Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
12x−15y7 =20z−12x9 =15y−20z11 =12x−15y+20z−12x+15y−20x7+9+11 =027 =0
=> 12x - 15y = 0 => 12x = 15y => x15 =y12 => x60 =y48
20z - 12x = 0 => 20z = 12x => x20 =z12 => x60 =z36
=> x60 =y48 =z36 =x+y+z60+48+36 =48144 =13
=> x = 1 . 60 : 3 = 20
y = 1 . 48 : 3 = 16
z = 1 . 36 : 3 = 12
12x-15y/7=20z-12x/9=15y-20z/11
=12x-15y+20z-12x+15y-20z/7+9+11=0/27=0
Nên ta có 12x=15y ;20z=12x;15y=20z
do đó 12x=15y=20z
Suy ra 12x/60 =15y/60=20z/60hay
x/5=y/4=z/3 và x+y+z=48
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU TA CÓ:
x/5=y/4=z/3=x+y+z/5+4+3=48/12=4
=>x=5.4=20
y=4.4=16
z=3.4=12
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)
⇒\(\left\{{}\begin{matrix}12x=15y\\20z=12x\\15y=20z\end{matrix}\right.\)
⇔\(12x=15y=20z\)⇒\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{5+4+3}=\dfrac{48}{12}=4\)
⇒\(\left\{{}\begin{matrix}x=5.4=20\\y=4.4=16\\z=3.4=12\end{matrix}\right.\)
\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)
\(12x-15y=0\Rightarrow4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\)
\(20z-12x=0\Rightarrow5z=3x\Rightarrow\frac{z}{3}=\frac{x}{5}\)
\(15y-20z=0\Rightarrow3y=4z\Rightarrow\frac{y}{4}=\frac{z}{3}\)
\(\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{5+4+3}=\frac{48}{12}=4\)
ta có;x=4x5=20
y=4x4=16
z=4x3=12
sde dQTYTWAYEGFSAYEFGEYSARR WAFWIUFB A RR qiiRY ii yÌU ẨU YIUWYR URH Y Y2QUR2QGyrg Y4
KQWFJ | Ị |
Ị | Ị |
Ị | Ị |
ỊIW | FU |
ÌUEI | F |
ỊU | ÌU |
I | ÌUI |
FUI | ÙI |
Ù | 8FU |
ÌU | ÌU |
Ì | ÌU |
ÌU | ÌU |
ÌU | Ì |
Ì | IUI |
I | |
I | I |
I | FI |
I | Ì |
Ì | ÙIU |
Ì | IUFI |
I | I |
I | |
IU | IU |
Ì | FIF |
IU | UI |
U | FJ |
JFI | FUFNUFYFFTCBBYY |
7 | |
7 | ỲB |
FYD | YC87BBDYBUDYYY |
Y | |
7FYTF7 | YB7BDYD7OYBE |
Y | 7 |
YD7DY7YB | 7 YB |
ED7 | YE7 |
YD87 | BEY |
7BE8 | YDU |
E7E | YEQY7 |
7YYE7 | YE7 |
YE | 7WY |
7 | 7WY |
7 | YWWY |
7 | |
78YW7 | Y 7W |
YW7 | ƯY |
7EY | 7EYE7BEY |
7EE7 | BYE |
7EY | E7 |
YE7Y 7 | Y |
7EYB | 7EY |
7EY | 7E |
(12x-15y)+ (20z-12x)+ (15y-20z)/7+9+11=0
=>12x=15y=>x=5/4y
=>15y=20z=>z=3/4y
x+y+z+48=> y+5/4y+3/4y=48=>y=16
x=5/4*16=20
z=48-20-16=12
\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20x}{7+9+11}=\frac{0}{27}=0\)
=> 12x - 15y = 0 => 12x = 15y => \(\frac{x}{15}=\frac{y}{12}\)=> \(\frac{x}{60}=\frac{y}{48}\)
20z - 12x = 0 => 20z = 12x => \(\frac{x}{20}=\frac{z}{12}\)=> \(\frac{x}{60}=\frac{z}{36}\)
=> \(\frac{x}{60}=\frac{y}{48}=\frac{z}{36}=\frac{x+y+z}{60+48+36}=\frac{48}{144}=\frac{1}{3}\)
=> x = 1 . 60 : 3 = 20
y = 1 . 48 : 3 = 16
z = 1 . 36 : 3 = 12
a/ xem lại đề
b/đặt: \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
\(\Rightarrow x=12k;y=9k;z=5k\)
\(\Rightarrow xyz=12k\cdot9k\cdot5k=540k^3=20\)
\(\Rightarrow k^3=\dfrac{1}{27}\Rightarrow k=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x=12k=12\cdot\dfrac{1}{3}=4\\y=9k=9\cdot\dfrac{1}{3}=3\\z=5k=5\cdot\dfrac{1}{3}=\dfrac{5}{3}\end{matrix}\right.\)
Vậy........
c/ Áp dụng t/c của dãy tỉ số = nhau có:
\(\dfrac{12x-15y}{7}=\dfrac{20z-12x}{9}=\dfrac{15y-20z}{11}=\dfrac{12x-15y+20z-12x+15y-20z}{7+9+11}=\dfrac{0}{27}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{12x-15y}{7}=0\\\dfrac{20z-12x}{9}=0\\\dfrac{15y-20z}{11}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}12x-15y=0\\20z-12x=0\\15y-20z=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}12x=15y\\20z=12x\\15y=20z\end{matrix}\right.\)=> \(12x=15y=20z\)
\(\Rightarrow\dfrac{x}{\dfrac{1}{12}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{20}}\)
A/dụng t/c của dãy tỉ số = nhau có:
\(\dfrac{x}{\dfrac{1}{12}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x+y+z}{\dfrac{1}{12}+\dfrac{1}{15}+\dfrac{1}{20}}=\dfrac{48}{\dfrac{1}{5}}=240\)
\(\Rightarrow\left\{{}\begin{matrix}x=240\cdot\dfrac{1}{12}=20\\y=240\cdot\dfrac{1}{15}=16\\z=240\cdot\dfrac{1}{20}=12\end{matrix}\right.\)
Vậy......
a) sai đề bn nhé:
\(\frac{x}{2}\) = \(\frac{y}{3}\); \(\frac{y}{4}\) = \(\frac{z}{5}\) và x2 - y2 = -16