Cho :
\(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(Q=\frac{1}{2019}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}+\frac{1}{2017}\)
TÍNH P VÀ Q
GIÚP MIK VỚI Ạ ~~
A=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
B=\(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}+\frac{1}{2017}\)
tính (A-B)^2017
1.So sánh:
\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2014}\) và \(4\)
2. Tính :
\(\left(1-\frac{1}{2}+\frac{1}{3}+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\)
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}\)
\(B=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\)
Tính \(\left(A^{2017}-B^{2017}\right)^{2018}\)
Ta có:
\(\Rightarrow A=B.\)
\(\Rightarrow A^{2017}=B^{2017}\)
\(\Rightarrow\left(A^{2017}-B^{2017}\right)^{2018}=\left(B^{2017}-B^{2017}\right)^{2018}=0^{2018}=0.\)
Vậy \(\left(A^{2017}-B^{2017}\right)^{2018}=0.\)
Chúc bạn học tốt!
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(B=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}+\frac{1}{2017}\)
Tính \(\left(A-B\right)^{2016^{2017}}\)
\(A=\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}+\frac{1}{2017}\right): \)\(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\text{đặt}k=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(K=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+....+\frac{1}{2017}\Rightarrow A=1\)
1. Cho A=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)và B=\(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}+\frac{1}{2018}\)Tính \(\left(\frac{A}{B}\right)^{2018}\)
2. Tìm x biết
a)\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
b)\(|x+2016|+|x+2017|+2018=3x\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Leftrightarrow x+2=41\)
\(\Leftrightarrow x=41-2\)
\(\Leftrightarrow x=39\)
???????????????????????????????????????????????????????
99% LÀ 39
CÒN LAI LÀ ĐÁP ÁN KHÁC
Cho A = 1 - \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\)
và B = \(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}+\frac{1}{2019}\)
Tính \(\text{(A - B - 1) }^{2019}\)
Ta có : \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2019}=B\)
\(\Rightarrow A-B-1=-1\)
\(\Rightarrow\left(A-B-1\right)^{2019}=-1\)
Cho A =\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{\text{4}}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
và B=\(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}+\frac{1}{2019}\)
Tính \(\left(A-B-1\right)^{2019}\)
A=(1+1/3+...+1/2019)-(1/2+1/4+...+1/2018)
A=(1+1/3+...+1/2019)+(1/2+1/4+...+1/2018)-(1/2+1/4+...+1/2018).2
A=(1+1/2+1/3+1/4+...+1/2019)-(1+1/2+...+1/1009)
A=1/1010+1/1011+...+1/2019
=) A=B
=) (A-B-1)^2019=-1
Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
B = \(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}+\frac{1}{2019}\)
Tính ( A - B - 1)2019
Ơ !!! Bài này giống bài 5 môn Toán thi cuối học kỳ 2 trường mình nè !!!
Bài làm
Ta có: \(A=\) \(1\) \(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(......\)\(+\)\(\frac{1}{2017}\)\(-\)\(\frac{1}{2018}\)\(+\)\(\frac{1}{2019}\)
\(\Rightarrow\) \(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-2\left(\frac{1}{2}+\frac{1}{4}+......+\frac{1}{2018}\right)\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2019}-\left(1+\frac{1}{2}+......+\frac{1}{1009}\right)\)
\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+......+\frac{1}{2019}\)
\(\Rightarrow A=B\)
Khi đó: (A - B - 1)2019 = -12019 = -1
Chúc bạn học tốt. K cho mk nhé! Thank you.