1: \(\left|x-3,5\right|>=0\forall x\)

\(\left|4,5-y\right|>=0\forall y\)

Do đó: \(\left|x-3,5\right|+\left|4.5-y\right|>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-3,5=0\\4,5-y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3,5\\y=4,5\end{matrix}\right.\)

2: \(\left\{{}\begin{matrix}\left|x+\dfrac{2}{3}\right|>=0\forall x\\\left|y-\dfrac{3}{4}\right|>=0\forall y\\\left|z-5\right|>=0\forall z\end{matrix}\right.\)

Do đó: \(\left|x+\dfrac{2}{3}\right|+\left|y-\dfrac{3}{4}\right|+\left|z-5\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+\dfrac{2}{3}=0\\y-\dfrac{3}{4}=0\\z-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{3}{4}\\z=5\end{matrix}\right.\)

3: \(\left|x-2\right|+\left|3-x\right|=0\)

=>|x-2|+|x-3|=0(1)

TH1: x<2

Phương trình (1) sẽ trở thành 2-x+3-x=0

=>5-2x=0

=>2x=5

=>x=2,5(loại)

TH2: 2<=x<3

Phương trình (1) sẽ trở thành x-2+3-x=0

=>1=0(loại)

TH3: x>=3

Phương trình (1) sẽ trở thành x-2+x-3=0

=>2x=5

=>x=2,5(loại)

Vậy: Phương trình vô nghiệm

4: \(\left\{{}\begin{matrix}\left|x-\dfrac{2}{3}\right|>=0\forall x\\\left|x+y+\dfrac{3}{4}\right|>=0\forall x,y\\\left|y-z-\dfrac{5}{6}\right|>=0\forall y,z\end{matrix}\right.\)

Do đó: \(\left|x-\dfrac{2}{3}\right|+\left|x+y+\dfrac{3}{4}\right|+\left|y-z-\dfrac{5}{6}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\x+y+\dfrac{3}{4}=0\\y-z-\dfrac{5}{6}=0\end{matrix}\right.\)

=>\(\begin{matrix}x=\dfrac{2}{3}\\y=-x-\dfrac{3}{4}=-\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{-17}{12}\\z=y-\dfrac{5}{6}=-\dfrac{17}{12}-\dfrac{5}{6}=-\dfrac{27}{12}=-\dfrac{9}{4}\end{matrix}\)

5: \(\left\{{}\begin{matrix}\left|x-\dfrac{2}{3}\right|>=0\forall x\\\left|xy-\dfrac{5}{8}\right|>=0\forall x,y\\\left|yz+\dfrac{3}{4}\right|>=0\forall y,z\end{matrix}\right.\)

Do đó: \(\left|x-\dfrac{2}{3}\right|+\left|xy-\dfrac{5}{8}\right|+\left|yz+\dfrac{3}{4}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\xy-\dfrac{5}{8}=0\\yz+\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\xy=\dfrac{5}{8}\\yz=-\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{5}{8}:x=\dfrac{5}{8}:\dfrac{2}{3}=\dfrac{5}{8}\cdot\dfrac{3}{2}=\dfrac{15}{16}\\z=-\dfrac{3}{4}:\dfrac{15}{16}=-\dfrac{3}{4}\cdot\dfrac{16}{15}=\dfrac{-48}{60}=-\dfrac{4}{5}\end{matrix}\right.\)

6: \(\left\{{}\begin{matrix}\left|xy+\dfrac{2}{3}\right|>=0\forall x,y\\\left|yz-\dfrac{8}{9}\right|>=0\forall y,z\\\left|xz+\dfrac{3}{4}\right|>=0\forall x,z\end{matrix}\right.\)

Do đó: \(\left|xy+\dfrac{2}{3}\right|+\left|yz-\dfrac{8}{9}\right|+\left|xz+\dfrac{3}{4}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}xy+\dfrac{2}{3}=0\\yz-\dfrac{8}{9}=0\\xz+\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=-\dfrac{2}{3}\\yz=\dfrac{8}{9}\\xz=-\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left(xyz\right)^2=-\dfrac{2}{3}\cdot\dfrac{8}{9}\cdot\dfrac{-3}{4}=\dfrac{1}{2}\cdot\dfrac{8}{9}=\dfrac{4}{9}\)

=>\(\left[{}\begin{matrix}xyz=\dfrac{2}{3}\\xyz=-\dfrac{2}{3}\end{matrix}\right.\)

TH1: xyz=2/3

=>\(\left\{{}\begin{matrix}z=\dfrac{xyz}{xy}=\dfrac{2}{3}:\dfrac{-2}{3}=-1\\x=\dfrac{xyz}{yz}=\dfrac{2}{3}:\dfrac{8}{9}=\dfrac{2}{3}\cdot\dfrac{9}{8}=\dfrac{18}{24}=\dfrac{3}{4}\\y=\dfrac{xyz}{xz}=\dfrac{2}{3}:\dfrac{-3}{4}=\dfrac{2}{3}\cdot\dfrac{4}{-3}=-\dfrac{8}{9}\end{matrix}\right.\)

TH2: xyz=-2/3

=>\(\left\{{}\begin{matrix}z=\dfrac{xyz}{xy}=-\dfrac{2}{3}:\dfrac{-2}{3}=1\\x=\dfrac{xyz}{yz}=-\dfrac{2}{3}:\dfrac{8}{9}=\dfrac{-2}{3}\cdot\dfrac{9}{8}=\dfrac{-18}{24}=\dfrac{-3}{4}\\y=\dfrac{xyz}{xz}=\dfrac{-2}{3}:\dfrac{-3}{4}=\dfrac{-2}{3}\cdot\dfrac{4}{-3}=\dfrac{8}{9}\end{matrix}\right.\)

a: Xét ΔBAD và ΔBED có

BA=BE

\(\widehat{ABD}=\widehat{EBD}\)

BD chung

Do đó: ΔBAD=ΔBED
=>DA=DE

b: Ta có: BA=BE

=>B nằm trên đường trung trực của AE(1)

Ta có: DA=DE

=>D nằm trên đường trung trực của AE(2)

Từ (1),(2) suy ra BD là đường trung trực của AE

ΔBAD=ΔBED

=>\(\widehat{BAD}=\widehat{BED}\)

=>\(\widehat{BED}=90^0\)

=>DE\(\perp\)BC tại E

Ta có: ΔDEC vuông tại E

=>DC>DE

mà DE=DA

nên DC>DA

=>AD<DC

c: Ta có: BA+AF=BF

BE+EC=BC

mà BA=BE và BF=BC

nên AF=EC

Xét ΔDAF vuông tại A và ΔDEC vuông tại E có

DA=DE

AF=EC

Do đó: ΔDAF=ΔDEC

=>\(\widehat{ADF}=\widehat{EDC}\)

=>\(\widehat{ADF}+\widehat{ADE}=180^0\)

=>F,D,E thẳng hàng

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:

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

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

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