S= 2/3.4/5......4046/4047 so sánh S mũ 2 với 1/2024
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Bài 1:
a: Trên cùng một nửa mặt phẳng bờ chứa tia Ox, ta có: \(\widehat{xOy}< \widehat{xOz}\) nên tia Oy nằm giữa hai tia Ox và Oz
=>\(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\)
=>\(\widehat{yOz}+60^0=90^0\)
=>\(\widehat{yOz}=30^0\)
b: \(\widehat{x'Oz}+\widehat{xOz}=180^0\)(hai góc kề bù)
=>\(\widehat{x'Oz}+90^0=180^0\)
=>\(\widehat{x'Oz}=90^0\)
\(\widehat{x'Oy}+\widehat{xOy}=180^0\)(hai góc kề bù)
=>\(\widehat{x'Oy}+60^0=180^0\)
=>\(\widehat{x'Oy}=120^0\)
Bài 5:
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2024}{2025}\)
=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2024}{2025}\)
=>\(1-\dfrac{1}{x+1}=\dfrac{2024}{2025}\)
=>\(\dfrac{1}{x+1}=\dfrac{1}{2025}\)
=>x+1=2025
=>x=2024
\(C=\dfrac{4}{3\cdot5}+\dfrac{4}{5\cdot7}+\dots+\dfrac{4}{97\cdot99}\)
\(=2\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dots+\dfrac{2}{97\cdot99}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dots+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(=2\cdot\dfrac{32}{99}=\dfrac{64}{99}\)
\(D=\dfrac{18}{2\cdot5}+\dfrac{18}{5\cdot8}+\dots+\dfrac{18}{203\cdot206}\)
\(=6\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dots+\dfrac{3}{203\cdot206}\right)\)
\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dots+\dfrac{1}{203}-\dfrac{1}{206}\right)\)
\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)
\(=6\cdot\dfrac{51}{103}=\dfrac{306}{103}\)
Khi đó: \(\dfrac{C}{D}=\dfrac{\dfrac{64}{99}}{\dfrac{306}{103}}=\dfrac{3296}{15147}\)
mik làm bài 1 thôi nha,bài 1
a,có 4 kết quả có thể xảy ra đó là xanh,đỏ,tím,vàng
b,có 4 kết quả đó là lấy đc bóng vàng với vàng ,đỏ với đỏ ,xanh với xanh, tím với tím
tiếp bài 2
a, kết quả có thể xảy ra là ném bóng vô rổ và ném ko vô rổ
b,vì Hùng ném vô rổ đc 35 lần vậy số lần ném ko vô rổ là :
100 - 35 =65 ( lần )
xác xuất thực nghiệm của ném vô rổ là :
35 :100 =35/100
xác xuất thực nghiệm của ném ko vô rổ là :
65 :100=65/100
rùi nha mik giúp đến đây thôi
Câu 1:
a) A= \(\dfrac{-7}{13}.\dfrac{2}{11}+\dfrac{-7}{13}.\dfrac{9}{11}+1\dfrac{7}{13}\) b) B= \(2023\dfrac{1}{4}-25\%.\left(-1\right)^{2020}-1,8:\dfrac{3}{5}\)
A= \(\dfrac{-7}{13}.\dfrac{2}{11}+\dfrac{-7}{13}.\dfrac{9}{11}+\dfrac{20}{13}\) B=\(2023+\dfrac{1}{4}-\dfrac{25}{100}.1-1,8.\dfrac{5}{3}\)
A= \(\dfrac{-7}{13}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{20}{13}\) B=\(\dfrac{2023}{1}+\dfrac{1}{4}-\dfrac{1}{4}.1-\dfrac{9}{5}.\dfrac{5}{3}\)
