Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: x/3=y/-5 và x-y=32
=> x/3=y/4=x-y/3-(-5)=32/8=4
=> x=4.3=12
y=4.(-5)=-20
Vậy x=12
y=-20
x/2 = y/5
=> xy/10 = x/2 = y/5 = 10/10 = 1
=> x = 1x 2 = 2
y = 1 x 5 = 5
Đặt \(k=\frac{x}{2}=\frac{y}{5}\)
=> \(k^2=\frac{xy}{2.5}=\frac{xy}{10}=\frac{10}{100}=1\)
=> k = -1;1
+ k = -1 thì \(\frac{x}{2}=-1\Rightarrow x=-2\)
\(\frac{y}{5}=-1\Rightarrow y=-5\)
+ k = 1 thi \(\frac{x}{2}=1\Rightarrow x=2\)
\(\frac{y}{5}=1\Rightarrow y=5\)
Vậy .............................
a: Ta có: 2x/3=3y/4=4z/5
nên \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Đặt \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=k\)
=>x=3/2k; y=4/3k; z=5/4k
\(xy+yz-xz=32\)
\(\Leftrightarrow\dfrac{3}{2}k\cdot\dfrac{4}{3}k+\dfrac{4}{3}k\cdot\dfrac{5}{4}k-\dfrac{3}{2}k\cdot\dfrac{5}{4}k=32\)
\(\Leftrightarrow k^2\cdot\dfrac{43}{24}=32\)
\(\Leftrightarrow k^2=\dfrac{768}{43}\)
Trường hợp 1: \(k=\dfrac{16\sqrt{129}}{43}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{24\sqrt{129}}{43}\\y=\dfrac{64\sqrt{129}}{129}\\z=\dfrac{20\sqrt{129}}{43}\end{matrix}\right.\)
Trường hợp 2: \(k=-\dfrac{16\sqrt{129}}{43}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{24\sqrt{129}}{43}\\y=-\dfrac{64\sqrt{129}}{129}\\z=-\dfrac{20\sqrt{129}}{43}\end{matrix}\right.\)
b: Ta có: 4x=3y
nên x/3=y/4=k
=>x=3k; y=4k
\(x^2-xy+y^2=32\)
\(\Leftrightarrow9k^2-12k^2+16k^2=32\)
\(\Leftrightarrow13k^2=32\)
Trường hợp 1: \(k=\dfrac{32\sqrt{13}}{13}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{96\sqrt{13}}{13}\\y=\dfrac{128\sqrt{13}}{13}\end{matrix}\right.\)
Trường hợp 2: \(k=-\dfrac{32\sqrt{13}}{13}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{96\sqrt{13}}{13}\\y=-\dfrac{128\sqrt{13}}{13}\end{matrix}\right.\)
x:y=4:5
=>x/y=4/5
=>x/4=y/5
đặt x/4=y/5=k
ta có :x=4k
y=5k
=>x.y=4k.5k=20.k^2=5
=>k^2=1/4
=>k=1/2
=>x/4=1/2=>x=2
=>y/5=1/2=>y=5/2
Đặt x/3=y/2=k
=>x=3k; y=2k
Ta có: 2/x+5/y=32
\(\Leftrightarrow\dfrac{2}{3k}+\dfrac{5}{2k}=32\)
\(\Leftrightarrow\dfrac{4}{6k}+\dfrac{15}{6k}=32\)
=>6k=19/32
=>k=19/192
=>x=57/192; y=38/192=19/96
Có:
3.x = 2.y => x/2 = y/3
7.y = 5.z => y/5 = z/7
=> x/2 = y/3 ; y/5 = z/7
Có x/2 = y/3 => x/10 = y/15 (1)
y/5 = z/7 => y/15 = z/21 (2)
Từ (1) và (2) suy ra:
x/10 = y/15 = z/21 = x - y + z/10 - 15 + 21 = 32/16 = 2
=> * x/10 = 2 => x = 2.10 = 20
* y/15 = 2 => y = 2.15 = 30
* z/21 = 2 => z = 2.21 = 42
Vậy x = 20 ; y = 30 ; z = 42
Ủng hộ nha
a) Đặt \(\frac{x}{-3}=\frac{y}{5}=k\left(k\ne0\right)\)
\(\Rightarrow x=-3k\); \(y=5k\)
Ta có: \(xy=\left(-3k\right).5k=-15k^2=-\frac{5}{27}\)
\(\Rightarrow k^2=\frac{1}{81}\)\(\Rightarrow k=\pm\frac{1}{9}\)
+) Nếu \(k=\frac{-1}{9}\)\(\Rightarrow x=\left(\frac{-1}{9}\right).\left(-3\right)=\frac{1}{3}\); \(y=\frac{-1}{9}.5=\frac{-5}{9}\)
+) Nếu \(k=\frac{1}{9}\)\(\Rightarrow x=\frac{1}{9}.3=\frac{1}{3}\); \(y=\frac{1}{9}.5=\frac{5}{9}\)
Vậy \(x=\frac{1}{3}\); \(y=\frac{-5}{9}\)hoặc \(x=\frac{1}{3}\); \(y=\frac{5}{9}\)
a, Theo tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{4}=\frac{x+y}{3+4}=\frac{14}{7}=2\Rightarrow x=6;y=8\)
b, Theo tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{4}=\frac{x^2+y^2}{9+16}=\frac{25}{25}=1\Rightarrow x=3;y=4\)