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B=\(1+3^2+3^4+...+3^{100}\)
9B=\(3^2+3^4+...+3^{100}\)
9B-B=\(\left(3^2+3^4+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
8B=\(3^{102}-1\)
B=\(\left(3^{102}-1\right):8\)
C=\(1+5^3+5^6+...+5^{99}\)
125C=\(5^3+5^6+5^9+...+5^{102}\)
125C-C=\(\left(5^3+5^6+5^9+...+5^{102}\right)-\left(1+5^3+5^6+...+5^{99}\right)\)
124C=\(5^{102}-1\)
C=\(\left(5^{102}-1\right):124\)
Ta có:
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
\(...\)
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\) \(\left(1\right)\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
Đặt \(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{3}-\dfrac{1}{100}\)\(< \dfrac{1}{3}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
a. \(\frac{1}{5}+\frac{3}{4}+\frac{1}{10}\)
= \(\frac{4}{20}+\frac{15}{20}+\frac{2}{20}\)
= \(\frac{21}{20}\)
b. \(\frac{5}{6}-\frac{1}{3}+\frac{1}{6}\)
= \(\frac{5}{6}-\frac{2}{6}+\frac{1}{6}\)
= \(\frac{4}{6}=\frac{2}{3}\)
c. \(\frac{3}{8}-\frac{10}{2}:\frac{4}{5}\)
= \(\frac{3}{8}-\frac{50}{8}\)
= \(\frac{-47}{8}\)
a) \(\frac{1}{5}+\frac{3}{4}+\frac{1}{10}\)
= \(\frac{4+15+2}{20}\)
= \(\frac{21}{20}\)
b) \(\frac{5}{6}-\frac{1}{3}+\frac{1}{6}\)
= \(\frac{5-2+1}{6}\)
= \(\frac{4}{6}\)
c) \(\frac{3}{8}-\frac{10}{2}:\frac{4}{5}\)
= \(\frac{3}{8}-\frac{25}{4}\)
= \(-\frac{47}{8}\)
\(\frac{3}{5}-\frac{-7}{10}-\frac{13}{-20}=\frac{3}{5}+\frac{7}{10}+\frac{13}{20}=\frac{12}{20}+\frac{14}{20}+\frac{13}{20}=\frac{39}{20}\)
\(\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}-\frac{-1}{6}=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{6+(-4)+3+2}{12}=\frac{7}{12}\)
\(\frac{9}{4}.\frac{8}{27}.\frac{5}{7}=\frac{9.8.5}{4.27.7}=\frac{1.2.5}{1.3.7}=\frac{10}{21}\)
\(\frac{2}{5}.(\frac{2}{3}-\frac{1}{4})+\frac{1}{2}=\frac{2}{5}.(\frac{8}{12}-\frac{3}{12})+\frac{1}{2}=\frac{2}{5}.\frac{5}{12}+\frac{1}{2}=\frac{1}{6}+\frac{1}{2}=\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\)
\((\frac{1}{3}-\frac{1}{6}):(\frac{1}{3}+\frac{1}{6})=(\frac{2}{6}-\frac{1}{6}):(\frac{2}{6}+\frac{1}{6})=\frac{1}{6}:\frac{3}{6}=\frac{1}{6}.\frac{6}{3}=\frac{1.6}{6.3}=\frac{1.1}{1.3}=\frac{1}{3}\)
Hok tốt
1/3 - 1/4 = 4/12 - 3/12 = 1/12
2/7 . 14/5 - 1 = 2.14/7.5 - 1 =4/5 - 1 = -1/5
3/4 - 1/25 x 5 = 3/4 - 1.5/25.1 = 3/4 - 1/5 = 11/20
-8 . ( 6. 1/24 ) = -8 . ( 6.1/24.1 ) = -8 . 1/4 = -8 . 1/4.1 = -2
a) 4 + ( 5x + 2 ) : 3 = 58
( 5x + 2 ) : 3 = 58 - 4
( 5x + 2 ) : 3 = 54
( 5x + 2 ) = 54 . 3
( 5x + 2 ) = 162
5x = 162 - 2
5x = 160
x = 160 : 5
x = 32
a) x = 32.
b) x = 5.
c) a = -1;0;1. Riêng câu này thì mình chứ chắc đứng nha bạn.
câu a+b dùng quy tắc chuyển vế
c, 3.(1/2-x)-5.(x-1/10)=-7/4
=>(3.1/2-3x)-(5x-5.1/10)=-7/4
=>3/2-3x-5x+1/2=-7/4
=>(3/2+1/2)-(3x+5x)=-7/4
=> 2-8x=-7/4
=>8x=15/4
=>x=15/4:8
=>x=15/32
a) 2.(1/4 - 3x) = 1/5 - 4x
=> 1/2 - 6x = 1/5 -4x
=> -6x + 4x = 1/5 - 1/2
=> -2x = -3/10 = 3/20
b) 4.(1/3 - x) + 1/2 = 5/6 +x
=> 4/3 - 4x + 1/2 = 5/6 +x
=> -4x - x = 5/6 - 4/3 - 1/2
=> -5x = -1
=> x= 1/5
c) 3. (1/2 - x) -5. ( x - 1/10) = -7/4
=> 3/2 - 3x - 5x + 1/2 = -7/4
=> -3x - 5x = -7/4 - 3/2 - 1/2
=> -8x = -15/4
=> x = 15/32
1.714406324
tìm số tự nhiên có ba chữ số biết rằng số đó bằng thương của 1000 với tổng các chữ số của nó