Question

tìm x,y,z

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) và 2x+3y-z=50

Answer 1

Mình k ghi lại đề bài bạn nhé!

=>\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Ap dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{2x-2+3y-6-z+3}{4+9-4}=\dfrac{\left(2x+3y-z\right)-\left(2+6-3\right)}{9}\) = \(\dfrac{50-5}{9}=5\)

Suy ra: \(\dfrac{2x-2}{4}=5=>2x-2=20=>2x=22=>x=11\)

\(\dfrac{3y-6}{9}=5=>3y-6=45=>3y=51=>y=17\)

\(\dfrac{z-3}{4}=5=>z-3=20=>z=23\)

Vậy x=11; y=17 ; z=23

Answer 2

Có : \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

<=> \(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ,ta có :

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=\dfrac{\left(2x+3y-z\right)+\left(-2-6+3\right)}{9}=\dfrac{50-5}{9}=\dfrac{45}{9}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=2.5=10\\y-2=3.5=15\\z-3=4.5=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)