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\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(A=1-\frac{1}{2^9}\)
\(A=\frac{2^9-1}{2^9}\)
Vậy \(A=\frac{2^9-1}{2^9}\)
Chúc bạn học tốt ~
Câu a:
33 ⋮ (x+ 1)
(x+ 1) ∈ Ư(33) = {-33; -11; -3; -1; 1; 3; 11}
x ∈ {-34; -12; -4; -2; 0; 2; 10}
Vậy: {-34; -12; -4; -2; 0; 2; 10}
Câu b:
x ∈ ƯC(250; 48)
250 = 2.5^3; 48 = 2^4.3
ƯCLN(250; 48) = 2
x ∈ ƯC(2) = {-2; -1}
Vậy x ∈ {-2; -1}
a) Hình tròn tâm O,bán kính 3cm
a) Hình tròn tâm O,bán kính 3cm
a) Hình tròn tâm O,bán kính 3cm
a,15-3./2x-1/=6
3./2x-1/=15-6
3./2x-1/=9
2x-1=9:3=3
TH1:2x-1=3
2x=3+1=4
\(\Rightarrow x=2\)
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+...+\left(a+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Rightarrow12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)(1)
Ta có \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)=\frac{1}{2}\left(1-\frac{1}{25}\right)=\frac{1}{2}.\frac{24}{25}=\frac{12}{25}\)
Lại có \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}=\frac{3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)}{2}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}-\frac{1}{3^5}}{2}=\frac{1-\frac{1}{3^5}}{2}=\frac{1}{2}-\frac{1}{3^5.2}\)
Khi đó (1) <=> \(12a-\frac{12}{25}=11a+\frac{1}{2}-\frac{1}{3^5.2}\)
=> \(a=\frac{12}{25}+\frac{1}{2}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{486}=\frac{23764}{24300}\)
Gọi \(A=\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)\)
\(\Rightarrow A=12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{23.25}\right)\)
\(\Rightarrow A=12a+\left[\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{23.25}\right)\right]\)
\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]\)
\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{25}\right)\right]\)
\(\Rightarrow A=12a+\left(\frac{1}{2}.\frac{24}{25}\right)\)
\(\Rightarrow A=12a+\frac{12}{25}\)
Gọi \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow B=\frac{1}{1.3}+\frac{1}{3.3}+\frac{1}{9.3}+\frac{1}{27.3}+\frac{1}{81.3}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(\Rightarrow3B-B=1-\frac{1}{243}\)
\(\Rightarrow2B=\frac{242}{243}\)
\(\Rightarrow B=\frac{121}{243}\)
\(\Rightarrow A=11a+B\)
\(\Rightarrow12a+\frac{12}{25}=11a+\frac{121}{243}\)
\(\Leftrightarrow12a-11a=\frac{121}{243}-\frac{12}{25}\)
\(\Leftrightarrow a=\frac{109}{6075}\)
ƯCLN(a,b).BCNN(a,b) = ab = 3549
\(\frac{a}{b}=\frac{15}{35}=\frac{3}{7}\Rightarrow\frac{a}{3}=\frac{b}{7}\)mà \(\frac{a}{3}.\frac{b}{7}=\left(\frac{a}{3}\right)^2=\frac{ab}{21}=\frac{3549}{21}=169\)
=> \(\frac{a}{3}\in\left\{-13;13\right\}\)=> a\(\in\left\{-39;39\right\}\)=> b\(\in\left\{-91;91\right\}\).Vậy (a ; b) = (-39 ; -91);(39 ; 91)
Khi phân tích ra,anh sẽ thấy:
3549=a.b(em sẽ tắt lại)
\(\frac{a}{b}=\frac{15}{35}=\frac{3}{7}\)
a=3k
b=7p.
21.k.p=3549
k.p=169
k.p=13^2
k=p=13.
b=91
a=36.
Chúc anh học tốt^^