frac{2019}{2018}\)
Vậy \(M_{MAX}=2019\)tại \(x=\frac{2019}{2018}\)
18 tháng 1 2019
\(\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}\)
\(\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}\)
\(\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}\)
\(\Rightarrow5^x=3^{2x}\)
Mà \(\left(5;3\right)=1\)
\(\Rightarrow x=2x=0\)
XO
12 tháng 8 2019
\(A=\frac{2020}{2019}-\frac{2019}{2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018}{2019\times2018}-\frac{2019\times2019}{2019\times2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018-2019\times2019+1}{2019\times2018}\)
\(=\frac{\left(2019+1\right)\times\left(2019-1\right)-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-2019+2019-1-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-1-\left(2019\times2019-1\right)}{2019\times2018}\)
\(=\frac{0}{2019\times2018}\)
\(=0\)
Vậy A = 0
D
12 tháng 8 2019
ta có
A=2020*2018/2019*2018-2019*2019/2018*2019+1/2018*2019
=>A*(2018*2019)=2020*2018-2019*2019+1
=>A*(2018*2019)=(2019+1)*2018-(2018+1)*2019+1
=>A*(2018*2019)=(2019*2018+2018)-(2018*2019+2019)+1
=>A*(2018*2019)=2019*2018+2018-2018*2019-2019+1
=>A*(2018*2019)=2018-2019+1
=>A*(2018*2019)=2018+1-2019
=>A*(2018*2019)=0
=>A=0/(2018*2019)
=>A=0
9 tháng 10 2019
Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)
\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)
\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
S
30 tháng 10 2018
\(a+b=c+\frac{1}{2019}\Leftrightarrow a+b-c=\frac{1}{2019}\Leftrightarrow\frac{1}{a+b-c}=2019\)
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+2019\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=2019\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=\frac{1}{a+b-c}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b-c}+\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{a+b}{c\left(a+b-c\right)}\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)=\left(a+b\right)ab\)
\(\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)-ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ca+bc-c^2-ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a-c\right)-b\left(a-c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(c-b\right)\left(a-c\right)=0\)
=>a=-b hoặc c=b hoặc a=c
không mất tính tổng quát, giả sử a=-b, ta có:
\(P=\left(-b^{2019}+b^{2019}-c^{2019}\right)\left(-\frac{1}{b^{2019}}+\frac{1}{b^{2019}}-\frac{1}{c^{2019}}\right)=\left(-c\right)^{2019}\cdot\left(\frac{-1}{c}\right)^{2019}=1\)
tương tư với các trường hợp khác ta cũng có P=1
Vậy P=1
Y
16 tháng 6 2019
\(\frac{a^4}{2018}+\frac{b^4}{2019}=\frac{1}{4037}\)
\(\Leftrightarrow\frac{2019a^4+2018b^4}{2018\cdot2019}=\frac{a^2+b^2}{2018+2019}\)
\(\Leftrightarrow\left(2018+2019\right)\left(2019a^4+2018b^4\right)=2018\cdot2019\left(a^2+b^2\right)\)
\(\Leftrightarrow2019^2\cdot a^4+2018^2\cdot b^4+2018\cdot2019\cdot a^4+2018\cdot2019b^4=2018\cdot2019\cdot a^2+2018\cdot2019\cdot b^2\)
\(\Leftrightarrow2019^2\cdot a^4+2018^2\cdot b^4=2018\cdot2019\cdot a^2\left(1-a^2\right)+2018\cdot2019\cdot b^2\left(1-b^2\right)\)
\(\Leftrightarrow\left(2019a^2\right)^2+\left(2018b^2\right)^2=2\cdot2018\cdot2019\cdot a^2\cdot b^2\)
\(\Leftrightarrow\left(2019a^2-2018b^2\right)=0\)
\(\Leftrightarrow2019a^2=2018b^2\Leftrightarrow\frac{a^2}{2018}=\frac{b^2}{2019}=\frac{a^2+b^2}{2018+2019}=\frac{1}{4037}\)
\(\Rightarrow\frac{a^{2018}}{2018^{10009}}=\frac{b^{2018}}{2019^{1009}}=\frac{1}{4037^{1009}}\)
\(\Rightarrow P=\frac{2}{4037^{1009}}\)
11 tháng 5 2019
Ơ !!! Bài này giống bài 5 môn Toán thi cuối học kỳ 2 trường mình nè !!!
22 tháng 5 2019
đặt 22018 = a ; 32019 = b ; 52020 = c
Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
\(\Rightarrow A>1>\frac{3}{4}>B\)
22 tháng 5 2019
Mình chỉ biết cách tính B thôi, đây nhé:
B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)