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\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Áp dụng t.c của dãy tỉ só bằng nhau,ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y}{3+4}=\dfrac{16}{7}\)
=>\(x=\dfrac{16}{7}.3=\dfrac{48}{7}\)
\(y=\dfrac{16}{7}.4=\dfrac{64}{7}\)
\(z=\dfrac{16}{7}.5=\dfrac{80}{7}\)
Vậy...
Các câu sau tương tự
ADTC dãy tỉ số bằng nhau ta cs
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x^2+y^2}{2^2+3^2}=\frac{52}{13}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=4\\\frac{y}{3}=4\\\frac{z}{4}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=16\end{matrix}\right.\)
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\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+2y+3z}{2+2.3+3.4}=\frac{51}{20}\)
x=51/10
y=153/20
z=51/50
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\(\frac{z}{4}=\frac{y}{3}\Rightarrow\frac{z}{20}=\frac{y}{15}^{\left(1\right)}\)
\(\frac{x}{2}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{z}{20}^{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{x}{8}=\frac{z}{20}=\frac{y}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{8}=\frac{z}{20}=\frac{y}{15}=\frac{x+y+z}{8+20+15}=\frac{51}{43}\)
..... ( tới bước này bạn tự làm tiếp nhá )
\(\frac{z}{4}=\frac{y}{3}\Rightarrow\frac{z}{20}=\frac{y}{15}\)
\(\frac{x}{2}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{z}{20}\)
Suy ra: \(\frac{x}{8}=\frac{y}{15}=\frac{z}{20}=\frac{x+y+z}{8+15+20}=\frac{51}{43}\)
Vậy \(x=8.\frac{51}{43}=\frac{408}{43}\)
\(y=15.\frac{51}{43}=\frac{765}{43}\)
\(z=20.\frac{51}{43}=\frac{1020}{43}\)
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bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
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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: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)
=>...
Tìm x,y,z?
Có \(\dfrac{x}{4}=\dfrac{y}{3}\)
\(\dfrac{z}{5}=\dfrac{x}{2}\Leftrightarrow\dfrac{z}{10}=\dfrac{x}{4}\)
\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{10}\)
Áp dụng TCDTSBN có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{10}=\dfrac{x+y+z}{4+3+10}=\dfrac{51}{17}=3\)
Do đó \(\dfrac{x}{4}=3;\dfrac{y}{3}=3;\dfrac{z}{10}=3\)
\(\Rightarrow x=12;y=9;z=30\)