a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)

3) Với a = 3 ta có hpt:

\(\left\{{}\begin{matrix}2x+3y=5\\3x+2y=2\cdot3+1=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+9y=15\\6x+4y=14\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5y=1\\2x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{5}\\2x+\dfrac{3}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{5}\\2x=5-\dfrac{3}{5}=\dfrac{22}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=\dfrac{11}{5}\end{matrix}\right.\)

2: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\y\ne1\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}=1-\dfrac{1}{y-1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y-1=5\\x-2=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=2+\dfrac{5}{4}=\dfrac{8}{4}+\dfrac{5}{4}=\dfrac{13}{4}\end{matrix}\right.\left(nhận\right)\)

Bài 6:

\(a)P=\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+...+\dfrac{2}{33\cdot37}+\dfrac{2}{37\cdot41}\\ =\dfrac{1}{2}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{33\cdot37}+\dfrac{4}{37\cdot41}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{33}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{41}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{41}\right)\\ =\dfrac{1}{2}\cdot\dfrac{40}{41}\\ =\dfrac{20}{41}\\ b)Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\\ =\dfrac{6}{7}\cdot\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\\ =\dfrac{6}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\\ =\dfrac{6}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{121}\right)\\ =\dfrac{6}{7}\cdot\dfrac{119}{242}\\ =\dfrac{51}{121}\)

Bài 5:

a: Để A>0 thì \(\dfrac{2a-1}{-5}>0\)

=>2a-1<0

=>\(a< \dfrac{1}{2}\)

b: Để A<0 thì \(\dfrac{2a-1}{-5}< 0\)

=>2a-1>0

=>2a>1

=>\(a>\dfrac{1}{2}\)

c: Để A=0 thì \(\dfrac{2a-1}{-5}=0\)

=>2a-1=0

=>2a=1

=>\(a=\dfrac{1}{2}\)

Bài 6:

a: \(P=\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+...+\dfrac{2}{37\cdot41}\)

\(=\dfrac{2}{4}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{37\cdot41}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{37}-\dfrac{1}{41}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{41}\right)=\dfrac{1}{2}\cdot\dfrac{40}{41}=\dfrac{20}{41}\)

b: \(Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\)

\(=\dfrac{6}{7}\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\)

\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\)

\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{121}\right)=\dfrac{6}{7}\cdot\dfrac{119}{242}=\dfrac{51}{121}\)

Bài 5:

\(a)\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}:\dfrac{5}{3}-\dfrac{2}{7}:1\dfrac{2}{3}\\ =\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}\cdot\dfrac{3}{5}-\dfrac{2}{7}:\dfrac{5}{3}\\ =\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}\cdot\dfrac{3}{5}-\dfrac{2}{7}\cdot\dfrac{3}{5}\\ =\dfrac{3}{5}\cdot\left(\dfrac{6}{7}+\dfrac{3}{7}-\dfrac{2}{7}\right)\\ =\dfrac{3}{5}\cdot\dfrac{7}{7}\\=\dfrac{3}{5}\) 

\(b)\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\\ =\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)\\ =\dfrac{4}{9}:\dfrac{-9}{15}+\dfrac{4}{9}:\dfrac{-3}{22}\\ =\dfrac{4}{9}\cdot\dfrac{-5}{3}+\dfrac{4}{9}\cdot\dfrac{-22}{3}\\ =\dfrac{4}{9}\cdot\left(\dfrac{-5}{3}+\dfrac{-22}{3}\right)\\ =\dfrac{4}{9}\cdot\dfrac{-27}{3}\\ =\dfrac{4}{9}\cdot\left(-9\right)\\ =-4\)

Bài 4:

a: \(-1\dfrac{2}{5}\cdot x=\dfrac{2}{3}-\dfrac{4}{5}\)

=>\(x\cdot\dfrac{-7}{5}=\dfrac{10}{15}-\dfrac{12}{15}=-\dfrac{2}{15}\)

=>\(x=\dfrac{-2}{15}:\dfrac{-7}{5}=\dfrac{2}{15}\cdot\dfrac{5}{7}=\dfrac{2}{21}\)

b: \(\dfrac{2}{3}-4x=\dfrac{1}{2}-\dfrac{2}{5}\)

=>\(\dfrac{2}{3}-4x=\dfrac{5}{10}-\dfrac{4}{10}=\dfrac{1}{10}\)

=>\(4x=\dfrac{2}{3}-\dfrac{1}{10}=\dfrac{20}{30}-\dfrac{3}{30}=\dfrac{17}{30}\)

=>\(x=\dfrac{17}{30}:4=\dfrac{17}{120}\)

Bài 5:

a: \(\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}:\dfrac{5}{3}-\dfrac{2}{7}:1\dfrac{2}{3}\)

