Question

Tìm x,y,z biết

n) \(\frac{x+1}{3}=\frac{y+2}{4}=\frac{Z+3}{5}vàx+y+z=18\)\(\)

O)\(\frac{x}{-3}=\frac{y}{-5}=\frac{Z}{-4}và3Z-2x=36\)

P) \(\frac{X}{4}=\frac{Y}{3}\)VÀ \(x.y=12\)

 

 

Answer 1

\(\text{Áp dụng dãy tỉ lệ bằng nhau ta được:}\)

\(\frac{x+1}{3}=\frac{y+2}{4}=\frac{z+3}{5}=\frac{x+y+z+6}{3+4+5}=\frac{24}{12}=2\)

\(\Rightarrow\hept{\begin{cases}x=2.3-1=5\\y=2.4-1=7\\z=2.5-3=7\end{cases}}\)

Answer 2

o) \(\frac{x}{-3}=\frac{y}{-5}=\frac{z}{-4}=\frac{2x}{2.\left(-3\right)}=\frac{y}{-5}=\frac{3z}{3.\left(-4\right)}=\frac{2x}{-6}=\frac{y}{-5}=\frac{3z}{-12}\)
Áp dụng tính chất DTSBN:
\(\frac{x}{-3}=\frac{y}{-5}=\frac{z}{-4}=\frac{2x}{-6}=\frac{y}{-5}=\frac{3z}{-12}=\frac{3z-2x}{-12-\left(-6\right)}=\frac{36}{-6}=-6\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{-3}=-6\Rightarrow x=-3.\left(-6\right)=18\\\frac{y}{-5}=-6\Rightarrow y=-5.\left(-6\right)=30\\\frac{z}{-4}=-6\Rightarrow z=-4.\left(-6\right)=24\end{cases}}\)
Vậy x = 18, y = 30, z = 24
p) \(\frac{x}{4}=\frac{y}{3}\Rightarrow\left(\frac{x}{4}\right)^2=\left(\frac{y}{3}\right)^2=\frac{xy}{4.3}=\frac{12}{12}=1\)
\(\Rightarrow\hept{\begin{cases}\left(\frac{x}{4}\right)^2=1\Rightarrow\frac{x^2}{16}=1\Rightarrow x^2=1.16=16=4^2\\\left(\frac{y}{3}\right)^2=1\Rightarrow\frac{y^2}{9}=1\Rightarrow y^2=1.9=9=3^2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x\in\text{{}4;-4\\y\in\text{{}3;-3\end{cases}}\)Nhớ thêm dấu ''}'' ở đằng sau -4 và -3 nhé
Vậy ...

Answer 3

n) Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x+1}{3}=\frac{y+2}{4}=\frac{z+3}{5}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{3+4+5}=\frac{18+6}{12}=\frac{24}{12}=2\)

=>\(\frac{x+1}{3}=2=>x+1=2.3=6=>x=6-1=5\)

=>\(\frac{y+2}{4}=2=>y+2=4.2=8=>y=8-2=6\)

=>\(\frac{z+3}{5}=2=>z+3=2.5=10=>z=10-3=7\)

Vậy x=5;y=6;z=7