Tìm x, y, z biết rằng:
a) \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) và 2x + 3y - z = 50
b) \(\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
c) 10x = 6y và \(2x^2-y^2=-28\)
Tìm x , y ,z :
a, \(\dfrac{x+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b, 10x = 6y và \(2x^2-y^2=-28\)
c, \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
d, \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
1/ x\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\text{và}2x+3y-z=50\)
2/ x : y : z = 3 : 5 ; ( - 2 ) và 5x - y + 3z = -16
3/ 2x + 3y ; 7z = 5y và 3x - 7y + 5z = 30
4/ \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\text{và}x-y-z=38\)
4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)
Do đó: x=-16; y=-24; z=-30
Bài 1 : Tìm x,y,z biết :
a) 2x = 3y ; 5y = 7z và 3x - 7y + 5z = -30
b) 3x =5y ; 7y = 2z và x + y + z = 74
c) x : z = \(\dfrac{2}{3}\) : \(\dfrac{1}{2}\) ; z : y = 1 : \(\dfrac{4}{7}\) và y + z = 66
d) x : y : z = 3 : 4 : 5 và \(2x^2\) + \(2y^2\) - \(3z^2\) = -100
e) \(x:y:z\) = 2 : 5 : 6 và \(2x^2\) + \(4y^2\) - \(4z^2\) = -324
f) \(\dfrac{x-1}{2}\) = \(\dfrac{y-2}{3}\) = \(\dfrac{z-3}{4}\) và \(x-2y+3z=14\)
g)\(\dfrac{x-1}{2}\) = \(\dfrac{y+3}{4}\) =\(\dfrac{z-5}{6}\) và \(5z-3x-4y=50\)
h) \(\dfrac{x}{2}=\dfrac{y}{7}\) và \(xy=56\)
i)\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{xy}{200}\)
k) \(\dfrac{x-5}{6}=\dfrac{x+5}{18}\)
l) \(\dfrac{2x-11}{12}=\dfrac{x+5}{20}\)
1, x : y : z = 2 : 3 : 4 và x + y + z = 18
2, \(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}\) và 4x - 3y - 2z = 81
3, \(\dfrac{x}{3}=\dfrac{y}{2};\) 4y = 3z và x + y +z = 46
4, 5x = 3y; \(\dfrac{y}{z}=\dfrac{3}{2}\) và 2x + 3y -4z =34
1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)
⇒ x=4;y=6;z=8
\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)
\(2,\) Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)
\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)
\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)
Tìm x,y,z biết:
a)\(\dfrac{x-1}{2}\)=\(\dfrac{y-2}{3}\)=\(\dfrac{z-3}{4}\) và 2x+3y-z=50
b)\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)và xyz=810
a, Ta có :
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)
\(\Rightarrow x=11;y=17;z=23\)
b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)
\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)
\(\Rightarrow x=6;y=9;z=15\)
a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)
Áp dụng t/c dtsbn:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)
b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
xyz = 810
=> 2k.3k.5k = 810
=> k = 3
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)
a) Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
nên \(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
mà 2x+3y-z=50
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)
Do đó:
\(\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Ta có: xyz=810
\(\Leftrightarrow30k^3=810\)
\(\Leftrightarrow k^3=27\)
\(\Leftrightarrow k=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=6\\z=5k=5\cdot3=15\end{matrix}\right.\)
Tìm x,y,z biết:
a) 3x=2y, 7y=5z và x-y+z=32
b) \(\dfrac{x}{2}\)=\(\dfrac{y}{3}\) và x.y=24
c)\(\dfrac{x-1}{2}\)=\(\dfrac{y-2}{3}\)=\(\dfrac{z-3}{4}\) và 2x+3y-z=50
d)\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\) và x.y.z=810
Tìm x,y,z biết :
1) \(\dfrac{x}{-7}=\dfrac{y}{4}\) và \(2x-3y=-78\)
2) \(\dfrac{x}{y}=\dfrac{9}{7};\dfrac{y}{z}=\dfrac{7}{3}\) và \(x-y+z=-15\)
1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)
=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)
2. Ta có:
- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
1) Rút gọn bt:
(x+y+z)3+(x-y-z)3+(y-x-z)3+(z-y-x)3
2)Tìm x,y,z t/m: 9x2+y2+2z2-18x+4z-6y+20=0
3)Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\)=1 và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\)=0 . CMR:
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)=1
Bài 1 Tìm x, y, z
a)\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b)\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)và x+y+z=49
c)\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-4}{4}\) và 2x+3y-z=50
d)\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và xyz=810
Giải cụ thể giúp mình nhé!!!
d) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(xyz=810\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
=> \(x=2k\) ; \(y=3k\) ; \(z=5k\)
Thay \(x=2k;y=3k;z=5k\) vào \(xyz=810\) ta được
\(2k.3k.5k=810\)
\(30k=810\)
\(k^3=27\)
=> k = 3
=> \(x=2.3=6\)
=> \(y=3.3=9\)
=> \(z=5.3=15\)
a) Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{2x+2y+2z}{x+y+z}=\dfrac{2\cdot\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\)
\(\Rightarrow\dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\)
\(\Rightarrow\dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\)
\(\Rightarrow\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow y+z=\dfrac{1}{2}-x\)
Thay vào \(y+z+1=2x\) ; ta có :
\(\dfrac{1}{2}-x+1=2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+z=\dfrac{1}{2}-y\)
Thay vào \(x+z+2=2y\) ; ta có :
\(\dfrac{1}{2}-y+2=2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+y=\dfrac{1}{2}-z\)
Thay vào \(x+y-3=2z\) ; ta có :
\(\dfrac{1}{2}-z-3=2z\Rightarrow3z=\dfrac{-5}{2}\Rightarrow z=\dfrac{-5}{6}\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)