Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bạn tham khảo tại đây:
https://hoc24.vn/cau-hoi/giup-minh-voiiiii-minh-cam-on-tim-xy-biet-dfracx4-dfrac2y13-dfracx-2y-1y-voi-y-0.4107067269450
a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)
\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)
Đến đây tự làm tiếp nhé
b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)
=> x = 75, y = 50, z = 30
c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)
\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)
\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)
\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)
=> x=... , y=... , z=...
d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)
Ta có: xy = 90 => 2k.5k = 90 => 10k2 = 90 => k2 = 9 => k = 3 hoặc -3
Với k = 3 => x = 6, y = 15
Với k = -3 => x = -6, y = -15
Vậy...
e, Tương tự câu d
b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)
a) Theo bài ra ta có : \(x+y+z=49\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\\ =\dfrac{12x+12y+12z}{18+16+15}\\ =\dfrac{12\left(x+y+z\right)}{49}\\ =\dfrac{12\cdot49}{49}\\ =12\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{12x}{18}=12\Rightarrow12x=216\Rightarrow x=18\\\dfrac{12y}{16}=12\Rightarrow12y=192\Rightarrow y=16\\\dfrac{12z}{15}=12\Rightarrow12z=180\Rightarrow z=15\end{matrix}\right.\)
\(\text{Vậy }x=18\\ y=16\\ z=15\)
b) Theo bài ra ta có : \(2x+3y-z=50\)
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \Rightarrow\dfrac{2\left(x-1\right)}{4}=\dfrac{3\left(y-2\right)}{9}=\dfrac{z-3}{4}\\ \Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{2x-2}{4}=\dfrac{3y-2}{9}=\dfrac{z-3}{4}=\\ \dfrac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\\ =\dfrac{2x-2+3y-6-z+3}{9}\\ =\dfrac{\left(2x+3y-z\right)-\left(2+6-3\right)}{9}\\ =\dfrac{50-5}{9}\\ =\dfrac{45}{9}\\ =5\\ \)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x-2}{4}=5\Rightarrow2x-2=20\Rightarrow2x=22\Rightarrow x=11\\\dfrac{3y-6}{9}=5\Rightarrow3y-6=45\Rightarrow3y=51\Rightarrow y=17\\\dfrac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow z=23\end{matrix}\right.\)
\(\text{Vậy }x=11\\ y=17\\ z=23\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}\) = \(\dfrac{y}{4}\) = \(\dfrac{z}{3}\) = \(\dfrac{x+2y+3z}{5+8+9}\) = \(\dfrac{x+2y+3z}{22}\)
\(\dfrac{x}{5}\)= \(\dfrac{y}{4}\) = \(\dfrac{z}{3}\) = \(\dfrac{x-2y+3z}{5-8+9}\) = \(\dfrac{x-2y+3z}{6}\)
=> \(\dfrac{x+2y+3z}{22}\) = \(\dfrac{x-2y+3z}{6}\)
=> \(\dfrac{x+2y+3z}{x-2y+3z}\) = \(\dfrac{22}{6}\) =\(\dfrac{11}{3}\)
\(\frac{x-1}{2}=\frac{y-2}{3}\Rightarrow\frac{3\left(x-1\right)}{2}=y-2\Rightarrow y=\frac{3\left(x-1\right)}{2}+2=\frac{3\left(x-1\right)+4}{2}\)(1)
\(\frac{x-1}{2}=\frac{z-3}{4}\Rightarrow\frac{4\left(x-1\right)}{2}=z-3\Rightarrow z=\frac{4\left(x-1\right)}{2}+3=\frac{4\left(x-1\right)+6}{2}\)(2)
Từ (1) và (2) => 2x+3y-z=\(2x+3\left(\frac{3\left(x-1\right)+4}{2}\right)-\frac{4\left(x-1\right)+6}{2}=50\)
\(\Rightarrow\frac{4x}{2}+\frac{9\left(x-1\right)+12}{2}-\frac{4\left(x-1\right)+6}{2}=50\)
\(\Rightarrow\frac{4x+9x-9+12-4x+4-6}{2}=50\)
\(\Rightarrow9x+1=100\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=11\)
