a: Các biến cố không thể là C

Biến cố chắc chắn là D

b: Biến cố ngẫu nhiên là A,B

A: "Mai lấy được bút màu đỏ"

n(A)=1

\(P_A=\dfrac{n\left(A\right)}{n\left(\Omega\right)}=\dfrac{1}{3}\)

B: "Mai lấy được bút màu xanh"

=>n(B)=2

=>\(P_B=\dfrac{n\left(B\right)}{n\left(\Omega\right)}=\dfrac{2}{3}\)

tk ạ

a) Biến cố không thể là C: "Mai lấy được chiếc bút màu đen", vì trong hộp chỉ có hai chiếc bút màu xanh và một chiếc bút màu đỏ.

Biến cố chắc chắn là "Mai lấy được chiếc bút màu đỏ hoặc màu xanh", vì không có trường hợp nào khác xảy ra.

b) Xác suất của các biến cố:

P(A) = 1/3 (vì trong hộp có một chiếc bút màu đỏ và hai chiếc bút màu xanh)
P(B) = 2/3
P(C) = 0
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 1/3 + 2/3 - 0 = 1
Chú ý rằng P(A ∪ B) là biến cố "Mai lấy được chiếc bút màu đỏ hoặc màu xanh", và P(A ∩ B) = 0 vì không thể lấy được cả hai chiếc bút cùng lúc.

\(B=\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{1000^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{1000}-1\right)\cdot\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{1000}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot....\cdot\dfrac{1001}{1000}\)

\(=\dfrac{-1}{1000}\cdot\dfrac{1001}{2}=\dfrac{-1001}{2000}\)

Thanks

Bài 3:

a: \(3\in N;3\in Z;3\in Q\)

b: \(10\in N;10\in Z;10\in Q\)

c: \(-\dfrac{3}{7}\in Q\)

d: \(-2\in Z;-2\in Q\)

Bài 2:

\(\dfrac{-3}{5}=\dfrac{-27}{45};\dfrac{-5}{9}=\dfrac{-25}{45};-1\dfrac{2}{3}=\dfrac{-5}{3}=\dfrac{-75}{45};0,5=\dfrac{22,5}{45};\dfrac{10}{9}=\dfrac{50}{45}\)

mà -75<-27<-25<22,5<50

nên \(-1\dfrac{2}{3}< -\dfrac{3}{5}< -\dfrac{5}{9}< \dfrac{1}{2}< \dfrac{10}{9}\)

Bài 1:

a: \(\dfrac{9}{70}=\dfrac{9\cdot3}{70\cdot3}=\dfrac{27}{210};\dfrac{5}{42}=\dfrac{5\cdot5}{42\cdot5}=\dfrac{25}{210}\)

mà 27>25

nên \(\dfrac{9}{70}>\dfrac{5}{42}\)

b: \(\dfrac{-4}{27}=\dfrac{-4\cdot7}{27\cdot7}=\dfrac{-28}{189};\dfrac{10}{-63}=\dfrac{-10}{63}=\dfrac{-10\cdot3}{63\cdot3}=\dfrac{-30}{189}\)

mà -28>-30

nên \(\dfrac{-4}{27}>\dfrac{10}{-63}\)

c: \(\dfrac{999}{556}=1+\dfrac{443}{556};\dfrac{1000}{557}=1+\dfrac{443}{557}\)

mà \(\dfrac{443}{556}>\dfrac{443}{557}\)

nên \(\dfrac{999}{556}>\dfrac{1000}{557}\)

d: \(\dfrac{-2}{15}< 0;\dfrac{-10}{-11}=\dfrac{10}{11}>0\)

Do đó: \(\dfrac{-2}{15}< \dfrac{-10}{-11}\)

a: Xét ΔAMC và ΔDMB có

MA=MD

\(\widehat{AMC}=\widehat{DMB}\)(hai góc đối đỉnh)

MC=MB

Do đó: ΔAMC=ΔDMB

b: ΔAMC=ΔDMB

=>\(\widehat{MAC}=\widehat{MDB}\)

=>AC//DB

mà AC\(\perp\)AB

nên BD\(\perp\)BA

c: Xét ΔCAB vuông tại A và ΔDBA vuông tại B có

BA chung

AC=BD

Do đó: ΔCAB=ΔDBA

=>CB=DA

mà \(AM=\dfrac{1}{2}AD\)

nên \(AM=\dfrac{1}{2}CB\)

Xét ΔABE có: \(\widehat{BAE}+\widehat{ABE}+\widehat{AEB}=180^o\)

\(\Rightarrow90^o+x+x=180^o\Rightarrow2x=180^o-90^o=90^o\)

\(\Rightarrow x=\dfrac{90^o}{2}=45^o\) 

Xét ΔABC có: \(\widehat{ABC}+\widehat{BAC}+\widehat{CAB}=180^o\)

\(\Rightarrow\left(x+y\right)+90^o+30^o=180^o\)

\(\Rightarrow\left(x+y\right)+120^o=180^o\)

\(\Rightarrow45^o+y=180^o-120^o\)

\(\Rightarrow45^o+y=60^o\)

