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a) Có x:y:z=3:5:6
\(\Rightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{6}\)
Đặt \(k=\frac{x}{3}=\frac{y}{5}=\frac{z}{6}\)
\(\Rightarrow x=3k\)
\(\Rightarrow y=5k\)
\(\Rightarrow z=6k\)
Thay vào \(\frac{2x-3y+4z}{x-11y-4z}=\frac{2.3k-3.5k+4.6k}{3k-11.5k-4.6k}\)\(=\frac{k.\left(2.3-3.5+4.6\right)}{k.\left(3-11.5-4.6\right)}=\frac{k.15}{k.\left(-76\right)}=\frac{15}{-76}\)
b) Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{1+2y}{18}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}\)\(=\frac{2+8y}{18+6x}=\frac{2.\left(1+4y\right)}{2.\left(9+3x\right)}=\frac{1+4y}{9+3x}\)
\(\Rightarrow\frac{1+4y}{9+3x}=\frac{1+4y}{24}\Rightarrow9+3x=24\Rightarrow x=5\)
Theo đề bài, ta có:
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{2x}{6}\)=\(\dfrac{3y}{15}\)=\(\dfrac{4z}{24}\)
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{x}{3}\)=\(\dfrac{11y}{55}\)=\(\dfrac{4z}{24}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{2x}{6}\)=\(\dfrac{3y}{15}\)=\(\dfrac{4z}{24}\)= \(\dfrac{2x-3y+4z}{6-15+24}\)=\(\dfrac{2x-3y+4z}{15}\)(*)
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{x}{3}\)=\(\dfrac{11y}{55}\)=\(\dfrac{4z}{24}\)=\(\dfrac{x-11y-4z}{3-55-24}\)=\(\dfrac{x-11y-4z}{-76}\)(**)
Từ (*) và (**) suy ra:
\(\dfrac{2x-3y+4z}{15}\)=\(\dfrac{x-11y-4z}{-76}\)=\(\dfrac{2x-3y+4z}{x-11y-4z}\)=\(\dfrac{15}{-76}\)
=> m=\(\dfrac{15}{-76}\)
Vậy m=\(\dfrac{15}{-76}\)
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
Xét \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) (1)
Thay (1) vào P
=> P = \(\dfrac{3k+2.4k+3.5k}{2.5k+3.4k+4.5k}+\dfrac{2.5k+3.4k+4.5k}{3.3k+4.4k+5.5k}\) + \(\dfrac{3.3k+4.4k+5.5k}{4.3k+5.4k+6.5k}\)
=> P = \(\dfrac{26k}{42k}+\dfrac{42k}{50k}\) + \(\dfrac{50k}{62k}\)
=> P = \(\dfrac{13}{21}+\dfrac{21}{25}+\dfrac{25}{31}\approx2,265499232\)
2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+6y}{18+6x}=\dfrac{2\left(1+4y\right)}{2\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)
⇒ \(\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{28}\)
⇒\(9+3x=28\)
⇒\(3x=19\)
⇒\(x=\dfrac{19}{3}\)
bạn thay vào là tìm được y
\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)
\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)
Bài 1:
Ta có: \(3x=2y\)
nên \(\dfrac{x}{2}=\dfrac{y}{3}\)
mà x+y=-15
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)
Vậy: (x,y)=(-6;-9)
Bài 2:
a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)
mà x+y-z=20
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)
Vậy: (x,y,z)=(40;30;50)
a: Vì \(2.8\cdot0.4=1.4\cdot0.8\)
nên 2,8/0,8=1,4/0,4; 2,8/1,4=0,8/0,4; 0,8/2,8=0,4/1,4; 1,4/2,8=0,4/0,8
b: Vì x,y,z tỉ lệ với 3;5;6 nên x/3=y/5=z/6=k
=>x=3k; y=5k; z=6k
\(M=\dfrac{2x-3y+4z}{x-11y-4z}=\dfrac{6k-15k+24k}{3k-55k-24k}=\dfrac{-15}{76}\)