HH
1
K
Khách
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16 tháng 1 2019
2.a)\(\dfrac{3\text{x}-2}{2}\)=\(\dfrac{1-2\text{x}}{3}\)
<=>\(\dfrac{9\text{x}-6}{6}\)=\(\dfrac{2-4\text{x}}{6}\)
<=>9x-6=2-4x
<=>9x+4x=2+6
<=>13x=8
<=>x=\(\dfrac{8}{13}\)
16 tháng 1 2019
1.a)2(x-0,5)+3=0,25(4x-1)
<=>2x-1+3=x-1phần4
<=>2x-x=-1/4+1-3
<=>x=-3/4
9 tháng 4 2020
a)
\(\frac{7}{x-5}-2=\frac{3}{5-x}\\ \Leftrightarrow\frac{-7}{5-x}-2-\frac{3}{5-x}=0\\ \Leftrightarrow\frac{-7}{5-x}-\frac{10-2x}{5-x}-\frac{3}{5-x}=0\\ \Leftrightarrow\frac{-7-10+2x-3}{5-x}=0\\ \Leftrightarrow\frac{2x-20}{5-x}=0\\ \Rightarrow2x-20=0\\ \Rightarrow x=10\)
b)
\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}\\ \Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{2x-4}{\left(x+1\right)\cdot\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\cdot\left(x-2\right)}-\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{2x-4-x-1-3x+11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{6-2x}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Rightarrow6-2x=0\\ \Rightarrow x=3\)
c)
\(\frac{1}{x}-\frac{x+2}{x-2}=\frac{2}{x\cdot\left(2-x\right)}\\ \Leftrightarrow\frac{1}{x}-\frac{x-2}{2-x}-\frac{2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{2-x}{x\cdot\left(2-x\right)}-\frac{x^2-2x}{x\cdot\left(2-x\right)}-\frac{2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{2-x-x^2+2x-2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{x-x^2}{x\cdot\left(2-x\right)}=0\\ \Rightarrow x-x^2=0\\ \Rightarrow x\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
DT
2
11 tháng 4 2018
\(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\Leftrightarrow\dfrac{x^2-2x-4}{x^2-2x-3}-1>0\)
\(\Leftrightarrow\dfrac{x^2-2x-4-x^2+2x+3}{x^2-3x+x-3}>0\)
\(\Leftrightarrow\dfrac{-1}{\left(x-3\right)\left(x+1\right)}>0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x>-1\end{matrix}\right.\end{matrix}\right.\)
TH1 : vô lý
Vậy \(-1< x< 3\) thì \(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
LD
31 tháng 1 2019
<=> \(\dfrac{x+2}{x-2}\)-\(\dfrac{1}{x}\)=\(\dfrac{2}{x\left(x-2\right)}\)
<=> \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
ok, ở đây đã có mẫu chung rồi, em cứ vậy làm tiếp thôi :D
S
31 tháng 1 2019
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (ĐKXĐ: \(x\ne0;x\ne2\))
\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x+2-2=0\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow S=\left\{-1\right\}\)
HT
2
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2019
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+...+\dfrac{x-2016}{1}=2016\)
\(\Leftrightarrow\dfrac{x-1}{2016}-1+\dfrac{x-2}{2015}-1+\dfrac{x-3}{2014}-1+...+\dfrac{x-2016}{1}-1=0\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}+...+\dfrac{x-2017}{1}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\right)=0\)
\(\Leftrightarrow x-2017=0\) (do \(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\ne0\))
\(\Rightarrow x=2017\)
3 tháng 3 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)=\frac{7}{15}\)
Ta có: \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\Leftrightarrow\frac{7}{15}x=\frac{3}{5}x\)
\(\Leftrightarrow\frac{2}{15}x=0\Leftrightarrow x=0\)
Tập nghiệm: \(S=\left\{0\right\}\)
19 tháng 1 2019
\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)
\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)
Mấy câu khác tương tự :v
2 tháng 1 2023
b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
=>123-x=0
=>x=123
c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2017}-1\right)=\left(\dfrac{x-1}{2018}-1\right)+\left(\dfrac{x}{2019}-1\right)\)
=>x-2019=0
=>x=2019
Ẩn của phương trình đâu vậy em?