\(\frac{x+2}{x+3}< \frac{x+4}{x+5}\) Tìm x, ai giải đc, tik liền
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
6 tháng 2 2018
a, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
9 tháng 4 2017
a) (2x+3)-(5x-17)=9
2x+3-5x+17=9
-3x+20=9
-3x=9-20=-11
=>x=11/3
b) \(\left(\frac{1}{5}+\frac{4}{5}x\right)-\left(\frac{2}{5}x+\frac{1}{9}\right)=\frac{3}{7}\)
=> \(\frac{1}{5}+\frac{4}{5}x-\frac{2}{5}x-\frac{1}{9}=\frac{3}{7}\)
=> \(\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{4}{5}x-\frac{2}{5}x\right)=\frac{3}{7}\)
\(\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\)
\(\frac{2}{5}x=\frac{3}{7}-\frac{4}{45}=\frac{107}{315}\)
x=107/315:2/5=107/126
NX
31 tháng 7 2018
\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)
<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)
<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)
<=> \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{6}\) : \(\frac{5}{12}\)
,<=> \(\left(x-2\frac{1}{5}\right)\)= 2
<=. x = 2 + \(\frac{11}{5}\)
<=> x = \(\frac{21}{5}\)
DN
15 tháng 7 2018
a) \(\frac{3x+2}{-4x+5}=-\frac{4}{3}\left(ĐKXĐ:x\ne\frac{5}{4}\right)\)
\(\Rightarrow3\left(3x+2\right)=-4\left(-4x+5\right)\)
\(\Leftrightarrow9x+6=16x-20\)
\(\Leftrightarrow7x=26\)
\(\Leftrightarrow x=\frac{26}{7}\)
b) \(\frac{2\left|x\right|+5}{-4x+3}=-\frac{5}{4}\)(Thôi bài sau tự tìm đkxđ nhá)
\(\Rightarrow8\left|x\right|+20=20x-15\)
\(\Leftrightarrow8\left|x\right|-20x+35\)\(\left(1\right)\)
TH1: Nếu \(x\ge0\)thì \(\left(1\right)\Leftrightarrow8x-20x+35=0\Leftrightarrow x=\frac{35}{12}\left(tm\right)\)
TH2: Nếu \(x< 0\)thì \(\left(1\right)\Leftrightarrow-8x-20x+35=0\Leftrightarrow x=\frac{35}{28}\left(ktm\right)\)
Vậy x=35/12
c)\(\frac{2x+1}{5}=\frac{3}{2x-1}\)
\(\Rightarrow4x^2-1=15\)
\(\Leftrightarrow4x^2=16\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
d)\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow0,5x=1\)
\(\Leftrightarrow x=2\)
e) \(\frac{\left|6x+1\right|}{4}=\frac{2}{4}\)
\(\Leftrightarrow\left|6x+1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}6x+1=2\\6x+1=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{1}{2}\end{cases}}}\)
g)\(\frac{\left|3x-5\right|}{3}=\frac{\left|x\right|}{2}\)
\(\Leftrightarrow\frac{\left|3x-5\right|}{\left|x\right|}=\frac{3}{4}\)
\(\Leftrightarrow\left|\frac{3x-5}{x}\right|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x-5}{x}=\frac{3}{4}\\\frac{3x-5}{x}=-\frac{3}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{20}{9}\\x=\frac{4}{3}\end{cases}}}\)
Mỏi tay quá, xin tý cho sảng khoái nào!!
\(\)
LN
9 tháng 4 2017
(4/3-1/4.X-5/12)-2.X=8/3
(16-3.x-5)-24.X=32
16-3X-5-24X=32
11-27X=32
x=-7/9
9 tháng 4 2017
\(\left(\frac{3}{2}-\frac{1}{4}.x-\frac{5}{12}\right)-2.x=\frac{8}{5}:\frac{3}{5}\)
\(\left(\frac{13}{12}-\frac{1}{4}.x\right)-2.x=\frac{8}{3}\)
\(\frac{13}{12}-x.\left(\frac{1}{4}-2\right)=\frac{8}{3}\)
\(x.\left(\frac{-1}{4}\right)=\frac{13}{12}-\frac{8}{3}=-\frac{19}{12}\)
\(x=\frac{-19}{12}:\frac{-1}{4}=\frac{19}{3}\)
20 tháng 8 2018
a)
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)
=> \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
Đến đây dễ rồi
20 tháng 8 2018
b)
\(\left(\frac{x}{3}\right)^2=\frac{x}{3}\cdot\frac{x}{3}=\frac{x}{3}\cdot\frac{y}{4}=\frac{xy}{3\cdot4}=\frac{48}{12}=4=\left(\pm2\right)^2\)
TH1 : \(\frac{x}{3}=\frac{y}{4}=2\)
Sau đó tìm x và y
TH2 : \(\frac{x}{3}=\frac{y}{4}=-2\)
Sau đó lại tìm x và y
Sau cùng kết luận
Học tốt
23 tháng 4 2017
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(x=2002\)
Vậy x = 2002
\(\frac{x+2}{x+3}< \frac{x+4}{x+5}\)
<=> \(\left(x+2\right)\left(x+5\right)< \left(x+3\right)\left(x+4\right)\)
<=> \(x^2+7x+10< x^2+7x+12\)
<=> \(x^2-x^2+7x-7x+10-12< 0\)
???
x = -3 ; x = -4/8/9