Cho A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
a, Tính A
b, So sánh A với 1
Giúp với , mik cần gấp
Bài 1
tính: \(\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{10}}\)
Bài 2
So sánh A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2017^2}\)
Với 1
Mau mau giúp mik nkoa chiều mai mik nộp r ^^
Bài 1
Nhân 2 vào biểu thức
Rút gọn và trừ đi 1 lần nó
còn lại \(\frac{1}{2}_{ }-\frac{1}{2^{10}}\)
\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{10}}\)
cho A=1+\(\frac{1}{3}\)+\(\frac{1}{5}\)+.....+\(\frac{1}{2017}\)
B=\(\frac{1010}{1.2017}+\frac{1010}{2.2016}+...+\frac{1010}{2017.1}\)
so sánh A và B
giúp mik với, mik tick cho, đang cần gấp
Bài 1.So sánh A và B biết: A=\(\frac{10^{17}+1}{10^{18}+1}\) B=\(\frac{10^{18}+1}{10^{19}+1}\)
Bài 2.So sánh S=\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)với 4
Bài 3.Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)Chứng minh rằng A<\(\frac{3}{4}\)
Bài 4.
a)Tính nhanh tổng sau:A=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
b)Tìm x biết:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1008}{2017}\)
mn giúp mk nha mk đang cần gấp
ai nhanh mk sẽ tick cho
tk mn
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
1. Tìm x:
x(x - 2) + 3(x+ 5) + 4x - 15 = 0
2. Cho a + b + c = 2017 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=2017\)
Tính S = \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Giúp mik nha, mik đang cần gấp. Thanks!!!
1) x(x-2) + 3(x+5) + 4x -15 =0
=> x\(^2\) - 2x + 3x + 15 + 4x - 15 = 0
=> ( x\(^2\) -2x + 3x + 4x ) + 15 - 15 = 0
=> x \(^2\) -2x+3x+4x = 0
=> x(x-2+3+4)=0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2+3+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}}\)
2) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=2017\)
\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=2017.2017\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=2017^2\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}=2017^2\)
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{c}{a+b}\right)=2017^2\)
\(\Rightarrow\left(1+\frac{c}{a+b}\right)+\left(1+\frac{a}{b+c}\right)+\left(1+\frac{c}{a+b}\right)=2017^2\)
\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2-3\)
xin lỗi mik xin đc sửa lại 3 dòng cuối vì mik ghi nhầm :
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{b}{a+c}\right)=2017^2\)
\(\Rightarrow\left(1+\frac{c}{a+b}\right)+\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{a+c}\right)=2017^2\)
\(\Rightarrow3+\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=2017^2\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2-3\)
Cho A = \(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)
So sánh A với \(\frac{-1}{2}\)
Giúp gấp vs!!!!
\(A=\frac{-3}{4}.\frac{-8}{9}......\frac{-9999}{1000}\)
\(=-\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{99.101}{100.100}\)
\(=-\frac{1.2.3...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)
\(=-\frac{1}{100}.\frac{101}{2}=-\frac{101}{200}< \frac{-100}{200}=\frac{-1}{2}\)
VẬY \(A< \frac{-1}{2}\)
cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2016^2}-1\right)\left(\frac{1}{2017^2}-1\right)\)và b=-1/2
Hãy so sánh A với B
Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)
\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)
\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)
\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)
\(A=\frac{1.2018}{2017.2}\)
\(A=\frac{1009}{2017}\)
Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)
\(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)
Vậy A>B
Thực hiện phép tính:
a, A= \(\frac{1}{6}.\left(-2\frac{3}{5}\right)+1\frac{2}{3}.\left(\frac{-13}{5}\right)\)
b, B=\(\frac{5}{7}:\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{5}{7}:\left(\frac{1}{5}-\frac{1}{6}\right)\)
c, C= \(2^3+3.\left(\frac{2017}{2018}\right)^0.\left(\frac{1}{2}\right)^{-2}:4+[\left(-2\right)^2:\frac{1}{2}].\left(\frac{-1}{2}\right)^3\)
giúp vs, mik cần gấp
A= E387E4837
B = 883433
C = UỲUWFHQWURY48E3947
Cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4034},B=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{4033}\). So sánh A/B với \(1\frac{2017}{2018}\)
Cho D=\(\frac{\left(2!\right)^2}{1^2}+\frac{\left(2!\right)^2}{3^2}+...+\frac{\left(2!\right)^2}{2015^2}\)
So sánh D với 6
Mình đang cần gấp. Giúp MIk nha!