Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)
Lập bảng xét dấu:
| x | -2 | \(\dfrac{1}{2}\) | |||
| x + 2 | - | 0 | + | + | |
| x - \(\dfrac{1}{2}\) | - | - | 0 | + |
TH : Xét x < -2
Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
-x - 2 -x + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
- 2x - 2 + \(\dfrac{1}{2}\)= \(\dfrac{3}{4}\)
-2x = 2\(\dfrac{1}{4}\)
=> x = \(-1\dfrac{1}{8}\) ( loại )
TH 2: \(-2\le x< \dfrac{1}{2}\)
Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
=> \(2,5=\dfrac{3}{4}\) ( loại )
TH3 : \(x\ge\dfrac{1}{2}\)
x+ 2 + x - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
2x + 1,5 = \(\dfrac{3}{4}\)
x = -0,375( loại )
vậy ....
b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)
c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)
TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)
\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)
TH2 : \(x-1< 0\Rightarrow x< 1\)
=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)
Vậy x = 1
b: Ta có: x/y=7/9
nên x/7=y/9
=>x/49=y/63
Ta có: y/z=7/3
nên y/7=z/3
=>y/63=z/27
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{49}=\dfrac{y}{63}=\dfrac{z}{27}=\dfrac{x-y+z}{49-63+27}=\dfrac{-15}{13}\)
Do đó: x=-735/13; y=-945/13; z=-405/13
c: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x+5y-2z}{2\cdot7+5\cdot20-2\cdot32}=\dfrac{100}{50}=2\)
Do đó: x=14; y=40; z=64
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=3\)
Do đó: x=24; y=15; z=6
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
a, \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
Vậy x = 23
b, \(\dfrac{x+4}{20}=\dfrac{5}{5x+4}\)
\(\Rightarrow\left(x+4\right)\left(5x+4\right)=100\)
\(\Rightarrow5x^2+24x+16=100\)
sai đề à?
c, \(\dfrac{7}{x+1}=\dfrac{x+1}{9}\)
\(\Rightarrow\left(x+1\right)^2=63\)
\(\Rightarrow\left[{}\begin{matrix}x+1=\sqrt{63}\\x+1=-\sqrt{63}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{63}-1\\x=-\sqrt{63}-1\end{matrix}\right.\)
Vậy...
a, \(2,5:7,5=x:\dfrac{3}{5}\)
\(\Leftrightarrow x:\dfrac{3}{5}=2,5:7,5\)
=> \(x.7,5=\dfrac{3}{5}.2,5\) => \(x=\dfrac{1,5}{7,5}=\dfrac{1}{5}\)
b, \(2\dfrac{2}{3}:x=1\dfrac{7}{9}\)
=> \(x=2\dfrac{2}{3}:1\dfrac{7}{9}\)
=> \(x=\dfrac{3}{2}\)
c, \(\dfrac{5}{6}:x=20:3\)
=> \(x.20=\dfrac{5}{6}.3\) => \(x=\dfrac{2,5}{20}=\dfrac{1}{8}\)
3, Tìm x, biết:
a) 2,5 : 7,5 = x : \(\dfrac{3}{5}\)
=> \(x:\dfrac{3}{5}=\dfrac{1}{3}\)
=> \(x=\dfrac{1}{3}.\dfrac{3}{5}=>x=\dfrac{1}{5}\)
b) \(2\dfrac{2}{3}\) : x = \(1\dfrac{7}{9}\)
=> \(\dfrac{8}{3}:x=\dfrac{16}{9}\)
=> \(x=\dfrac{8}{3}:\dfrac{16}{9}=>x=\dfrac{3}{2}\)
c) \(\dfrac{5}{6}\) : x = 20 : 3
=> \(x=\dfrac{5}{6}:\dfrac{20}{3}=>x=\dfrac{1}{8}\)
a, Theo bài ra ta có:
\(M=\dfrac{2007}{1}+1+\dfrac{2006}{2}+1+.......+\dfrac{2}{2006}+1+\dfrac{1}{2007}+1-2007\)
( Ta thêm 1 vào mỗi một số hạng trong M nên phải bớt đi 2017 vì có 2017 số hạng ) ;'
\(=>M=2008+\dfrac{2008}{2}+\dfrac{2008}{3}+......+\dfrac{2008}{2007}+\dfrac{2008}{2007}-2007\)
\(=>M=\dfrac{2008}{2}+\dfrac{2008}{3}+\dfrac{2008}{4}+.....+\dfrac{2008}{2006}+\dfrac{2008}{2007}+1\)
Ta thấy xuất hiện 2008 chung nên đặt ra ngoài ta có:
\(=>M=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(=>M:N=2008\)
Câu b đợi 1 chút nha.......
b, \(M=\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{31.33}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{31.33}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{33}\)
\(N=\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+...+\dfrac{12}{31.33.35}\)
\(=3\left(\dfrac{4}{11.13.15}+\dfrac{4}{13.15.17}+...+\dfrac{4}{31.33.35}\right)\)
\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.17}+...+\dfrac{1}{31.33}-\dfrac{1}{33.35}\right)\)
\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{33.35}\right)\)
\(=\dfrac{92}{5005}\)
\(\Rightarrow M:N=\dfrac{1}{33}:\dfrac{92}{5005}=\dfrac{455}{276}\)
Vậy...
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)
\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)
\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)
Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)
\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)
\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)
\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)
\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)
\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)
\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)
\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
a) \(x-\dfrac{3}{5}=\dfrac{2}{7}\)
\(\Rightarrow x=\dfrac{2}{7}+\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{10}{35}+\dfrac{21}{35}\)
\(\Rightarrow x=\dfrac{31}{35}\)
b) \(x+\dfrac{20}{11\cdot13}+\dfrac{20}{13\cdot15}+...+\dfrac{20}{53\cdot55}=\dfrac{3}{11}\)
\(\Rightarrow x+10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
\(\Rightarrow x+10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Rightarrow x+10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Rightarrow x+10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
\(\Rightarrow x+\dfrac{40}{55}=\dfrac{3}{11}\)
\(\Rightarrow x=\dfrac{3}{11}-\dfrac{40}{55}\)
\(\Rightarrow x=\dfrac{-25}{55}=\dfrac{-5}{11}\)