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Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Rightarrow x=-4k;y=-7k;z=3k\) (1)
Thay (1) vào A , ta được
\(A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{2\left(-4k\right)-3\left(-7k\right)-6.3k}\)
\(\Rightarrow A=\dfrac{8k+\left(-7k\right)+15k}{-8k+21k+\left(-18k\right)}\)
\(\Rightarrow A=\dfrac{k[8+\left(-7\right)+15]}{k[-8+21+\left(-18\right)]}\)
\(\Rightarrow A=\dfrac{16k}{-5k}\)
\(\Rightarrow A=\dfrac{16}{5}\)
Vậy \(A=\dfrac{16}{5}\)
\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{-2\left(-4k\right)-7k+5.3k}{2.\left(-4k\right)-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-5k}=-\dfrac{16}{5}\)
Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Rightarrow x=-4k;y=-7k;z=3k\)(1)
Thay (1) vào ta có :
\(A=\dfrac{-2x+y+5z}{2x-3y-6z}=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{2.\left(-4k\right)-3.\left(-7k\right)-6.3k}=\dfrac{8k+-7k+15k}{\left(-8k\right)-\left(-27k\right)-18k}=\dfrac{k\left(8+-7+15\right)}{k\left(-8+27-18\right)}=\dfrac{16}{17}\)
a. Có \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\) => \(\dfrac{x}{4}=\dfrac{3x}{9}=\dfrac{4z}{36}\) và x-3y+4z=62
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{4}=\dfrac{3y}{9}=\dfrac{4z}{36}\)= \(\dfrac{x-3y+4z}{4-9+36}=\dfrac{62}{31}=2\)
=> x=8
3y=18=>y=6
4z=72=>z=18
Vậy x=8 ; y=6 ; z=18
b, Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\\ =\dfrac{2x+3y-5z}{4+9-20}=\dfrac{-21}{-7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot4=12\end{matrix}\right.\\ vậy...\)
Câu c bạn làm tương tự nhé!
d, Ta có : \(\left|x+y-z\right|=95\Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\)
\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\\ =\dfrac{x+y-z}{15+10-6}=\dfrac{x+y-z}{19}\\ \Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=15\cdot5=75\\y=10\cdot5=50\\z=6\cdot5=30\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5\cdot15=-75\\y=-5\cdot10=-50\\z=-5\cdot6=-30\end{matrix}\right.\end{matrix}\right.\)
Vậy...
Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\left(k\ne0\right)\)
=>\(\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
Ta có P =\(\dfrac{-2\cdot\left(-4k\right)+\left(-7k\right)+5\cdot3k}{2\cdot\left(-4k\right)-3\left(-7k\right)-6\left(3k\right)}\)=\(\dfrac{8k+\left(-7k\right)+15k}{-8k+21k-18k}\)=
\(\dfrac{k\cdot\left(8+\left(-7\right)+15\right)}{k.\left(-8+21-18\right)}=\dfrac{-16}{5}\)
Vậy P= \(\dfrac{-16}{5}\)
Theo đề ta có:
\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}\)
Đặt k cho biểu thức trên
=>\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}\) =k
=> \(\left[{}\begin{matrix}\dfrac{x}{-4}=k\\\dfrac{y}{-7}=k\\\dfrac{z}{3}=k\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-4\right).k\\y=\left(-7\right).k\\z=3.k\end{matrix}\right.\)
Thay \(\left[{}\begin{matrix}x=\left(-4\right).k\\y=\left(-7\right).k\\z=3.k\end{matrix}\right.\) vào biểu thức \(P=\dfrac{-2x+y+5z}{2x-3y-6z}\)
Ta được:
\(P=\dfrac{-2.\left(-4.k\right)+\left(-7.k\right)+5\left(3.k\right)}{2\left(-4.k\right)-3\left(-7.k\right)-6\left(3.k\right)}\)
=> \(P=\dfrac{8.k+\left(-7.k\right)+15.k}{-8.k+21.k-18.k}\)
=> \(P=\dfrac{k.\left(8+-7+15\right)}{k.\left(-8+21-18\right)}\)
=> P= \(-\dfrac{16}{5}\)
Vậy:....................
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\Rightarrow x=15k;y=20k;z=24k\)
\(M=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186}{245}\)
a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
⇒2x = 3.30 = 90 ⇒ x = 45
3y = 3.60 = 180 ⇒ y = 60
z = 3.28 = 84
Ý b) có gì đó sai sai ?
c)Ta có :
\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
⇒x = 5.15 = 75
y = 5.10 = 50
z = 5.6 = 30
d)Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)
⇒ x = 2k ; y = 3k ; z = 5k
⇒ xyz = 2k.3k.5k = 30k3 = 810
⇒ k = 3 Vậy x = 3.2 = 6; y = 3.3 = 9; z = 3.5 = 15
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 : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
Đặt :
\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
Thay vào A ta được :
\(A=\dfrac{-2.-4k-7k+5.3k}{2.-4k+3.7k-6.3k}\)
\(=\dfrac{8k-7k+15k}{-8k+21k-18k}=\dfrac{16k}{-5k}=-\dfrac{16}{5}\)
Vậy..