Giải PT:
a)\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-295}{21}+\frac{x-166}{23}=0\)
b)\(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)
c)\(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^2-1}\)
d)2x\(^2\)-5x-3=0
K
Khách
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Những câu hỏi liên quan
N
30 tháng 12 2018
\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+16}{84}+1\right)+\left(\frac{x+17}{83}+1\right)+\left(\frac{166}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(x+100\right)=0\Rightarrow x=-100\left(\text{vì }\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)
S
31 tháng 12 2018
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=-10\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=-10\)
\(.....................\)
đến đây thì dễ rồi :)
C
5
D
4 tháng 2 2018
\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right).\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow x-258=0\)
\(\Leftrightarrow x=258\)
4 tháng 2 2018
\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\text{Mà }\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\ne0\text{ nên }x-258=0\Leftrightarrow x=258\)
1 tháng 4 2020
Giải các pt sau:
a) (x+4)(2x-3)=0
TH1: x+4=0 => x=-4
TH2 : 2x-3=0 => 2x=3 =>x=3/2
TT
1
TN
13 tháng 11 2016
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10\)
\(\Leftrightarrow\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10-1-2-3-4\)
\(\Leftrightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-166}{23}-4\right)=10-1-2-3-4\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{20}+\frac{x-258}{21}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\right)=0\)
\(\Leftrightarrow x-258=0\).Do \(\frac{1}{17}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\ne0\)
\(\Leftrightarrow x=258\)
8 tháng 7 2017
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
TH
3
22 tháng 2 2017
Giải:
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10\)
\(\Rightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-166}{23}-4\right)\)
\(=10-1-2-3-4=0\)
\(\Rightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Rightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Rightarrow x-258=0\)
\(\Leftrightarrow x=258\)
22 tháng 2 2017
\(\frac{\text{x−241}}{17}+\frac{220}{19}+\frac{x−195}{21}+\frac{x−166}{23}=10\)
\(\Rightarrow\left[\frac{\left(x-241\right)}{17-1}\right]+\left[\frac{\left(x-220\right)}{19-2}\right]+\left[\frac{\left(x-195\right)}{21-3}\right]+\left[\frac{\left(x-166\right)}{23-4}\right]=10-1-2-3-4\)
\(\left(\text{Cộng 2 vế cho -1 - 2 - 3 - 4}\right)\)
\(\Rightarrow\frac{\left(x-258\right)}{17}+\frac{\left(x-258\right)}{19}+\frac{\left(x-258\right)}{21}+\frac{\left(x-258\right)}{23}=0\)
\(\Rightarrow\left(x-258\right).\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Rightarrow x-258=0\Rightarrow x=258\)
d, 2x2-5x-3 = 0
\(\Leftrightarrow\)2x2-6x+x-3= 0
\(\Leftrightarrow\)(2x2-6x) +(x-3) = 0
\(\Leftrightarrow\)2x(x-3) + (x-3) = 0
\(\Leftrightarrow\)(x-3) (2x+1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S =\(\left\{3;\frac{-1}{2}\right\}\)