\(=\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}\cdot\dfrac{3}{5}-\dfrac{2}{7}:\dfrac{7}{3}\)

\(=\dfrac{18+9}{35}-\dfrac{2}{7}\cdot\dfrac{3}{7}=\dfrac{27}{35}-\dfrac{6}{49}=\dfrac{159}{245}\)

b: \(\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\)

\(=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)\)

\(=\dfrac{4}{9}:\dfrac{-9}{15}-\dfrac{4}{9}:\dfrac{-3}{22}\)

\(=\dfrac{4}{9}\cdot\dfrac{-5}{3}+\dfrac{4}{9}\cdot\dfrac{22}{3}=\dfrac{-20+88}{27}=\dfrac{68}{27}\)

a: \(0,5^{1000}=\left(0,5^5\right)^{200}=\left(0,03125\right)^{200}\)

mà \(0,03125< 0,625\)

nên \(0,5^{1000}< 0,625^{200}\)

b: \(\left(-32\right)^{27}=-32^{27}< 0;\left(-27\right)^{32}>0\)

Do đó: \(\left(-32\right)^{27}< \left(-27\right)^{32}\)

c: \(A=2+2^2+...+2^{2022}\)

=>\(2A=2^2+2^3+...+2^{2023}\)

=>\(2A-A=2^2+2^3+...+2^{2023}-2-2^2-...-2^{2022}\)

=>\(A=2^{2023}-2=B-2\)

=>A<B

Toán ạ

Bài 2: Viết các tích sau dưới dạng luỹ thừa, của một số tự nhiên.

a; a⁴.a⁶ = a⁴⁺⁶ = a¹⁰

b; (a⁵)⁷ = a⁵⁺⁷ = a¹² 

c; (a³)⁴.a⁹ = a¹².a⁹ = a¹²⁺⁹ = a²¹

d; (2³)⁵.(2³)⁴ = 2¹⁵.2¹² = 2¹⁵⁺¹² = 2²⁷

\(-\left|\dfrac{1}{2}x-2\right|+\dfrac{-2}{3}=\dfrac{-6}{5}\\ \Rightarrow-\left|\dfrac{1}{2}x-2\right|=\dfrac{-6}{5}+\dfrac{2}{3}\\ \Rightarrow-\left|\dfrac{1}{2}x-2\right|=-\dfrac{8}{15}\\ \Rightarrow\left|\dfrac{1}{2}x-2\right|=\dfrac{8}{15}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-2=\dfrac{8}{15}\\\dfrac{1}{2}x-2=-\dfrac{8}{15}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{38}{15}\\\dfrac{1}{2}x=\dfrac{22}{15}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{76}{15}\\x=\dfrac{44}{15}\end{matrix}\right.\)

#$\mathtt{Toru}$

\(-\left|\dfrac{1}{2}x-2\right|+\left(-\dfrac{2}{3}\right)=-\dfrac{6}{5}\\ =>-\left|\dfrac{1}{2}x-2\right|=-\dfrac{6}{5}+\dfrac{2}{3}\\ =>-\left|\dfrac{1}{2}x-2\right|=-\dfrac{8}{15}\\ =>\left|\dfrac{1}{2}x-2\right|=\dfrac{8}{15}\)

TH1: \(\dfrac{1}{2}x-2=\dfrac{8}{15}\left(x\ge4\right)\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{8}{15}+2=\dfrac{38}{15}\\ \Rightarrow x=\dfrac{38}{15}\cdot2=\dfrac{76}{15}\left(tm\right)\)

TH2: \(\dfrac{1}{2}x-2=-\dfrac{8}{15}\left(x< 4\right)\)

\(\Rightarrow\dfrac{1}{2}x=-\dfrac{8}{15}+2=\dfrac{22}{15}\\ \Rightarrow x=\dfrac{22}{15}\cdot2=\dfrac{44}{15}\left(tm\right)\)

1) D

2) D

3)C

4) D

5) C

6) C

7) D

8) B

Bài này dễ mà bn,chúc bn học tốt ^^

Tom always _____ breakfast before going to work.   A. have   B. is having C. had D. has

My English teacher _____ Mr Tuan Anh.  A. are   B. have been  C. were   D. is

Our lesson _____ at 7.30 a.m every morning.  A. begin  B. beginning    C. begins   D. began

I think Mary ____ a job.  A. have  B. will has   C. had   D. will have

Chelsea _____ next Sunday.  A. wins   B. will won   C. will win   D. won

. Do you like ________? – No, I don’t. A. going camping B. go camping  C. to go camping  D. go to camping

I am very fond of _______.  A. cook  B. cooks  C. cooking  D. to cook

 I hope everything _____ fine. A. will  B. won’t are  C. won’t  D. will be

 After breakfast, I and my friend ____ to the park every Sunday.   A. goes B. go  C. is going  D. went