\(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\Rightarrow\frac{2x}{4}=\frac{2y}{3}=\frac{3z}{4}=\frac{2\left(x+y+x\right)+z}{4+3+4}=\frac{2.145+z}{11}\)

\(\Rightarrow\frac{3z}{4}=\frac{290+z}{11}\Rightarrow z=10\)

Từ đó tìm ra x,y thông qua biểu thức \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}=\frac{3.10}{4}=\frac{15}{2}\)

Theo bài ra ta cs 

\(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)

\(\Rightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}\)và \(x+y+z=145\)

ADTC dãy tỉ số bằng nhau ta cs 

\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x+y+z}{2+\frac{3}{2}+\frac{4}{3}}=\frac{145}{\frac{29}{6}}=30\)

\(\hept{\begin{cases}\frac{x}{2}=30\\\frac{y}{\frac{3}{2}}=30\\\frac{z}{\frac{4}{3}}=30\end{cases}\Rightarrow\hept{\begin{cases}x=60\\y=45\\z=40\end{cases}}}\)

 x/2=2y/3=3z/4

=> x= 4y/3 ; z= 8y/9

Co x+y+z=145

=> 4y/3+y+8y/9=145

=>12y/9+9y/9+8y/9=145

=> 29y= 145*9

=> y= (29*5*9)/29= 45

=> x=60

=> Z=40

Vay x= 60 ; y=45 ; z=40

nhầm rồi

1cong1=

Ta có: \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}.\)

\(\Rightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}\) và \(x+y+z=145.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x+y+z}{2+\frac{3}{2}+\frac{4}{3}}=\frac{145}{\frac{29}{6}}=30.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=30\Rightarrow x=30.2=60\\\frac{y}{\frac{3}{2}}=30\Rightarrow y=30.\frac{3}{2}=45\\\frac{z}{\frac{4}{3}}=30\Rightarrow z=30.\frac{4}{3}=40\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(60;45;40\right).\)

Chúc bạn học tốt!

x¹⁵=x

=> x = 1 hoặc x = 0.

Vậy x = 1 hoặc x  = 0.

X=1,65 nha bạn

x = 1,65

X = 1,65

\(\dfrac{x}{y}=\dfrac{2}{5}=\dfrac{x}{2}=\dfrac{y}{5}\)

Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=k2\\y=k5\end{matrix}\right.\)

mà \(xy=40\)

\(\Rightarrow2k.5k=40\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{5}=4\\\dfrac{x}{2}=\dfrac{y}{5}=-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=8;y=20\\x=-8;y=-20\end{matrix}\right.\)