Tìm x,y:
\(\dfrac{x-9}{3}\)=\(\dfrac{x+y}{13}\)=\(\dfrac{xy}{200}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
3 tháng 9 2021
a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)
\(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)
b)Ta có: \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)
c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)
Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)
12 tháng 8 2017
Theo đề bài ta có :
\(\dfrac{x-y}{3}=\dfrac{x+y}{13}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{x-y+x+y}{3+13}=\dfrac{2x}{16}=\dfrac{x}{8}\)
\(\dfrac{x}{8}=\dfrac{xy}{200}\Leftrightarrow\) \(\dfrac{x}{xy}=\dfrac{8}{200}\Rightarrow\) \(\dfrac{1}{y}=\dfrac{1}{25}\) \(\Rightarrow y=25\)
Thay y = 25 vào biểu thức ta có :
\(\dfrac{x-25}{3}=\dfrac{x+25}{13}\)
\(\Leftrightarrow\) \(13x-325=3x+75\)
\(\Leftrightarrow13x-3x=75+325\)
\(\Leftrightarrow10x=400\)
\(\Rightarrow x=40\)
Vậy \(x=40\) ; \(y=25\)
MH
10 tháng 1 2022
\(\left(\dfrac{1}{3}.x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\left(\dfrac{1}{3}.x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)
10 tháng 1 2022
b: \(f\left(x\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)
14 tháng 5 2022
a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)
b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)
c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)
=1/3
d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)
23 tháng 11 2021
\(a,\dfrac{1}{3x-3y}=\dfrac{x-y}{3\left(x-y\right)^2};\dfrac{1}{x^2-2xy+y^2}=\dfrac{3}{3\left(x-y\right)^2}\\ b,\dfrac{3}{x^2-3x}=\dfrac{6}{2x\left(x-3\right)};\dfrac{5}{2x-6}=\dfrac{5x}{2x\left(x-3\right)}\\ c,\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{3-x}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{x^2-9}=\dfrac{1}{\left(x-3\right)\left(x+3\right)}\)
\(d,\dfrac{1}{x^2+xy}=\dfrac{xy-y^2}{xy\left(x+y\right)\left(x-y\right)};\dfrac{1}{xy-y^2}=\dfrac{x^2+xy}{xy\left(x-y\right)\left(x+y\right)};\dfrac{2}{y^2-x^2}=\dfrac{-2xy}{xy\left(x-y\right)\left(x+y\right)}\)
DD
2
22 tháng 10 2021
Áp dụng BĐT cosi:
\(A=\left(3x+\dfrac{3}{x}\right)+\left(\dfrac{4}{9}y+\dfrac{4}{y}\right)+\left(2x+y\right)\\ A\ge2\sqrt{\dfrac{9x}{x}}+2\sqrt{\dfrac{16y}{9y}}+5\\ A\ge2\cdot3+2\cdot\dfrac{4}{3}+5=\dfrac{41}{3}\)
Vậy \(A_{min}=\dfrac{41}{3}\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{3}{x}\\\dfrac{4y}{9}=\dfrac{4}{y}\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
đề có sai ko bạn
bạn coi kĩ lại đi
cái chỗ \(x-9\) ấy hay là x-y vậy? bạn coi kĩ lại giúp mk