Cho ba so x, y, z biet \(x-y+z=3\)va \(\frac{x}{2}=\frac{y}{1}=\frac{z}{4}\). Khi do y= ?
Tim ba so x, y, z biet \(\frac{y+z+1}{x}=\frac{x+ z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
Suy ra
\(x+y+z=\frac{1}{2}\)(1)
\(y+z+1=2x\)(2)
\(x+z+2=2y\)(3)
\(x+y-3=2z\)(4)
(2)-(1) ta có
\(1-x=2x-\frac{1}{2}\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)
\(x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\Leftrightarrow y+z=\frac{1}{2}-\frac{1}{2}=0\)
\(y=-z\)
\(x+z+2=\frac{1}{2}+2-y==\frac{5}{2}-y\)
\(\frac{\frac{5}{2}-y}{y}=\frac{5}{2y}-1=2\Leftrightarrow\frac{5}{2y}=3\Leftrightarrow y=\frac{5}{6}\)
\(z=-\frac{5}{6}\)
Tim cac so x,y,z biet \(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)va x2 va y2 va z2
Cho ba so x,y,z khac 0 thoa man dieu kien \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\).Khi do B=\(\left(1+\frac{x}{y}\right)+\left(1+\frac{y}{z}\right)+\left(1+\frac{z}{x}\right)\)Co gia tri bang
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+x}=2\)
ta có:\(B=\left(1+\frac{x}{y}\right)+\left(1+\frac{y}{z}\right)+\left(1+\frac{z}{x}\right)=3+\frac{x+y+z}{y+z+x}=3+1=4\)
B có giá trị bằng 4
tim x,y,z khi
\(\frac{x}{7}=\frac{y}{3}va\)x-24=y
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{2}\)va y-x=48
\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\)va x-y- z=28
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}\)va 2x+3-z=-14
Mình làm 1 phép thôi nha những phép còn lại bạn tự nghĩ nhé !
\(\frac{x}{7}=\frac{y}{3}\) và \(x-24=y\)'
Ta có : \(x-24=y\) hay cũng có thể viết \(x-y=24\)
Ta lại có : \(\frac{x}{7}=\frac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau nên ta được :
\(\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{24}{4}=6\) ( vì \(x-y=24\) )
\(\Rightarrow\frac{x}{7}=6\Rightarrow x=6\cdot7\Rightarrow x=42\)
\(\Rightarrow\frac{y}{3}=6\Rightarrow y=6\cdot3\Rightarrow y=18\)
Vậy \(x=42\) và \(y=18\)
cho x,y,z thoa man \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\) va x-y=15
Khi do x-y-z=.......................?
\(\frac{x}{12}=\frac{y}{9}=\frac{z}{8}\) => \(\frac{x-y}{12-9}=\frac{15}{3}=5\)=> x = 60 ; y= 45 ; z=40 => x-y-z = 60-45-40 =-25
\(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\Leftrightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}\)
Áp dụng t/c dãy tỉ số=nhau:
\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
=>x/2=30=>x=60
y/3/2=30=>y=30.3/2=45
z/4/3=30=>z=30.4/3=40
Vậy x-y-z=60-45-40=-25
C, CHO 7X=3Y VA X -Y =16
D, CHO \(\frac{X}{2}=\frac{Y}{3}=\frac{Z}{4}\)VA A +2B -3C = -20
E, CHO X :Y :Z =7:4:2 VA X- 3Z =9
F,CHO \(\frac{X}{Y}=\frac{7}{10};\frac{Y}{Z}=\frac{10}{3}\)VA X+Y+Z=120
G,CHO 3X=4Y=5Z VA X-Y-Z=-42
C, CHO 7X=3Y VA X -Y =16
=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)
=> \(\hept{\begin{cases}x=-4.3\\y=-4.7\end{cases}\Rightarrow\hept{\begin{cases}x=-12\\y=-28\end{cases}}}\)