Vì \(y=\frac{3\left(x-1\right)+4}{2}=\frac{3\left(11-1\right)+4}{2}=\frac{34}{2}=17\Leftrightarrow y=17\)
Vì \(z=\frac{4\left(x-1\right)+6}{2}=\frac{4\left(11-1\right)+6}{2}+\frac{46}{2}=23\Leftrightarrow z=23\)
Vậy x=11
y=17
z=23
\(\Rightarrow\frac{2\left(x-1\right)}{2.2}=\frac{3\left(y-2\right)}{3.3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng t/c dãy tỉ số = nhau
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{50-2-6+3}{9}=\frac{45}{9}=5\)
\(\Rightarrow\hept{\begin{cases}\frac{x-1}{2}=5\Rightarrow x-1=10\Rightarrow x=11\\\frac{y-2}{3}=5\Rightarrow y-2=15\Rightarrow y=17\\\frac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow z=23\end{cases}}\)
a: \(\dfrac{2x-y}{3x+2y}=\dfrac{5}{2}\)
\(\Leftrightarrow15x+10y=4x-2y\)
=>11x=-12y
=>\(\dfrac{x}{-12}=\dfrac{y}{11}\)
Đặt \(\dfrac{x}{-12}=\dfrac{y}{11}=k\)
=>x=-12k; y=11k
\(P=\dfrac{5x+4y}{25x-y}=\dfrac{5\cdot\left(-12k\right)+4\cdot11k}{25\cdot\left(-12k\right)-11k}=\dfrac{16}{311}\)
b: \(\dfrac{x-5y}{x-3y}=\dfrac{4}{3}\)
=>4x-12y=3x-15y
=>x=-3y
\(\Leftrightarrow\dfrac{x}{-3}=\dfrac{y}{1}=k\)
=>x=-3k; y=k
\(P=\dfrac{x^3+2y^3}{x^3-y^3}=\dfrac{-27k^3+2k^3}{-27k^3-k^3}=\dfrac{-25}{-28}=\dfrac{25}{28}\)
a) Giải
Vì \(5x=2y=3z\)
\(\Rightarrow\dfrac{5x}{30}=\dfrac{2y}{30}=\dfrac{3z}{30}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{x+y-z}{6+15-10}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=3\Rightarrow x=18\\\dfrac{y}{15}=3\Rightarrow y=45\\\dfrac{z}{10}=3\Rightarrow z=30\end{matrix}\right.\)
Vậy \(x=18,\) \(y=45\) hoặc \(z=30.\)
c) Giải
(Vì mk bt bạn bấm nhầm nên đề bị sai, mk sửa 7 \(\rightarrow\) y do trên bàn phím, 7 với y ở vị trí gần nhau mà 2 với y ở cách xa nhau nên sửa như vậy nhé)
Vì \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Rightarrow\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{4-6+12}=\dfrac{x-1-2y+4+3z-9}{10}\)
\(=\dfrac{\left(x-2y+3z\right)-\left(1-4+9\right)}{10}=\dfrac{14-6}{10}=\dfrac{4}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{4}{5}\Rightarrow x=\dfrac{13}{5}\\\dfrac{y-2}{3}=\dfrac{4}{5}\Rightarrow y=\dfrac{22}{5}\\\dfrac{z-3}{4}=\dfrac{4}{5}\Rightarrow z=\dfrac{31}{5}\end{matrix}\right.\)
Vậy \(x=\dfrac{13}{5},\) \(y=\dfrac{22}{5}\) và \(z=\dfrac{31}{5}.\)
c) Giải
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Mà \(x^2+2y^2-z^2=-12\)
\(\Rightarrow\left(2k\right)^2+2\left(3k\right)^2-\left(5k\right)^2=-12\)
\(\Rightarrow4.k^2+18.k^2-25.k^2=-12\)
\(\Rightarrow\left(-3\right)k^2=-12\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
\(\circledast k=-2\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-6\\z=-10\end{matrix}\right.\)
\(\circledast k=2\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=10\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-4;y=-6;z=-10\\x=4;y=6;z=10\end{matrix}\right..\)
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\)\(\dfrac{x}{3}=\dfrac{2y}{8}\)\(\Rightarrow\dfrac{x+2y}{3+8}=\dfrac{66}{11}=6\)
\(\dfrac{x}{3}=6\Rightarrow x=18\)
\(\dfrac{y}{4}=6\Rightarrow y=24\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
`x/3=y/4=(x+2y)/(3+2.4)=66/11=6`
`=>x=6.3=18`
`y=6.4=24`.