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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 \(\frac{1}{17}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\ne0\)
\(\Leftrightarrow x=258\)
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 :)
PN
3 tháng 1 2016
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10\)
\(\Leftrightarrow\) \(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}-10=0\)
\(\Leftrightarrow\) \(\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-166}{23}-4\right)=0\)
\(\Leftrightarrow\) \(\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-166}{23}=0\)
\(\Leftrightarrow\) \(\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{21}\right)=0\)
\(\Leftrightarrow\) \(x-258=0\) \(\Leftrightarrow\) \(x=258\)
17 tháng 1 2017
Ta có: \(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}=10\)
\(\Rightarrow\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}-10=0\)
\(\Leftrightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-170}{23}-4\right)=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow\left(x-258\right)=0\)
\(\Rightarrow x=258\)
Vậy x=258
N
30 tháng 12 2018
\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+16}{84}+1\right)+\left(\frac{x+17}{83}+1\right)+\left(\frac{166}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(x+100\right)=0\Rightarrow x=-100\left(\text{vì }\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)
NQ
1
12 tháng 9 2019
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-163}{23}=10\)
\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow x=258\)
Vậy \(x=258\)
Chúc bạn học tốt !!!
H
1
NT
Nguyễn thành Đạt
CTVHS
30 tháng 1 2023
\(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)
\(\Leftrightarrow\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}-10=0\)
\(\Leftrightarrow(\dfrac{x-241}{17}-1)+(\dfrac{x-220}{19}-2)+(\dfrac{x-195}{21}-3)+(\dfrac{x-166}{23}-4)=0\)
\(\Leftrightarrow\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\)
\(Do\) \(\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)\ne0\) \(nên\) \(để\) \(gt=0\)
\(\Leftrightarrow x-258=0\)
\(\Leftrightarrow x=258\)
\(Vậy...\)
1 tháng 5 2019
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 =\(\left\{3;\frac{-1}{2}\right\}\)
T
0
\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right).\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow x-258=0\)
\(\Leftrightarrow x=258\)
\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\text{Mà }\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\ne0\text{ nên }x-258=0\Leftrightarrow x=258\)