bạn viết lại đề đi đè gì mà sai hết
tim x , y, z biet,\(\frac{z}{4}=\frac{y}{3};\frac{x}{2}=\frac{z}{5}\) va x+y+z=51
\(\frac{z}{4}=\frac{y}{3}\Rightarrow\frac{z}{20}=\frac{y}{15}^{\left(1\right)}\)
\(\frac{x}{2}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{z}{20}^{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{x}{8}=\frac{z}{20}=\frac{y}{15}\)
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 \(x=8.\frac{51}{43}=\frac{408}{43}\)
\(y=15.\frac{51}{43}=\frac{765}{43}\)
\(z=20.\frac{51}{43}=\frac{1020}{43}\)
tim x, y, z biet
1. \(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}\)
2.\(\frac{2x+2}{3}=\frac{3y-1}{4}=\frac{4x+2}{5}\)va x+y+z=7
1) Áp dụng tích chất dãy tỉ số bằng nhau ta có:
\(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y-x+y}{2015-2017}=\frac{2y}{-2}\)
\(=-y\)
\(\Rightarrow xy=-2016y;x+y=-2015y;\)
\(x-y=-2017y\)
\(\Rightarrow-2016y-xy=0\)
\(\Rightarrow y\left(-2016-x\right)=0\)
\(\Rightarrow\orbr{\orbr{\begin{cases}y=0\\-2016-x=0\end{cases}\Rightarrow}}\orbr{\begin{cases}y=0\\x=-2016\end{cases}}\)
\(+) \)\(y=0\Rightarrow0+x=-2015.0=0\Rightarrow x=0\)
\(+) \)\(x=-2016\Rightarrow-2016-y=-2017y\Rightarrow-2016\)
Vậy +) x=y=0
+) x=-2016;y=1
2) Có: \(\frac{2x+2}{3}=\frac{x+1}{1,5};\frac{4z+2}{5}=\frac{z+0,5}{1,25};\frac{3y-1}{4}=\frac{y-\frac{1}{3}}{\frac{4}{3}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+1}{1,5}=\frac{y-\frac{1}{3}}{\frac{4}{3}}=\frac{z+0,5}{1,25}=\frac{x+y+z+\left(1-\frac{1}{3}+0,5\right)}{1,5+\frac{4}{3}+1,25}=\frac{7+\frac{7}{6}}{\frac{49}{12}}=2\)
Suy ra: \(x+1=2.1,5=3\Rightarrow x=2\)
\(y-\frac{1}{3}=2.\frac{4}{3}=\frac{8}{3}\Rightarrow y=3\)
\(z+0,5=2.1,25=2,5\Rightarrow z=2\)
Vậy x=2;y=3;z=2.
Câu 1 :
Áp dụng t/c dãy TSBN ta có : \(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y+x-y}{2015+2017}=\frac{x}{2016}\)
\(\Rightarrow\frac{xy}{2016}=\frac{x}{2016}\)=> xy=x => xy-x=0 => x(y-1)=0 => x=0 hoặc y=1
+) Nếu x=0 => \(\frac{0+y}{2015}=\frac{0.y}{2016}\Rightarrow\frac{y}{2015}=0\Rightarrow y=0\)
+) Nếu y=1 => \(\frac{x+1}{2015}=\frac{x.1}{2016}\)=> 2016(x+1)=2015x => 2016x+2016 = 2015x => x=-2016
Vậy ...
Câu 2 :
Áp dụng t/c dãy TSBN ta có : \(\frac{2x+2}{3}=\frac{3y-1}{4}=\frac{4z+2}{5}=\frac{6.\left(2x+2\right)+4.\left(3y-1\right)+3.\left(4z+2\right)}{3.6+4.4+5.3}\)
\(=\frac{12\left(x+y+z\right)+14}{49}=\frac{12.7+14}{49}=2\)
Từ \(\frac{2x+2}{3}=2\Rightarrow2x+2\Rightarrow6\Rightarrow2x=4\Rightarrow x=2\)
Tương tự tìm đc y=3 và z=2
Vậy ...
Tim x,y,z biet
a,5x= 8y= 20z va x-y-z = 3
b,\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)va -x+y+z =120
c,\(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)va x . y . z =20
d,\(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)va \(^{x^2+y^2-z^2}\)=585
nguyen tran phuong vy: vt sai kìa, phải là I don't know
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
\(\Rightarrow x=165;y=20;z=25\)