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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 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:....................
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
a,3x=2y;7y=5z
=>\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta co:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\\ \Rightarrow x=2.10=20\\ y=2.15=30\\ z=2.21=42\)
Các câu sau tương tự
b,\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\),\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\) và 2x-3y+z=6
Từ đề bài ta có:
\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)\(\Rightarrow\)\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)\(\Rightarrow\)\(\dfrac{y}{12}\)=\(\dfrac{z}{20}\)(2)
từ (1) và (2)\(\Rightarrow\)\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{20}\)\(\Rightarrow\)\(\dfrac{2x}{18}\)=\(\dfrac{3y}{36}\)=\(\dfrac{z}{20}\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\dfrac{2x}{18}\)=\(\dfrac{3y}{36}\)=\(\dfrac{z}{20}\)=\(\dfrac{2x-3y+z}{18-36+20}\)=\(\dfrac{6}{2}\)=3
\(\Rightarrow\)x=3.9=27
y=3.12=36
z=3.20=60
Vậy.....
chúc bạn học tốt,nhớ tick cho mình nha
Ta có : \(\dfrac{x}{y}=\dfrac{3}{4};\dfrac{y}{z}=\dfrac{5}{6}\)
\(\Rightarrow4x=3y;6y=5z\)
\(\Leftrightarrow8x=6y=5z.\)
\(\Rightarrow y=\dfrac{8x}{6}=\dfrac{4}{3}x\)
Thay vào A ta có :
\(A=\dfrac{2x+3y+5z}{y+5z}=\dfrac{2x+4x+8x}{\dfrac{4}{3}x+8x}=\dfrac{3}{2}.\)
a) Ta có:
\(x+y+z=49\Rightarrow12x+12y+12z=588\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12x+12y+12z}{18+16+15}=\dfrac{588}{49}=12\)
\(\Rightarrow\left\{{}\begin{matrix}x=12.3:2\\y=12.4:3\\z=12.5:4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)
a/ Do \(x+y=22\Rightarrow y=22-x\)
\(\Rightarrow\dfrac{4+x}{7+22-x}=\dfrac{4}{7}\Leftrightarrow\dfrac{4+x}{29-x}=\dfrac{4}{7}\)
\(\Leftrightarrow7\left(4+x\right)=4\left(29-x\right)\Leftrightarrow28+7x=116-4x\)
\(\Leftrightarrow11x=88\Rightarrow x=8\)
\(\Rightarrow y=22-x=14\)
b/ \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow y=\dfrac{4x}{3}\)
\(\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow z=\dfrac{6y}{5}\) \(\Rightarrow z=\dfrac{6}{5}\left(\dfrac{4x}{3}\right)=\dfrac{8x}{5}\)
Vậy \(M=\dfrac{2x+3y+4z}{3x+4y+5z}=\dfrac{2x+3.\dfrac{4x}{3}+4.\dfrac{8x}{5}}{3x+4.\dfrac{4x}{3}+5.\dfrac{8x}{5}}\)
\(\Rightarrow M=\dfrac{x\left(2+4+\dfrac{32}{5}\right)}{x\left(3+\dfrac{16}{3}+8\right)}=\dfrac{\dfrac{62}{5}}{\dfrac{49}{3}}=\dfrac{186}{245}\)
Câu a:
Ta có: \(x+y=22\Rightarrow y=22-x\)
\(\Rightarrow\dfrac{4+x}{7+22-x}=\dfrac{4}{7}\Leftrightarrow\dfrac{4+x}{29-x}=\dfrac{4}{7}\)
\(\Leftrightarrow7\left(4+x\right)=4\left(29-x\right)\Leftrightarrow28+7x=116-4x\)
\(\Leftrightarrow11x=88\Rightarrow x=8\)
\(\Rightarrow y=22-x=22-8=14\)
Vậy \(x=8,y=14\)
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..