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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:
\(x+y+z=49\)
\(\Rightarrow30k+16k+15k=49\)
\(\Rightarrow61k=49\)
\(\Rightarrow k=\dfrac{49}{61}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{49}{61}.30=\dfrac{1470}{61}\\y=\dfrac{49}{61}.16=\dfrac{784}{61}\\z=\dfrac{49}{61}.15=\dfrac{735}{61}\end{matrix}\right.\)
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
5x/2=7z/3
nên 15x=14z
=>x/14=z/15
3x=5y nên x/5=y/3
=>x/70=y/42=z/45
Đặt x/70=y/42=z/45=k
=>x=70k; y=42k; z=45k
Tacó: xz=47250
=>3150k2=47250
=>k2=15
TH1: \(k=\sqrt{15}\)
\(x=70\sqrt{15};y=42\sqrt{15};z=45\sqrt{15}\)
TH2:
\(k=-\sqrt{15}\)
\(x=-70\sqrt{15};y=-42\sqrt{15};z=-45\sqrt{15}\)
\(2x=3y=4z\)
\(\Leftrightarrow\dfrac{2x}{12}=\dfrac{3y}{12}=\dfrac{4z}{12}\)
\(\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}\)
Đặt :
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=4k\\z=3k\end{matrix}\right.\)
\(2x^2-3z^2=1125\Leftrightarrow2.\left(6k\right)^2-3.\left(3k\right)^2=1125\Leftrightarrow72k^2-27k^2=1125\)
\(\Leftrightarrow45k^2=1125\)
\(\Leftrightarrow k^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)
Với \(k=5\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6.5=30\\y=4.5=20\\z=3.5=15\end{matrix}\right.\)
Với \(k=-5\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6.\left(-5\right)=-30\\y=4.\left(-5\right)=-20\\z=3.\left(-5\right)=-15\end{matrix}\right.\)
Vậy ...
Bạn tham khảo:
Giả sử:\(\hept{\begin{cases}xyz-x=1945\left(1\right)\\xyz-y=1975\left(2\right)\\xyz-z=1995\left(3\right)\end{cases}}\)với \(x,y,z\in N\)
Tứ \(\left(1\right)\Rightarrow x\left(yz-1\right)=1945\)là số lẻ \(\Rightarrow x\)lẻ
Từ \(\left(2\right)\Rightarrow y\left(xz-1\right)=1975\)là số lẻ \(\Rightarrow y\)lẻ
Từ \(\left(3\right)\Rightarrow z\left(xy-1\right)=1995\)là số lẻ \(\Rightarrow z\)lẻ
Nên \(x,y,z\)là số lẻ
\(\Rightarrow x,y,z-x\)là số chẵn khác 1945
Vậy không tồn tại \(x,y,z\in N\)thỏa mãn \(\left(1\right),\left(2\right),\left(3\right)\).
\(+)3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)
\(+)7y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{7}\)
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{x}{2}=\dfrac{5y}{15};\dfrac{3y}{15}=\dfrac{z}{7}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y-z}{10+15-21}=\dfrac{32}{16}=2\)
Suy ra: \(\dfrac{x}{10}=2\Rightarrow x=20\)
\(\dfrac{y}{15}=2\Rightarrow y=30\)
\(\dfrac{z}{21}=3\Rightarrow z=63\)
Vậy \(x=20;y=30;z=63\)
Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow7x-5x=21+25\)
\(\Leftrightarrow2x=46\)
\(\Rightarrow x=46:2=23\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Rightarrow x^2=\left(\pm8\right)^2\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(7x-21=5x+25\)
\(7x-5x+25=21\)
\(2x+25=21\)
\(2x=-4\Rightarrow x=-2\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x+1\right)\left(x-1\right)\)
\(63=x\left(x-1\right)+1\left(x-1\right)\)
\(63=x^2-x+x-1\)
\(x^2=63+1=64\)
\(x=\left\{\pm8\right\}\)
c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)
\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)
\(x\left(x+4\right)+4\left(x+4\right)=40\)
\(x^2+4x+4x+16=40\)
\(x^2+8x=40-16=24\)
\(x\left(x+8\right)=24\)
\(x\in\left\{\varnothing\right\}\)
d) \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+3\right)\)
\(x\left(x-2\right)+2\left(x-2\right)=x\left(x+3\right)-1\left(x+3\right)\)
\(x^2-2x+2x-4=x^2+3x-x-3\)
\(\)\(x^2-4=x^2+2x-3\)
\(\Leftrightarrow x^2-x^2-2x+3=4\)
\(-2x+3=4\)
\(-2x=1\)
\(x=-\dfrac{1}{2}\)
\(\dfrac{2x}{5}=\dfrac{3y}{2}=\dfrac{5z}{7}\)
\(\Leftrightarrow28x=105y=50z\)
hay x/75=y/20=z/42
Đặt x/75=y/20=z/42=k
=>x=75k; y=20k; z=42k
Ta có: xyz=504000
\(\Leftrightarrow k^3\cdot63000=504000\)
\(\Leftrightarrow k=2\)
=>x=150; y=40; z=84