\(\Rightarrow y=60^o-45^o=15^o\)

∆ABE vuông tại A (gt)

⇒ ∠ABE + ∠AEB = 90⁰

⇒ x + x = 90⁰

⇒ x = 90⁰ : 2

= 45⁰

∆ABC vuông tại A (gt)

⇒ ∠ABC + ∠ACB = 90⁰

⇒ ∠ABC = 90⁰ - ∠ACB

= 90⁰ - 30⁰

= 60⁰

⇒ y = ∠ABC - x

= 60⁰ - 45⁰

= 15⁰

a) 

\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,3\right)x-8=\left(\dfrac{3}{5}+0,415+\dfrac{1}{200}\right):0,01\\ \Rightarrow\left(\dfrac{1}{12}+\dfrac{19}{6}-\dfrac{303}{10}\right)x-8=\left(\dfrac{3}{5}+0,415+\dfrac{1}{200}\right)\cdot100\\ \Rightarrow-\dfrac{541}{20}x-8=\dfrac{51}{50}\cdot100\\ \Rightarrow-\dfrac{541}{20}x-8=102\\ \Rightarrow-\dfrac{541}{20}x=110\\ \Rightarrow x=110:-\dfrac{541}{20}\\\Rightarrow x=-\dfrac{2200}{541}\approx4,067\)

b) 

\(\left(\dfrac{1}{12}+2\dfrac{1}{12}-10,75\right)x-7=\left(\dfrac{2}{5}+\dfrac{3}{8}+0,225\right):0,1\\ \Rightarrow\left(\dfrac{1}{12}+\dfrac{25}{12}-\dfrac{43}{4}\right)x-7=\left(\dfrac{2}{5}+\dfrac{3}{8}+\dfrac{9}{40}\right)\cdot10\\ \Rightarrow-\dfrac{103}{12}x=1\cdot10\\ \Rightarrow x=10:-\dfrac{103}{12}\\ \Rightarrow x=-\dfrac{120}{103}\\ \Rightarrow x\approx-1,165\) 

c) 

\(\left(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\right)\left(x-3\right)=\dfrac{21}{13}\\ \Rightarrow\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\right)\left(x-3\right)=\dfrac{21}{13}\\ \Rightarrow\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\left(x-3\right)=\dfrac{21}{13}\\\Rightarrow \dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\left(x-3\right)=\dfrac{21}{13}\\ \Rightarrow\dfrac{84}{305}\left(x-3\right)=\dfrac{21}{13}\\ \Rightarrow x-3=\dfrac{21}{13}:\dfrac{84}{305}\\ \Rightarrow x-3=\dfrac{305}{52}\\ \Rightarrow x=\dfrac{305}{52}+3\\ \Rightarrow x=\dfrac{461}{52}\\ \Rightarrow x\approx8,865\)

Bài 10:

\(A=\dfrac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot3^4}{\left(2^2\right)^6\cdot3^6+\left(2^3\right)^4\cdot3^5}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\\ =\dfrac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}\\ =\dfrac{3-1}{3\cdot\left(3+1\right)}\\ =\dfrac{2}{3\cdot4}\\ =\dfrac{1}{6}\)

\(B=\dfrac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\\ =\dfrac{2\cdot3\cdot5\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{2\cdot3^3\cdot2^{14}\cdot3^{14}\cdot\left(3^2\right)^7-2^2\cdot3\cdot\left(2^3\right)^5\cdot7^5\cdot}\\ =\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{2^{15}\cdot3^{17}\cdot3^{14}-2^{17}\cdot7^5\cdot3}\\ =\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{2^{15}\cdot3^{31}-2^{17}\cdot7^5\cdot3}\\ =\dfrac{5\cdot\left(3^{30}\cdot2^{15}-2^{17}\cdot7^5\right)}{3\cdot\left(2^{15}\cdot3^{30}-2^{17}\cdot7^5\right)}\\ =\dfrac{5}{3}\)

Bài 8:

\(\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\y+z=\dfrac{1}{3}\\x+z=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{3}-z+\dfrac{1}{6}-z=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-2z=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2z=0\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=0\\y=\dfrac{1}{3}-0=\dfrac{1}{3}\\x=\dfrac{1}{6}-0=\dfrac{1}{6}\end{matrix}\right.\)

Bài 5:

\(A=\left(\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}\right):\left(\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}\right)\)

\(=\dfrac{15-4+1}{10}:\dfrac{9-4+1}{12}\)

\(=\dfrac{12}{10}\cdot\dfrac{12}{6}=\dfrac{6}{5}\cdot2=\dfrac{12}{5}\)

\(S=1+2+2^2+2^3+...+2^{2009}+2^{2008}\\ 2S=2\left(1+2+2^2+...+2^{2008}\right)\\ 2S=2+2^2+2^3+...+2^{2009}\\ 2S-S=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+...+2^{2008}\right)\\ S=2^{2009}-1\)

\(-\dfrac{25}{20}>0\)

\(\dfrac{20}{25}>0\)

\(\Rightarrow-\dfrac{25}{20}< \dfrac{20}{25}\)