Giúp mình với??:(
Tìm x; y; z biết :
1) x/2 = y/3 ; y/4 = z/5 và x – y + z = 10
2) 4x = 3y ; 7y = 5z và 2x + 3y - z= 136
3) x-3/5 = y-5/1 = z+3/7 và 3x + 5y - 7z = 100
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\(\Leftrightarrow164-4\left(x-5\right)=80\\ \Leftrightarrow4\left(x-5\right)=84\\ \Leftrightarrow x-5=21\Leftrightarrow x=26\)
x x 3 + : 0,5=12,8
x x (3 + 2)=12,8
x x 5=12,8
x = 12,8 :5
x = 2,56
`5(x+2) -2x=18`
`=> 5x+10 -2x=18`
`=> 3x+10=18`
`=>3x=18-10`
`=>3x=8`
`=>x=8/3`
=>5x+10-2x=18
=>3x+10=18
=>3x=18-10=8
=>x=\(\dfrac{8}{3}\)
=>(x-2023)[(x-2023)^21-1]=0
=>x-2023=0 hoặc x-2023=1
=>x=2023 hoặc x=2024
\(2x-49=5.32\\ \Leftrightarrow2x-49=160\\ \Leftrightarrow2x=209\\ \Leftrightarrow x=\dfrac{209}{2}\)
\(200-\left(2x+6\right)=43\\ \Leftrightarrow2x+6=157\\ \Leftrightarrow2x=151\\ \Leftrightarrow x=\dfrac{151}{2}\)
\(135-5\left(x+4\right)=35\\ \Leftrightarrow5\left(x+4\right)=100\\ \Leftrightarrow x+4=20\\ \Leftrightarrow x=16\)
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)