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a) ADTCDTSBN
có: \(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3.\)
=> x/2 = 3 => x = 6
y/3 = 3 => y = 9
z/4 = 3 => z = 12
KL:...
b,c làm tương tự nha
d) ta có: \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{2x}{10}\)
ADTCDTSBN
có: \(\frac{2x}{10}=\frac{y}{-6}=\frac{z}{7}=\frac{2x+y-z}{10+\left(-6\right)-7}=\frac{49}{-3}\)
=>...
e) ADTCDTSBN
có: \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}=\frac{x+1+y+2+z+3}{2+3+4}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{9}\)
\(=\frac{21+6}{9}=\frac{27}{9}=3\)
=>...
g) ta có: \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)
mà xy = 12 => 4k.3k = 12
12.k2 = 12
k2 = 1
=> k = 1 hoặc k = -1
=> x = 4.1 = 4
y = 3.1 = 3
x=4.(-1) = -4
y=3.(-1) = -3
KL:...
h) ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)
ADTCDTSBN
có: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)
=>...
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\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}=\dfrac{4a-3b+2c}{4-6+6}=\dfrac{36}{4}=9\\ \Rightarrow\left\{{}\begin{matrix}a=9\\b=18\\c=27\end{matrix}\right.\\ \dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{16}=\dfrac{x-y+z}{10-15+16}=\dfrac{-49}{11}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{490}{11}\\y=-\dfrac{735}{11}\\z=-\dfrac{784}{11}\end{matrix}\right.\)
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1.
\(\frac{x}{2}=\frac{y}{3}=>\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{7}=>\frac{y}{15}=\frac{z}{21}\)
=>\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)
=> x=2x10=20
y=2x15=30
z=2x21=42
\(\frac{x}{3}=\frac{y}{4}\)\(\Rightarrow y=\frac{4x}{3}\)
\(\Rightarrow\)x2 . y = 36
hay x2 . \(\frac{4x}{3}\)= 36
\(\frac{4}{3}x^3=36\)
\(x^3=27\)
\(\Rightarrow\)x = 3
Thay x = 3 vào x2 . y = 36 ta được :
32 . y = 36
y = 36 : 32 = 4
Vậy x = 3 ; y = 4
Đặt x/3=y/4=k => x=3k,y=4k
x2y = 36
=> (3k)2.4k = 36
=> 9k24k = 36
=> 36k3 = 36
=> k3 = 1
=> k = 1
=> x=3,y=4