Bài 1:
a: Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=k\)
=>x=15k; y=20k; z=28k
2x+3y-z=31
=>\(2\cdot15k+3\cdot20k-28k=31\)
=>30k+60k-28k=31
=>62k=31
=>\(k=\dfrac{1}{2}\)
\(x=15\cdot\dfrac{1}{2}=\dfrac{15}{2};y=20\cdot\dfrac{1}{2}=10;z=28\cdot\dfrac{1}{2}=14\)
b: Đặt \(\dfrac{x+3}{5}=\dfrac{y-2}{3}=\dfrac{z-1}{7}=k\)
=>\(\left\{{}\begin{matrix}x+3=5k\\y-2=3k\\z-1=7k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5k-3\\y=3k+2\\z=7k+1\end{matrix}\right.\)
3x+5y-7z=32
=>\(3\left(5k-3\right)+5\left(3k+2\right)-7\left(7k+1\right)=32\)
=>15k-9+15k+10-49k-7=32
=>-19k-6=32
=>-19k=38
=>k=-2
\(\left\{{}\begin{matrix}x=5\cdot\left(-2\right)-3=-10-3=-13\\y=3\cdot\left(-2\right)+2=-6+2=-4\\z=7\cdot\left(-2\right)+1=-14+1=-13\end{matrix}\right.\)
c: 5x=8y=20z
=>\(\dfrac{5x}{40}=\dfrac{8y}{40}=\dfrac{20z}{40}\)
=>\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}\)
mà x-y-z=3
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=\dfrac{3}{1}=3\)
=>\(x=3\cdot8=24;y=3\cdot5=15;z=3\cdot2=6\)
d: \(\dfrac{x}{10}=\dfrac{y}{5}\)
=>\(\dfrac{x}{2}=\dfrac{y}{1}\)
=>\(\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{5}=k\)
=>x=4k; y=2k; z=5k
\(2x+3y+4z=330\)
=>\(2\cdot4k+3\cdot2k+4\cdot5k=330\)
=>8k+6k+20k=330
=>\(k=\dfrac{165}{17}\)
\(x=4\cdot\dfrac{165}{17}=\dfrac{660}{17};y=2\cdot\dfrac{165}{17}=\dfrac{330}{17};z=5\cdot\dfrac{165}{17}=\dfrac{825}{17}\)
e: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
=>\(\dfrac{2x}{3}\cdot60=\dfrac{3y}{4}\cdot60=\dfrac{4z}{5}\cdot60\)
=>\(40x=45y=48z\)
=>\(\dfrac{40x}{720}=\dfrac{45y}{720}=\dfrac{48z}{720}\)
=>\(\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}\)
Đặt \(\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}=k\)
=>x=18k; y=16k; z=15k
\(x^2+y^2-z^2=1420\)
=>\(\left(18k\right)^2+\left(16k\right)^2-\left(15k\right)^2=1420\)
=>\(324k^2+256k^2-225k^2=1420\)
=>\(k^2=4\)
=>\(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
TH1: k=2
=>\(x=18\cdot2=36;y=16\cdot2=32;z=15\cdot2=30\)
TH2: k=-2
\(x=18\cdot\left(-2\right)=-36;y=16\cdot\left(-2\right)=-32;z=15\cdot\left(-2\right)=-30\)
Bài 2:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk; c=dk
\(\dfrac{2a+7b}{2a-7b}=\dfrac{2bk+7b}{2bk-7b}=\dfrac{b\left(2k+7\right)}{b\left(2k-7\right)}=\dfrac{2k+7}{2k-7}\)
\(\dfrac{2c+7d}{2c-7d}=\dfrac{2dk+7d}{2dk-7d}=\dfrac{d\left(2k+7\right)}{d\left(2k-7\right)}=\dfrac{2k+7}{2k-7}\)
Do đó: \(\dfrac{2a+7b}{2a-7b}=\dfrac{2c+7d}{2c-7d}\)
