Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\left\{{}\begin{matrix}x\ne2\\y\ne1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}-\dfrac{2}{x-2}+\dfrac{3}{y-1}=2-1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}=1-\dfrac{1}{y-1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-1=5\\x-2=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=2+\dfrac{5}{4}=\dfrac{8}{4}+\dfrac{5}{4}=\dfrac{13}{4}\end{matrix}\right.\left(nhận\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\)
\(=\dfrac{6}{7}\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\)
\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\)
\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{121}\right)=\dfrac{6}{7}\cdot\dfrac{119}{242}=\dfrac{51}{121}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\)
\(=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)\)
\(=\dfrac{4}{9}:\dfrac{-9}{15}-\dfrac{4}{9}:\dfrac{-3}{22}\)
\(=\dfrac{4}{9}\cdot\dfrac{-5}{3}+\dfrac{4}{9}\cdot\dfrac{22}{3}=\dfrac{-20+88}{27}=\dfrac{68}{27}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\left|\dfrac{1}{2}x-2\right|+\dfrac{-2}{3}=\dfrac{-6}{5}\\ \Rightarrow-\left|\dfrac{1}{2}x-2\right|=\dfrac{-6}{5}+\dfrac{2}{3}\\ \Rightarrow-\left|\dfrac{1}{2}x-2\right|=-\dfrac{8}{15}\\ \Rightarrow\left|\dfrac{1}{2}x-2\right|=\dfrac{8}{15}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-2=\dfrac{8}{15}\\\dfrac{1}{2}x-2=-\dfrac{8}{15}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{38}{15}\\\dfrac{1}{2}x=\dfrac{22}{15}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{76}{15}\\x=\dfrac{44}{15}\end{matrix}\right.\)
#$\mathtt{Toru}$
\(-\left|\dfrac{1}{2}x-2\right|+\left(-\dfrac{2}{3}\right)=-\dfrac{6}{5}\\ =>-\left|\dfrac{1}{2}x-2\right|=-\dfrac{6}{5}+\dfrac{2}{3}\\ =>-\left|\dfrac{1}{2}x-2\right|=-\dfrac{8}{15}\\ =>\left|\dfrac{1}{2}x-2\right|=\dfrac{8}{15}\)
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: \(\dfrac{1}{2}x-2=-\dfrac{8}{15}\left(x< 4\right)\)
\(\Rightarrow\dfrac{1}{2}x=-\dfrac{8}{15}+2=\dfrac{22}{15}\\ \Rightarrow x=\dfrac{22}{15}\cdot2=\dfrac{44}{15}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chữ số 6 có giá trị là 60
=>6 là chữ số hàng chục
5 không nằm ở hàng đơn vị
mà 5 không là chữ số hàng chục
nên 5 là chữ số hàng trăm
=>Chữ số hàng đơn vị là 7
Vậy: Số cần tìm là 567
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Để x là số hữu tỉ dương thì:
\(\dfrac{13-n}{-5}>0\)
Mà: `-5<0`
`=>13-n<0`
`=>n>13`
b) Để x là số hữu tỉ âm thì:
`(13-n)/-5<0`
Mà: `-5<0`
`=>13-n>0`
`=> n<13`
c) Đê x không phải số hữu tỉ âm cũng không phải số hữu tỉ dương thì:
\(x=0=>\dfrac{13-n}{-5}=0\\ =>13-n=0\\ =>n=13\)
Bài 2:
Để \(\dfrac{m+2}{5};\dfrac{m-5}{-6}\) đều là các số dương thì
\(\left\{{}\begin{matrix}\dfrac{m+2}{5}>0\\\dfrac{m-5}{-6}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+2>0\\m-5< 0\end{matrix}\right.\)
=>-2<m<5
mà m nguyên
nên \(m\in\left\{-1;0;1;2;3;4\right\}\)
Bài 3:
Để \(\dfrac{1-m}{-13};\dfrac{5-m}{3}\) đều là các số âm thì
\(\left\{{}\begin{matrix}\dfrac{1-m}{-13}< 0\\\dfrac{5-m}{3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m-1}{13}< 0\\\dfrac{m-5}{3}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m-1< 0\\m-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 1\\m>5\end{matrix}\right.\)
=>\(m\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(-\dfrac{25}{20}< 0;0< \dfrac{20}{25}\)
Do đó: \(-\dfrac{20}{25}< \dfrac{20}{25}\)
b: \(\dfrac{15}{21}=\dfrac{15:3}{21:3}=\dfrac{5}{7};\dfrac{21}{49}=\dfrac{21:7}{49:7}=\dfrac{3}{7}\)
mà 5>3
nên \(\dfrac{15}{21}>\dfrac{21}{49}\)
c: \(\dfrac{-19}{49}=\dfrac{-19\cdot47}{49\cdot47}=\dfrac{-893}{49\cdot47}\)
\(\dfrac{-23}{47}=\dfrac{-23\cdot49}{47\cdot49}=\dfrac{-1127}{47\cdot49}\)
mà -893>-1127
nên \(-\dfrac{19}{49}>-\dfrac{23}{47}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}\\ =\dfrac{n-1}{n}\)
\(B=\dfrac{49}{2\cdot9}+\dfrac{49}{9\cdot16}+\dfrac{49}{16\cdot23}+...+\dfrac{49}{65\cdot72}\\ =7\cdot\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+\dfrac{7}{16\cdot23}+...+\dfrac{7}{65\cdot72}\right)\\ =7\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\\ =7\cdot\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\\ =7\cdot\dfrac{35}{72}\\ =\dfrac{245}{72}\)
\(E=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{95\cdot99}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{33}{99}-\dfrac{1}{99}=\dfrac{32}{99}\)
\(D=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+\dfrac{1}{19\cdot25}+\dfrac{1}{25\cdot31}+\dfrac{1}{31\cdot37}\)
\(=\dfrac{1}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+...+\dfrac{6}{31\cdot37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)=\dfrac{1}{6}\cdot\dfrac{36}{37}=\dfrac{6}{37}\)
\(C=\dfrac{3}{1\cdot3}+\dfrac{3}{3\cdot5}+...+\dfrac{3}{49\cdot51}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{49\cdot51}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{51}\right)=\dfrac{3}{2}\cdot\dfrac{50}{51}=\dfrac{1}{17}\cdot25=\dfrac{25}{17}\)
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: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)