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\(\dfrac{55-x}{1963}\) + \(\dfrac{50-x}{1968}\) + \(\dfrac{45-x}{1973}\) + \(\dfrac{40-x}{1978}\) + 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
\(\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+4=0\)
\(\Rightarrow\text{ }\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+1+1+1+1=0\)
\(\Rightarrow\text{ }\left(\dfrac{55-x}{1963}+1\right)+\left(\dfrac{50-x}{1968}+1\right)+\left(\dfrac{45-x}{1973}+1\right)+\left(\dfrac{40-x}{1978}+1\right)=0\)
\(\Rightarrow\text{ }\dfrac{2018-x}{1963}+\dfrac{2018-x}{1968}+\dfrac{2018-x}{1973}+\dfrac{2018-x}{1978}=0\)
\(\Rightarrow\text{ }\left(2018-x\right)\left(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\right)=0\)
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, \(x=2018.\)
Câu 2:
\(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\)
\(\Leftrightarrow\left(\dfrac{x-10}{30}-3\right)+\left(\dfrac{x-14}{43}-2\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-148}{8}+6\right)=0\)
=>x-100=0
hay x=100
Đề sai sửa lại và làm:
Ta có:
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=-4\)
\(\Leftrightarrow\left(\dfrac{315-x}{101}+1\right)+\left(\dfrac{313-x}{103}+1\right)+\left(\dfrac{311-x}{105}+1\right)+\left(\dfrac{309-x}{107}+1\right)=0\)
\(\Leftrightarrow\dfrac{416-x}{101}+\dfrac{416-x}{103}+\dfrac{416-x}{105}+\dfrac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\right)=0\)
\(\Leftrightarrow416-x=0\)
\(\Leftrightarrow x=416\)
VẬY....
Lời giải:
PT \(\Leftrightarrow \frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=4\)
\(\Leftrightarrow \frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=4\)
\(\Leftrightarrow (416-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=4\)
\(\Rightarrow 416-x=\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
\(\Rightarrow x=416-\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
Lời giải:
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=4\)
\(\Leftrightarrow \frac{315-x}{101}-1+\frac{313-x}{103}-1+\frac{311-x}{105}-1+\frac{309-x}{107}-1=0\)
\(\Leftrightarrow \frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow (416-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra $416-x=0$
\(\Rightarrow x=416\)
=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)
=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)
=>x=-36
a, \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{10}{21}\)
(\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\) = \(\dfrac{10}{21}\)
- \(\dfrac{5}{21}\) \(\times\) \(x\) = \(\dfrac{10}{21}\)
\(x\) = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))
\(x\) = -2
b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)
\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))
\(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)
\(x\) = - \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)
\(x\) = - \(\dfrac{13}{6}\)
c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)
3\(x\) - \(\dfrac{3}{2}\) - 5\(x\) - 3 = - \(x\) + \(\dfrac{1}{5}\)
- \(x\) + 5\(x\) - 3\(x\) = - \(\dfrac{3}{2}\) - 3 - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{47}{10}\)
\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)
\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)
`(x+98)/2+(x+45)/55+(x-1)/101=(x+5)/95+(x+30)/70+(x-3)/103`
`=>((x+98)/2+1)+((x+45)/55+1)+((x-1)/101+1)=((x+5)/95+1)+((x+30)/70+1)+((x-3)/103+1)`
`=>(x+100)/2+(x+100)/55+(x+100)/101=(x+100)/95+(x+100)/70+(x+100)/103`
`=>(x+100)/2+(x+100)/55+(x+100)/101-(x+100)/95-(x+100)/70-(x+100)/103=0`
`=>(x+100)(1/2+1/55+1/10-1/96-1/70-1/103)=0`
`=>x+100=0`
`=>x=-100`
\(=\left(\dfrac{x+98}{2}+1\right)+\left(\dfrac{x+45}{55}+1\right)+\left(\dfrac{x-1}{101}+1\right)=\left(\dfrac{x+5}{95}+1\right)+\left(\dfrac{x+30}{70}\right)+\left(\dfrac{x-3}{103}\right)\)
\(=\dfrac{x+100}{2}+\dfrac{x+100}{ }+\dfrac{x+100}{ }\)