A= \(\dfrac{-7}{13}-\dfrac{11}{11}+\dfrac{20}{13}\) B=\(\dfrac{2023}{1}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{3.1}{1.1}\)
A= \(\dfrac{-7}{13}-1+\dfrac{20}{13}\) B= \(\dfrac{2023}{1}+\dfrac{1}{4}-\dfrac{1}{4}-3\)
A= \(\dfrac{-20}{13}+\dfrac{20}{13}\) B=\(\dfrac{8093}{4}-\dfrac{1}{4}-3\)
A= 0 B= 2023 -3
B=2020
Câu 2:
a) \(\left|2x-\dfrac{4}{3}\right|.\dfrac{4}{15}-75\%=9\dfrac{1}{4}\)
\(\left|2x-\dfrac{4}{3}\right|.\dfrac{4}{15}-\dfrac{75}{100}=\dfrac{37}{4}\)
\(\left|2x-\dfrac{4}{3}\right|.\dfrac{4}{15}\) \(=\dfrac{37}{4}+\dfrac{75}{100}\)
\(\left|2x-\dfrac{4}{3}\right|.\dfrac{4}{15}\) \(=\dfrac{37}{4}+\dfrac{3}{4}\)
\(\left|2x-\dfrac{4}{3}\right|.\dfrac{4}{15}\) \(=10\)
\(2x-\dfrac{4}{3}\) \(=10:\dfrac{4}{15}\)
\(2x-\dfrac{4}{3}\) \(=\dfrac{10}{1}.\dfrac{15}{4}\)
\(2x-\dfrac{4}{3}\) \(=\dfrac{75}{2}\)
\(2x\) \(=\dfrac{75}{2}+\dfrac{4}{3}\)
\(2x\) = \(\dfrac{225}{6}+\dfrac{8}{6}\)
2x = \(\dfrac{233}{6}\)
x = \(\dfrac{233}{6}:2\)
x =\(\dfrac{233}{6}.\dfrac{1}{2}\)
x = \(\dfrac{233}{12}\)
Vậy x = ..........
\(10A=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
\(10B=\dfrac{10^{10}+10}{10^{10}+1}=1+\dfrac{9}{10^{10}+1}\)
\(10^{11}+1>10^{10}+1\)
=>\(\dfrac{9}{10^{11}+1}< \dfrac{9}{10^{10}+1}\)
=>\(\dfrac{9}{10^{11}+1}+1< \dfrac{9}{10^{10}+1}+1\)
=>10A<10B
=>A<B
A = \(\dfrac{10^{10}+1}{10^{11}+1}\) < \(\dfrac{10^{10}+1+9}{10^{11}+1+9}\) = \(\dfrac{10^{10}+10}{10^{11}+10}\) = \(\dfrac{10.\left(10^9+1\right)}{10.\left(10^{10}+1\right)}\) = B
Vậy A < B
Lời giải:
$A=\frac{13}{20}+\frac{8}{21}+\frac{17}{22}< \frac{13}{20}+\frac{8}{20}+\frac{17}{20}=\frac{38}{20}< \frac{40}{20}=2$
Ta có đpcm.
\(M=\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+...+\dfrac{1}{n\left(n+4\right)}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{n\left(n+4\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{n+4}\right)=\dfrac{1}{4}\cdot\dfrac{n+4-1}{n+4}=\dfrac{n+3}{4\left(n+4\right)}\)
\(S=\dfrac{2}{3}\times\dfrac{4}{5}\times...\times\dfrac{4046}{4047}\)
\(S< \dfrac{3}{4}\times\dfrac{5}{6}\times...\times\dfrac{4047}{4048}\)
\(S^2< \dfrac{2}{3}\times\dfrac{4}{5}\times...\times\dfrac{4046}{4047}\times\left(\dfrac{3}{4}\times\dfrac{5}{6}\times...\times\dfrac{4047}{4048}\right)\)
\(S^2< \dfrac{2\times3\times4\times5\times...\times4046\times4047}{3\times4\times5\times6\times...\times4047\times4048}\)
\(S^2< \dfrac{2}{4048}\)
⇒ \(S^2< \dfrac{1}{2024}\)