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\(\dfrac{2x}{5}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
\(\Rightarrow\dfrac{2}{5}x=\dfrac{3}{4}y=\dfrac{4}{5}z\)
\(\Rightarrow\dfrac{2}{5}x.\dfrac{1}{12}=\dfrac{3}{4}y.\dfrac{1}{12}=\dfrac{4}{5}z.\dfrac{1}{12}\)
\(\Rightarrow\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}\)
Đặt \(\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}=k\Rightarrow\left\{{}\begin{matrix}x=30k\\y=16k\\z=15k\end{matrix}\right.\). Ta có:
\(x+y+z=49\)
\(\Rightarrow30k+16k+15k=49\)
\(\Rightarrow61k=49\)
\(\Rightarrow k=\dfrac{49}{61}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{49}{61}.30=\dfrac{1470}{61}\\y=\dfrac{49}{61}.16=\dfrac{784}{61}\\z=\dfrac{49}{61}.15=\dfrac{735}{61}\end{matrix}\right.\)
a) Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{x-3y+4z}{4-3.3+4.9}=\dfrac{63}{31}=2\)
\(\Rightarrow x=8\)
\(\Rightarrow y=6\)
\(\Rightarrow z=18\)
b. c. Xem lại đề.
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
\(\dfrac{3}{2}x=\dfrac{4}{3}y=\dfrac{5}{4}z\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}\\ \Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{2y}{\dfrac{3}{2}}=\dfrac{z}{\dfrac{4}{5}}=\dfrac{x-2y+z}{\dfrac{2}{3}-\dfrac{3}{2}+\dfrac{4}{5}}=-\dfrac{16}{-\dfrac{1}{30}}=480\)
suy ra: \(x=\dfrac{480}{\dfrac{2}{3}}=720\\ 2y=\dfrac{480}{\dfrac{3}{2}}=320\Rightarrow y=160\\ z=\dfrac{480}{\dfrac{4}{5}}=600\)
Ta có:
\(\dfrac{3}{2}x=\dfrac{4}{3}y=\dfrac{5}{4}z\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}=\dfrac{2y}{\dfrac{3}{2}}=\dfrac{x-2y+z}{\dfrac{2}{3}-\dfrac{3}{2}+\dfrac{4}{5}}=\dfrac{-16}{\dfrac{-1}{30}}=480\)
\(\Rightarrow\left\{{}\begin{matrix}x=320\\y=360\\z=384\end{matrix}\right.\)
Vậy...
Lời giải:
Ta có:
$A> \frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=\frac{x+y+z}{x+y+z}=1(1)$
Mặt khác:
$\frac{x}{x+y}-\frac{x+z}{x+y+z}=\frac{-yz}{(x+y)(x+y+z)}<0$ với mọi $x,y,z$ nguyên dương.
$\Rightarrow \frac{x}{x+y}< \frac{x+z}{x+y+z}$
Hoàn toàn tương tự:
$\frac{y}{y+z}< \frac{x+y}{x+y+z}$
$\frac{z}{z+x}< \frac{z+y}{z+y+x}$
Cộng các BĐT trên lại ta có:
$A< \frac{x+y}{x+y+z}+\frac{y+z}{x+y+z}+\frac{z+x}{x+y+z}=2(2)$
Từ $(1); (2)\Rightarrow 1< A< 2$ nên $A$ không thể có giá trị nguyên.
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) => \(\left(\dfrac{x}{3}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{5}\right)^2\)
=> \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}=\dfrac{2x^2+y^2-z^2}{2.9+16-25}=\dfrac{9}{18+16-25}=\dfrac{9}{9}=1\)
=> \(\left\{{}\begin{matrix}\dfrac{x^2}{9}=1\Rightarrow\dfrac{x}{3}=1\Rightarrow x=3\\\dfrac{y^2}{16}=1\Rightarrow\dfrac{y}{4}=1\Rightarrow y=4\\\dfrac{z^2}{25}=1\Rightarrow\dfrac{z}{5}=1\Rightarrow z=5\end{matrix}\right.\)
Vậy x = 3, y = 4, z = 5
Đặt x/3=y/4=z/5=k
=>x=3k; y=4k; z=5k
Ta có: \(2x^2+y^2-z^2=9\)
\(\Leftrightarrow18k^2+16k^2-25k^2=9\)
\(\Leftrightarrow9k^2=9\)
\(\Leftrightarrow k^2=1\)
TH1: k=1
=>x=3; y=4; z=5
TH2: k=-1
=>x=-3; y=-4; z=-5
