\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(x^2+y^2+2z^2=108\)
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MH
17 tháng 9 2021
1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)
⇒ x=4;y=6;z=8
17 tháng 9 2021
\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)
\(2,\) Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)
\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)
\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)
ND
3
HL
24 tháng 10 2017
Ta có: \(\dfrac{x}{2}=\dfrac{x^2}{4}\) ; \(\dfrac{y}{3}=\dfrac{y^2}{9}\) ; \(\dfrac{z}{4}=\dfrac{2z^4}{32}\)
Ta có: \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}=\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{108}{27}=4\)
\(x=4.2=8\)
\(y=4.3=12\)
\(z=4.32=128\)
Vậy 3 số cần tìm là:
x = 8; y = 12; z=128
VT
1 tháng 2 2023
a,Áp sụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\\\Rightarrow x=-3.3=-9\\ \Rightarrow y=-3.5=-15\\ \Rightarrow z=-3.7=-21 \)
1 tháng 2 2023
a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x}{9}=\dfrac{2z}{14}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\) (Vì 3x-2z=15)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-3\\\dfrac{y}{5}=-3\\\dfrac{z}{7}=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-21\end{matrix}\right.\)
Vậy ...
b) Ta có: \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x-3y}{10-9}=\dfrac{100}{1}=100\) (Vì 2x-3y=100)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=100\\\dfrac{y}{3}=100\\\dfrac{z}{2}=100\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=500\\y=300\\z=200\end{matrix}\right.\)
Vậy ...
c) Ta có: \(\dfrac{x}{-3}=\dfrac{y}{-5}=\dfrac{z}{-4}=\dfrac{3z}{-12}=\dfrac{2x}{-6}=\dfrac{3z-2x}{\left(-12\right)-\left(-6\right)}=\dfrac{36}{-18}=-2\) (Vì 3z-2x=36)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-3}=-2\\\dfrac{y}{-5}=-2\\\dfrac{z}{-4}=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\\z=8\end{matrix}\right.\)
Vậy ...
DQ
1
12 tháng 8 2022
Sửa đề: \(x^2-y^2+2z^2=108\)
Đặt x/2=y/3=z/5=k
=>x=2k; y=3k; z=5k
Ta có: \(x^2-y^2+2z^2=108\)
\(\Leftrightarrow4k^2-9k^2+50k^2=108\)
=>45k2=108
=>k2=12/5
TH1: \(k=\dfrac{2\sqrt{3}}{\sqrt{5}}\)
=>\(x=\dfrac{4\sqrt{3}}{\sqrt{5}};y=\dfrac{6\sqrt{3}}{\sqrt{5}};z=2\sqrt{15}\)
TH2: \(k=-\dfrac{2\sqrt{3}}{\sqrt{5}}\)
=>\(x=-\dfrac{4\sqrt{3}}{\sqrt{5}};y=-\dfrac{6\sqrt{3}}{\sqrt{5}};z=-2\sqrt{15}\)
M
1
26 tháng 9 2018
theo bài ra ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
dựa vào tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\\ =\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}\\ =\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{108}{27}=4\)=>x=4.2=8
=>y=4.3=12 =>z=4.4=16 vậy x,y,z lần lượt là 8;12;16
16 tháng 10 2021
1) \(\dfrac{x}{3}=\dfrac{y}{4}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)
\(\Rightarrow xy=12k^2=192\Rightarrow k=\pm4\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm12\\y=\pm16\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-12\\y=-16\end{matrix}\right.\end{matrix}\right.\)
2) Áp dụng t/c dtsbn:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{-90}{9}=-10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-10\right).2=-20\\y=\left(-10\right).3=-30\\z=\left(-10\right).5=-50\end{matrix}\right.\)
3) Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2z}{10}=\dfrac{3x+y-2z}{9+8-10}=\dfrac{14}{7}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.8=16\\z=2.5=10\end{matrix}\right.\)
2 tháng 3 2023
`P=x^3/(x+y)+y^3/(y+z)+z^3/(z+x)`
`=x^4/(x^2+xy)+y^4/(y^2+yz)+z^4/(z^2+zx)`
Ad bđt cosi-swart:
`P>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+xy+yz+zx)`
Mà `xy+yz+zx<=x^2+y^2+z^2)`
`=>P>=(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2))=(x^2+y^2+z^2)/2=3/2`
Dấu "=" xảy ra khi `x=y=z=1`
`Q=(x^3+y^3)/(x+2y)+(y^3+z^3)/(y+2z)+(z^3+x^3)/(z+2x)`
`Q=(x^3/(x+2y)+y^3/(y+2z)+z^3/(z+2x))+(y^3/(x+2y)+z^3/(y+2z)+x^3/(z+2x))`
`Q=(x^4/(x^2+2xy)+y^4/(y^2+2yz)+z^4/(z^2+2zx))+(y^4/(xy+2y^2)+z^4/(yz+2z^4)+x^4/(xz+2x^2))`
Áp dụng BĐT cosi-swart ta có:
`Q>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+2xy+2yz+2zx)+(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2)+xy+yz+zx))`
Mà`xy+yz+zx<=x^2+y^2+z^2`
`=>Q>=(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))+(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2)^2)/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2))/3=2`
Dấu "=" xảy ra khi `x=y=z=1.`
26 tháng 7 2017
a) Ta có: \(6x=4y=3z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{3z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-2}{-4}=\dfrac{1}{2}.\)
Với: \(\dfrac{x}{2}=\dfrac{1}{2}\Rightarrow x=1.\)
\(\dfrac{2y}{6}=\dfrac{y}{3}=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}.3=\dfrac{3}{2}.\)
\(\dfrac{3z}{12}=\dfrac{z}{4}=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}.4=\dfrac{4}{2}=2.\)
Vậy: \(x=1;y=\dfrac{3}{2};z=2.\)
NT
1
25 tháng 5 2022
Ta có \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2x-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}=\dfrac{12x-8y-12x+8y-6z}{29}\)
Do đó:
\(\dfrac{3x-2y}{4}=0\Rightarrow3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\left(1\right)\)
\(\dfrac{2z-4x}{3}=0\Rightarrow2z=4x\Rightarrow\dfrac{x}{2}=\dfrac{z}{4}\left(2\right)\)
Từ (1) và (2) suy ra \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\). Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\Rightarrow x=4;y=6;z=8\)
Theo đề : \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(x^2+y^2+2z^2=108\)
\(\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\left(\dfrac{z}{4}\right)^2\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=2.\left(\dfrac{z}{4}\right)^2=>\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}=\dfrac{x^2+y^2+2z^2}{4+9+32}=\dfrac{108}{45}=\dfrac{12}{5}\)
Với \(\dfrac{x^2}{2}=\dfrac{12}{5}\Rightarrow x^2=\dfrac{12}{5}.2=\dfrac{24}{5}\Rightarrow x=\dfrac{2\sqrt{30}}{5}\)
\(\dfrac{y^2}{3}=\dfrac{12}{5}\Rightarrow y^2=\dfrac{12}{5}.3=\dfrac{36}{5}\Rightarrow y=\dfrac{6\sqrt{5}}{5}\)
\(\dfrac{2z^2}{4}=\dfrac{12}{5}\Rightarrow2z^2=\dfrac{12}{5}.4=\dfrac{48}{5}\Rightarrow z^2=\dfrac{24}{5}=>\dfrac{2\sqrt{30}}{5}\)
tks