Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
A=1.2.3+2.3.4+3.4.5+...+98.99.100
a, Vào câu hỏi tương tự nhé
b, Vì \(\hept{\begin{cases}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{cases}\Rightarrow\left|x+3\right|+\left|x+1\right|\ge0\Rightarrow3x\ge0\Rightarrow x\ge0}\)
=> x+3+x+1=3x
=> 2x+4=3x
=>x=4
c, \(\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|=\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\)
Có \(\left|4-x\right|\ge4-x;\left|10-x\right|\ge10-x;\left|x+990\right|\ge x+990;\left|x+1000\right|\ge x+1000\)
=>\(\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\)
=> \(2005\ge4-x+10-x+x+990+x+1000+\left|x+101\right|\)
=> \(2005\ge\left|x+101\right|+2004\)
=> \(\left|x+101\right|\le1\)
=> \(x+101\in\left\{-1;0;1\right\}\Rightarrow x\in\left\{-102;-101;-100\right\}\)
d, tương tự b
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
\(a,\frac{-1}{2}+\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}.\)
\(\Rightarrow\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}-\frac{-1}{2}=\frac{-7}{6}\)
\(\Rightarrow x-3=\frac{-7}{6}\cdot\frac{-1}{2}=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}+3=3\frac{7}{12}\)
\(b.2,25+\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}\)
\(\Rightarrow\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}-2,25=\frac{1}{4}\)
\(\Rightarrow x-5=\frac{3}{2}:\frac{1}{4}=6\)
\(\Rightarrow x=6+5=11\)
\(c,\left(\frac{1}{3}-x\right)^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=\frac{1}{2}\\\frac{1}{3}-x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\\x=\frac{1}{3}-\frac{-1}{2}=\frac{5}{6}\end{cases}}\)
\(d,\frac{3}{2}+\frac{x-1}{3}=1\)
\(\Rightarrow\frac{x-1}{3}=1-\frac{3}{2}=-\frac{1}{2}\)
\(\Rightarrow x-1=-\frac{1}{2}\cdot3=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{3}{2}+1=\frac{1}{2}\)
\(e,-\frac{6}{8}+\frac{x}{12}=\frac{5}{6}\)
\(\Rightarrow\frac{x}{12}=\frac{5}{6}-\frac{-6}{8}=\frac{19}{12}\)
\(\Rightarrow x=19\)
\(g,\frac{1}{2}-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}\)
\(\Rightarrow-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}\)
\(\Rightarrow x-2=\frac{-7}{6}:\frac{-1}{3}=\frac{7}{2}\)
\(\Rightarrow x=\frac{7}{2}+2=2\frac{7}{2}\)
\(h,\frac{5}{2}\left(x+1\right)-\frac{1}{2}=3\frac{1}{2}\)
\(\Rightarrow\frac{5}{2}\left(x+1\right)=3\frac{1}{2}-\frac{1}{2}=3\)
\(\Rightarrow x+1=3:\frac{5}{2}=\frac{6}{5}\)
\(\Rightarrow x=\frac{6}{5}-1=\frac{1}{5}\)
\(k,\frac{x}{3}-\frac{1}{2}=-2\left(x+1\right)+3\)
\(\Rightarrow x\cdot\frac{1}{3}-\frac{1}{2}=-2x-2+3\)
\(\Rightarrow\frac{1}{3}x+2x=-2+3+\frac{1}{2}\)
\(\Rightarrow\frac{7}{3}x=\frac{3}{2}\Rightarrow x=\frac{3}{2}:\frac{7}{2}=\frac{3}{7}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
Dài đấy :))
a) \(\left|x-1\right|-\left(-2\right)^3=9\cdot\left(-1\right)^{100}\)
\(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\cdot1\)
\(\Leftrightarrow\left|x-1\right|+8=9\)
\(\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)( ĐKXĐ : \(x\ne2\))
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x-2\right)^2=36\)
\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\left(tmđk\right)\)
c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)( ĐKXĐ : \(x\ne5\))
\(\Leftrightarrow\frac{x-5}{3}=\frac{-12}{-\left(x-5\right)}\)
\(\Leftrightarrow\frac{x-5}{3}=\frac{12}{x-5}\)
\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=3\cdot12\)
\(\Leftrightarrow\left(x-5\right)^2=36\)
\(\Leftrightarrow\left(x-5\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\left(tmđk\right)\)
d) \(8x-\left|4x+\frac{3}{4}\right|=x+2\)
\(\Leftrightarrow8x-x-2=\left|4x+\frac{3}{4}\right|\)
\(\Leftrightarrow7x-2=\left|4x+\frac{3}{4}\right|\)(*)
\(\left|4x+\frac{3}{4}\right|\ge0\Leftrightarrow4x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{16}\)
Vậy ta xét hai trường hợp sau :
1. \(x\ge-\frac{3}{16}\)
(*) <=>\(7x-2=4x+\frac{3}{4}\)
\(\Leftrightarrow7x-4x=\frac{3}{4}+2\)
\(\Leftrightarrow3x=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{12}\)(tmđk)
2. \(x< -\frac{3}{16}\)
(*) <=> \(7x-2=-\left(4x+\frac{3}{4}\right)\)
\(\Leftrightarrow7x-2=-4x-\frac{3}{4}\)
\(\Leftrightarrow7x+4x=-\frac{3}{4}+2\)
\(\Leftrightarrow11x=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{5}{44}\left(ktmđk\right)\)
Vậy x = 11/12
e) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)
\(\Leftrightarrow x+1=4040\)
\(\Leftrightarrow x=4039\)
\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)
TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)
TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)
\(\Rightarrow x\in\left\{2;3;-4\right\}\)
\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)
\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)
\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{998}{1000}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{1000}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{1000}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{1000}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{499}{1000}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{499}{1000}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1000}\)
=>x+1=1000
=>x=999