Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)..........\left(\frac{2017.2019+1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.............\frac{4072324}{2017.2019}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...................\frac{2018^2}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{\left(2.3.4..........2018\right).\left(2.3.4............2018\right)}{\left(1.2.3............2017\right).\left(3.4.5..........2019\right)}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2018.2}{1.2019}\right)=\frac{2018.2}{2.2019}=\frac{2018}{2019}\)
Vậy \(A=\frac{2018}{2019}\)
Chúc bn học tốt
\(A:\frac{1}{2}=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{2018^2}{2017.2019}\)
\(=\frac{2.2.3.3.4.4.....2018.2018}{1.3.2.4.3.5....2017.2019}\)
\(=\frac{2.3.4.....2018}{1.2.3.4.....2017}.\frac{2.3.4....2018}{3.4.5.....2019}\)
\(=2018.\frac{2}{2019}\)
\(=\frac{4036}{2019}\)
\(\Rightarrow A=\frac{4036}{2019}.\frac{1}{2}\)
\(A=\frac{2018}{2019}\)
\(A=\frac{1}{1.2}-x+\frac{1}{2.3}-x+...+\frac{1}{100.101}-x+100x\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}-100x+100x\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Thay x = 2016 vào biểu thức B ta có :
\(B=\frac{2016}{1.2}+\frac{2016}{2.3}+\frac{2016}{3.4}+...+\frac{2016}{2015.2016}\)
\(B=2016\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)
\(B=2016\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(B=2016\left(1-\frac{1}{2016}\right)\)
\(B=2016.\frac{2015}{2016}\)
\(B=2015\)
GTLN là 2016 nha bạn
Số hạng đầu tiên có [1/4-2n]2 luôn dương
=>-1/5[1/4-2n]2 luôn âm
..........
Quốc Huy phải giải rõ ra chứ.Như mình nè:
Ta có:[1/4-2n]^2>=0
suy ra;-1/5[1/4-2n]<=0 (1)
Lại có:|8x-1|>=0
suy ra : -|8x-1|<=0 (2)
Từ (1) và (2) suy ra:-1/5[1/4-2n]^2-|8x-1|<=0
suy ra:-1/5[1/4-2n]^2-|8x-1|+2016 <=2016
suy ra D<=2016
suy ra giá trị lớn nhất của D là 2016 khi 1/4-2n=0 và 8x-1=0
*Với 1/4-2n=0 suy ra 2n=1/4 suy ra n=1/4:2=1/4.1/2 suy ra n=1/8
*Với 8x-1=0 suy ra 8x=1 suy ra x=1/8
Vậy giá trị lớn nhất của D là 2016 khi n=1/8 và x=1/8
\(\left(-2\right).\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2013}\right)\)
\(=\left(-2\right).\left(\frac{-3}{2}\right)\left(-\frac{4}{3}\right)......\left(\frac{-2014}{2013}\right)\)
\(=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2014\right)}{2.3.....2013}\)
\(=\frac{2.3.4....2014\left(\text{Vì có 2014 thừa số âm }\right)}{2.3....2013}\)
\(=\frac{\left(2.3.4....2013\right).2014}{2.3....2013}\)
\(=2014\)
cái nài toán lớp 6 nha em
A=1/1-1/2+1/2-1/3+....+1/n-1/n+1
A=1-1/n+1
A=n/n+1
\(A=\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
\(A=\frac{n}{n+1}\)
~~~~~Hok tốt~~~~~
\(B=\left(-13\cdot\frac{2}{5}+\frac{-2}{9}\div2\frac{1}{2}+\frac{2}{5}\cdot\frac{11}{9}\right)\cdot2\frac{1}{2}\)
\(B=\left(13\cdot\frac{2}{5}+\frac{-2}{9}\cdot\frac{2}{5}+\frac{2}{5}\cdot\frac{11}{9}\right)\cdot\frac{5}{2}\)
\(B=\left[\frac{2}{5}\cdot\left(-13+\frac{-2}{9}+\frac{11}{9}\right)\right]\cdot\frac{5}{2}\)
\(B=\left[\frac{2}{5}\cdot\left(-12\right)\right]\cdot\frac{5}{2}\)
\(B=\frac{-24}{5}\cdot\frac{5}{2}=-12\)