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1)
Có:\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\\\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\end{cases}\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}}\)
Áp dụng tc của DTSBN có:
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x-y+z}{8-12+15}=\frac{33}{11}=3\) (vì x-y+z=33)
\(\Rightarrow\hept{\begin{cases}x=3.8=24\\y=3.12=36\\y=3.15=45\end{cases}}\)(tm)
Vậy.....................
2)
Có: \(\text{ x:y:z=2:3:4 }\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x}{2}=\frac{3y}{9}=\frac{2z}{8}\)
Áp dụng tc của DTSBN có:
\(\frac{x}{2}=\frac{3y}{9}=\frac{2z}{8}=\frac{x+3y-2z}{2+9-8}=\frac{3}{3}=1\)(vì x+3y-z=3)
\(\Rightarrow\hept{\begin{cases}x=2\\y=3\\z=4\end{cases}}\)(tm)
Vậy................
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\(\dfrac{x}{5}=\dfrac{4}{-3}\)
⇔\(-3.x=4.5\)
⇔\(-3x=20\)
⇔\(x=-\dfrac{20}{3}\)
b: Đặt \(\dfrac{x}{4}=\dfrac{y}{7}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)
Ta có: \(x^2-y^2=-33\)
\(\Leftrightarrow k^2=1\)
Trường hợp 1: k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=4\\y=7k=7\end{matrix}\right.\)
Trường hợp 2: k=-1
\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=-4\\y=7k=-7\end{matrix}\right.\)
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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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Cái x:3 hay.... là phân số nha các bạn tại mik ko bik gõ lm sao nên ghi đại
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
\(\Rightarrow\frac{2x}{6}=\frac{3y}{12}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{2x-3y-x}{6-12-5}=\frac{33}{-12}=\frac{11}{-4}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{11}{-4}\cdot3=\frac{33}{-4}\\y=\frac{11}{-4}\cdot4=-11\\z=\frac{11}{-4}\cdot5=\frac{55}{-4}\end{cases}}\)
vậy_
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{3}=\frac{z}{4};\frac{y}{2}=\frac{z}{3}\)
suy ra \(\frac{x.}{3.3}=\frac{z}{4.3};\frac{y}{2.4}=\frac{z}{3.4}\)
suy ra \(\frac{x}{9}=\frac{z}{12};\frac{y}{8}=\frac{z}{12}\)
suy ra \(\frac{x}{9}=\frac{z}{12}=\frac{y}{8}\)
áp dụng dãy tỉ số bằng nhau ta có
\(\frac{x-z-y}{9-12-8}=\frac{33}{-11}=-3\)
suy ra
x/3=z/4;y/2=z/3 và x-y-z=33
x/3=z/12;y/8=z/12 và x-y-z=33
x/3=y/8=z/12 và x--y-z=33
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
x/3=y/8=z/12=x-y-z/3-8-12=33/-17=-33/17
x/3=-33/17=>x=-33/17.3=>x=-99/17
y/8=-33/17=>y=-33/17.8=>y=-264/17
z/12=-33/17=>z=-33/17.12=>z=-396/17
Vậy x=-99/17
y=-264/17
z=-396/17
tk mk nha bn
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`@` `\text {Ans}`
`\downarrow`
`5,`
Ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{3}=\dfrac{2y}{10}=\dfrac{3z}{18}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3\cdot3=-9\\y=-3\cdot5=-15\\z=-3\cdot6=-18\end{matrix}\right.\)
Vậy, `x = -9; y = -15; z = -18.`
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\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)
c, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)
d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
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làm xong rồi nhấn gửi bỗng dưng lỗi mất hết luôn TTT^^^^^^TTTTT