1,a)Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{7}=\dfrac{14}{7}=2\)
\(=>\left\{{}\begin{matrix}\dfrac{x}{4}=2\\\dfrac{y}{3}=2\end{matrix}\right.=>\left\{{}\begin{matrix}x=8\\y=6\end{matrix}\right.\)
1,b)Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{19}=\dfrac{y}{21}=\dfrac{2x}{38}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)
\(=>\left\{{}\begin{matrix}\dfrac{x}{19}=2\\\dfrac{y}{21}=2\end{matrix}\right.=>\left\{{}\begin{matrix}x=38\\y=42\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=38\\y=42\end{matrix}\right.\)
1a
Áp dụng tc DTSBN:
\(\frac{x}{4}=\frac{y}{3}=\frac{x+y}{4+3}=\frac{14}{7}=2\)
==> x = 2.4=8 ; y=3.4=12
1b
Ta có: \(\frac{x}{19}=\frac{2x}{38}\)
Áp dụng...: \(\frac{2x}{38}=\frac{y}{21}=\frac{2x-y}{38-21}=\frac{34}{17}=2\)
==> x = 2.19=38 ; y=21.2=42
1c
Ta có: \(\frac{x}{19}=\frac{x^2}{361}\) ; \(\frac{y}{3}=\frac{y^4}{81}\)
AD....: \(\frac{x^2}{361}=\frac{y^4}{81}=\frac{x^2-y^4}{361-81}=\frac{4}{280}=\frac{1}{70}\)
==> x = 19.1/70=19/70
y=3.1/70=3/70
2a
Ta có: \(\frac{x}{10}=\frac{5x}{50}\);\(\frac{z}{24}=\frac{2z}{48}\)
AD....: \(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{48}=\frac{5x+y-2z}{50+6-48}=\frac{28}{8}=3,5\)
==> x=3,5.10=35 ; y=3,5.6=21 ; z=24.3,5=84
2b
Ta có: \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\Rightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)
AD...: \(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)
==> x=3.15=45 ; y=20.3=60 ; z=28.3=84