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Ta có: \(\dfrac{x}{3}\)=\(\dfrac{y}{4}\) ; \(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)
=>\(\dfrac{x}{15}\)=\(\dfrac{y}{20}\)=\(\dfrac{z}{24}\)=k
=>x=15k
y=20k
z=24k
Thế x=15k; y=20k; z=24k vào biểu thức A, ta có:
\(\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}\)=\(\dfrac{30k+60k+96k}{45k+60k+120k}\)=\(\dfrac{k.\left(30+60+96\right)}{k.\left(45+60+120\right)}\)=\(\dfrac{186}{225}\)=\(\dfrac{62}{75}\)
x/3=y/4 -> y=4x/3 (1)
y/5=z/6 -> y=5z/6 (2)
(1)+(2) -> x=5z/8 thay vào M=\(\dfrac{2.\dfrac{5z}{8}+3.\dfrac{5z}{6}+4z}{3.\dfrac{5z}{8}+4.\dfrac{5z}{6}+5z}\)=\(\dfrac{186}{245}\)
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 \(x=8,y=14\)
a)ta có 4+x/7+y=4/7
<=>7x+28=28+4y
<=> 7x=4y
lại có x+y=22
=>4/7y+y=22
<=>11/7y=22 <=> y=14
<=> x= 4/7*14=8
vậy x=8, y=14
b) Từ x/3=y/4 va y/5=z/6-->x/15=y/20=z/24 (1)
(1) = 2x/30=3y/60=4z/96=(2x+3y+4z)/186 (2) (t/c dãy tỉ số bằng nhau)
Ta lại có
(1) = 3x/45=4y/80=5z/120=(3x+4y+5z)/245 (3)(t/c dãy tỉ số bằng nhau)
Từ (2)(3) ta có(2x+3y+4z)/186=(3x+4y+5z)/245
Vậy M = (2x+3y+4z)/(3x+4y+5z)=186/245
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...
a) ta có : \(\dfrac{x}{2}\) = \(\dfrac{y}{3}\) = \(\dfrac{x}{16}=\dfrac{y}{24}\) ( 1)
\(\dfrac{y}{8}=\dfrac{z}{5}\) = \(\dfrac{y}{24}=\dfrac{z}{15}\) (2)
từ (1) và (2) , ta có : \(\dfrac{x}{16}=\dfrac{y}{24}=\dfrac{z}{15}\)
mà x - y + z = 35
theo tính chất của dãy tỉ số bằng nhau , ta có :
\(\dfrac{x}{16}=\dfrac{y}{24}=\dfrac{z}{15}=\dfrac{x-y+z}{16-24+15}=\dfrac{35}{7}=5\)
do đó : \(\dfrac{x}{16}=5\) => x = 5. 16 = 80
\(\dfrac{y}{24}=5\) => y = 5.24 = 120
\(\dfrac{z}{15}=5\) => z = 5.15 = 75
vậy x = 80
y = 120
z = 75
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
a) Ta có:
\(x+y+z=49\Rightarrow12x+12y+12z=588\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12x+12y+12z}{18+16+15}=\dfrac{588}{49}=12\)
\(\Rightarrow\left\{{}\begin{matrix}x=12.3:2\\y=12.4:3\\z=12.5:4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)
Giải:
Ta có:
\(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}.\)
\(\Rightarrow\dfrac{3\left(2x-4y\right)}{3^2}=\dfrac{2\left(4z-3x\right)}{2^2}=\dfrac{4\left(3y-2z\right)}{4^2}.\)
\(\Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}.\)
\(=\dfrac{6x-12y+8z-6x+12y-8z}{9+4+16}.\)
\(=\dfrac{\left(6x-6x\right)+\left(8z-8z\right)+\left(12y-12y\right)}{19}=0.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}.\\4z=3x\Rightarrow\dfrac{z}{3}=\dfrac{x}{4}.\\3y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{3}.\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{2x}{8}=\dfrac{y}{2}=\dfrac{z}{3}_{\left(1\right)}\) và \(2x-y+z=27_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ số bằng nhau có:
\(\dfrac{2x}{8}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3.\)
Từ đó: \(\left\{{}\begin{matrix}2x=3.8=24\Rightarrow x=12.\\y=3.2=6.\\z=3.3=9.\end{matrix}\right.\)
Vậy.....
\(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}\\ \Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}\\ =\dfrac{6x-12y+8z-6x+12y-8z}{9+4+16}=\dfrac{0}{29}=0\\ \Rightarrow2x=4y;4z=3x;3y=2z\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3\\ \Rightarrow x=12;y=6;z=9\)
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\Rightarrow x=15k;y=20k;z=24k\)
\(M=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186}{245}\)