mọi người giúp mình bài này với ạ
bài 1: tìm x biết:
\(\dfrac{55-x}{1963}\)+\(\dfrac{50-x}{1968}\) +\(\dfrac{45-x}{1973}\)+\(\dfrac{40-x}{1978}\)+4= 0
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BN
1
Những câu hỏi liên quan
TD
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 5 2023
\(\dfrac{55-x}{1963}\) + \(\dfrac{50-x}{1968}\) + \(\dfrac{45-x}{1973}\) + \(\dfrac{40-x}{1978}\) + 4 = 0
(1 + \(\dfrac{55-x}{1963}\) ) + ( 1 + \(\dfrac{50-x}{1968}\)) + (1+ \(\dfrac{45-x}{1973}\))+ (1 + \(\dfrac{40-x}{1978}\)) = 0
\(\dfrac{1963+55-x}{1963}\) + \(\dfrac{1968+50-x}{1968}\)+\(\dfrac{1973+45-x}{1973}\)+\(\dfrac{1978+40-x}{1978}\)=0
\(\dfrac{2018-x}{1963}\)+\(\dfrac{2018-x}{1968}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1978}\)=0
(2018 - \(x\))\(\times\)( \(\dfrac{1}{1963}\)+\(\dfrac{1}{1986}\)+\(\dfrac{1}{1973}\)+) =0
2018 \(-x\) = 0
\(x\) = 2018
DN
17 tháng 5 2023
\(\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+4=0\)
\(\Rightarrow\text{ }\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+1+1+1+1=0\)
\(\Rightarrow\text{ }\left(\dfrac{55-x}{1963}+1\right)+\left(\dfrac{50-x}{1968}+1\right)+\left(\dfrac{45-x}{1973}+1\right)+\left(\dfrac{40-x}{1978}+1\right)=0\)
\(\Rightarrow\text{ }\dfrac{2018-x}{1963}+\dfrac{2018-x}{1968}+\dfrac{2018-x}{1973}+\dfrac{2018-x}{1978}=0\)
\(\Rightarrow\text{ }\left(2018-x\right)\left(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\right)=0\)
Mà \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\)
\(\Rightarrow\text{ }2018-x=0\)
\(\Rightarrow\text{ }x=2018-0\)
\(\Rightarrow\text{ }x=2018\)
Vậy, \(x=2018.\)
1 tháng 12 2017
\(\dfrac{45-x}{1963}+\dfrac{40-x}{1968}+\dfrac{35-x}{1973}+\dfrac{30-x}{1978}+4=0\)
\(\Rightarrow\dfrac{45-x}{1963}+\dfrac{40-x}{1968}+\dfrac{35-x}{1973}+\dfrac{30-x}{1978}=-4\)
\(\Rightarrow\left(\dfrac{45-x}{1963}+1\right)+\left(\dfrac{40-x}{1968}+1\right)+\left(\dfrac{35-x}{1973}+1\right)+\left(\dfrac{30-x}{1978}+1\right)=-4+1+1+1+1\)
\(\Rightarrow\dfrac{2008-x}{1963}+\dfrac{2008-x}{1968}+\dfrac{2008-x}{1973}+\dfrac{2008-x}{1978}=0\)
Vì \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\) nên 2008 - x = 0
\(\Rightarrow x=2008\)
NV
1
19 tháng 6 2020
\(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)
\(\Leftrightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)
\(\Leftrightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)
\(\Leftrightarrow\left(2018-x\right).\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)
\(\Leftrightarrow2018-x=0\)
\(\Leftrightarrow x=2018\)
Vậy \(x=2018\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
19 tháng 6 2020
Dễ dàng :v
Có \(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)
\(\Rightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)
\(\Rightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)
\(\Rightarrow\left(2018-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)
Mà \(\Rightarrow\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)>0\Rightarrow2018-x=0\)
\(\Rightarrow x=2018-8=2018\)
Vậy x = 2018
PB
1
24 tháng 12 2017
Đề sai:
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\)
\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1983}+1\right)=0\)
\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\)
\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)
Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên \(2013-x=0\Leftrightarrow x=2013\)
US
2
DT
29 tháng 3 2018
45-x/1963+40-x/1968+35-x/1973+30-x/1978+4=0
45-x/1963+40-x/1968+35-x/1973+30-x/1978=-4
(45-x/1963+1)+(40-x/1968+1)+(35-x/1973+1)+(30-x/1978+1)=-4+1+1+1+1
2008-x/1963+2008-x/1968+2008-x/1973+2008-x/1978=0
(2008-x).(1/1963+1/1968+1/1973+1/1978)=0
Vì 1/1963+1/1968+1/1973+1/1978 khac o
2008-x=0
x=2008
k cho minh dau tien nha!
CE
0
14 tháng 7 2017
Các câu dễ tự làm :v
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1981}=-4\) (sau khi đã sửa đề)
\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1981}+1\right)=0\)\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1981}=0\)
\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1981}\right)=0\)
\(\Rightarrow2013-x=0\Rightarrow x=2013\)
\(1+5+9+13+17+.....+x=5050\)
Số các số hạng là:
\(\dfrac{x-1}{4}+1=\dfrac{1}{4}x+\dfrac{3}{4}\)
Như vậy có :
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\) số hạng
Theo đề bài ta có:
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\left(x+1\right)=5050\)
\(\Rightarrow\left(\dfrac{1}{4}x+\dfrac{3}{4}\right)\left(x+1\right)=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+\dfrac{1}{4}x+\dfrac{3}{4}x+\dfrac{3}{4}=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+x+\dfrac{3}{4}=10100\)
Kiệt sức.đến đây ko nghĩ nổi nx
MV
14 tháng 7 2017
a,
\(5^x+5^{x+2}=650\\ 5^x\left(1+5^2\right)=650\\ 5^x\cdot26=650\\ 5^x=25\\ 5^x=5^2\\ \Rightarrow x=2\)
Vậy \(x=2\)
b,
\(\left(x+2\right)^2=81\\ \Rightarrow\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
Vậy \(x=7\) hoặc \(x=-11\)
d,
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\\ \dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}+4=0\\ \dfrac{45-x}{1968}+1+\dfrac{40-x}{1973}+1+\dfrac{35-x}{1978}+1+\dfrac{30-x}{1983}+1=0\\ \dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\\ \left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)
Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên
\(2013-x=0\\ x=2013\)
Vậy \(x=2013\)
e,
\(\dfrac{1}{2016}:2015x=\dfrac{-1}{2015}\\ 2015x=\dfrac{-2015}{2016}\\ x=\dfrac{-1}{2016}\)
Vậy \(x=\dfrac{-1}{2016}\)
27 tháng 2 2018
196345−x+196840−x+197335−x+197830−x=−4
\left(\frac{45-x}{1963}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0(196345−x+1)+(196840−x+1)+(197335−x+1)+(197830−x+1)=0
\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=019632008−x+19682008−x+19732008−x+19782008−x=0
\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0(2008−x)(19631+19681+19731+19781)=0
=> 2008 - x = 0 ( vì 1/ 1963 + ... khác 0 )
=> x = 2008
Lời giải:
$\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0$
$\frac{55-x}{1963}+1+\frac{50-x}{1968}+1+\frac{45-x}{1973}+1+\frac{40-x}{1978}+1=0$
$\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0$
$(2018-x)(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978})=0$
$\Rightarrow 2018-x=0$
$\Rightarrow x=2018$.