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\(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)
\(\Leftrightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)
\(\Leftrightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)
\(\Leftrightarrow\left(2018-x\right).\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)
\(\Leftrightarrow2018-x=0\)
\(\Leftrightarrow x=2018\)
Vậy \(x=2018\)
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
a) Từ \(9x=3y=2z\) ta chia các vế cho 18 (là BCNN của 9, 3 và 2) ta được:
\(\frac{9x}{18}=\frac{3y}{18}=\frac{2z}{18}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{6}=\frac{z}{9}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{6}=\frac{z}{9}=\frac{x-y+z}{2-6+9}=\frac{50}{5}=10\)
=> \(\frac{x}{2}=10\Rightarrow x=10.2=20\)
\(\frac{y}{6}=10\Rightarrow y=10.6=60\)
\(\frac{z}{9}=10\Rightarrow z=10.9=90\)
b) Đặt \(k=\frac{x}{5}=\frac{y}{2}=\frac{z}{-3}\)
=> \(x=5k\) ; \(x=2k\) ; \(z=-3k\) (*)
Biết xyz = 240 => \(5k.2k.\left(-3k\right)=240\)
\(\Rightarrow-30k^3=240\)
\(\Rightarrow k^3=-8\)
\(\Rightarrow k=-2\)
Thay vào (*) ta được
\(x=5k=5.\left(-2\right)=-10\)
\(y=2k=-4\)
\(z=-3k=6\)
a)\(\hept{\begin{cases}9x=3y=2z\\x-y+z=50\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{9}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}\\x-y+z=50\end{cases}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{\frac{1}{9}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}=\frac{x-y+z}{\frac{1}{9}-\frac{1}{3}+\frac{1}{2}}=\frac{50}{\frac{5}{18}}=180\)
\(\Rightarrow\hept{\begin{cases}x=20\\y=60\\z=90\end{cases}}\)
b) Đặt \(\frac{x}{5}=\frac{y}{2}=\frac{z}{-3}=k\)
\(\Rightarrow\hept{\begin{cases}x=5k\\y=2k\\z=-3k\end{cases}}\)
xyz = 240 <=> 5k.2k.(-3)k = 240
<=> -30k3 = 240
<=> k3 = -8
<=> k3 = (-2)3
<=> k = -2
=> \(\hept{\begin{cases}x=-10\\y=-4\\z=6\end{cases}}\)
Đặt \(\frac{x}{5}=\frac{y}{2}=\frac{z}{-3}=k\) => x = 5k ; y = 2k ; z = -3k
Ta có :
xyz = 5k . 2k . (-3k) = -30k3 = 240
=> k3 = 240 : (-30) = -8 = -23
=> k = -2
Do đó x = -2 . 5 = -10 ;
y = -2 . 2 = -4 ;
z = -2 . (-3) = 6
Quả nhiên nói cục nam châm hút l-i-k-e là không sai
a) Ta đặt: \(\frac{x}{4}=\frac{y}{3}=\frac{z}{-2}=k\)
\(\Rightarrow x=4k;y=3k;z=-2k\)
\(\Rightarrow xyz=\left(4.3.-2\right).k^3\)
\(\Rightarrow xyz=\left(-24\right).k^3\)
\(\Rightarrow k^3=240:\left(-24\right)=-10\)
\(\Rightarrow\)(đề sai, không ra số tự nhiên)
1, \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)\(\Leftrightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=k\)\(\Leftrightarrow\hept{\begin{cases}x=2k\\y=\frac{3}{2}k\\z=\frac{4}{3}k\end{cases}}\)
Mà xyz = -108
\(\Leftrightarrow2k.\frac{3}{2}k.\frac{4}{3}k=-108\)
\(\Leftrightarrow4k^3=-108\)
<=> k3 = -27
<=> k = -3
\(\Leftrightarrow\hept{\begin{cases}x=2k=2.-3=-6\\y=\frac{3}{2}k=\frac{3}{2}.\left(-3\right)=\frac{-9}{2}\\z=\frac{4}{3}k=\frac{4}{3}.\left(-3\right)=-4\end{cases}}\)
2, \(\frac{x}{5}=\frac{y}{7}=\frac{z}{8}\)\(\Leftrightarrow\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}=\frac{2x+3y-4z}{10+21-32}=\frac{15}{-1}=-15\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=-15\\\frac{y}{7}=-15\\\frac{z}{8}=-15\end{cases}}\Rightarrow\hept{\begin{cases}x=-75\\y=-105\\z=-120\end{cases}}\)
3, 3x = 5y \(\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)\(\Leftrightarrow\frac{x}{55}=\frac{y}{33}\)
2y = 11z \(\Leftrightarrow\frac{y}{11}=\frac{z}{2}\) \(\Leftrightarrow\frac{y}{33}=\frac{z}{6}\)
\(\Rightarrow\frac{x}{55}=\frac{y}{33}=\frac{z}{6}\)\(\Rightarrow\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}=\frac{2x+5y-z}{110+165-6}=\frac{34}{269}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{55}=\frac{34}{269}\\\frac{y}{33}=\frac{34}{269}\\\frac{z}{6}=\frac{34}{269}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1870}{269}\\y=\frac{1122}{269}\\z=\frac{204}{269}\end{cases}}\)
4, \(\frac{x}{3}=\frac{2}{y}=\frac{z}{4}=k\)\(\Leftrightarrow\hept{\begin{cases}x=3k\\y=\frac{2}{k}\\z=4k\end{cases}}\)
Mà xyz = 240
<=> 3k . 2/k . 4k = 240
<=> 24k = 240
<=> k = 10
\(\Leftrightarrow\hept{\begin{cases}x=3k=3.10=30\\y=\frac{2}{k}=\frac{2}{10}=\frac{1}{5}\\z=4k=4.10=40\end{cases}}\)
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
Ta có : \(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)
\(\Leftrightarrow\frac{45-x}{1963}+1+\frac{40-x}{1968}+1+\frac{35-x}{1973}+1+\frac{30-x}{1978}=0\)
\(\Leftrightarrow\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=0\)
\(\Leftrightarrow\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)
Vì \(\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)\ne0\)
Nên : 2008 - x = 0
<=> x = 2008
Vậy x = 2008