x phần 4=y phần 3; y phần 2=z phần 3 và x+y+z =-6 08/09/2021 Bởi Reese x phần 4=y phần 3; y phần 2=z phần 3 và x+y+z =-6
$\Large\frac{x}{4}$ = $\Large\frac{y}{3}$ <=> $x$ = $\Large\frac{4y}{3}$ => $x+y+z =-6$<=> $\Large\frac{4y}{3}$$+y+z =-6$ <=> $\Large\frac{7y}{3}$$+z =-6$ (1).+) $\Large\frac{y}{2}$ = $\Large\frac{z}{3}$ <=> $y$ = $\Large\frac{2z}{3}$ (2) Thay (2) vào (1) ta có:$\Large\frac{\Large\frac{14z}{3}}{3}$$+z =-6$ $\Large\frac{14z}{9}$$+z =-6$ => z= $\Large\frac{-27}{7}$ => y= $\Large\frac{-18}{7}$ => x= $\Large\frac{-24}{7}$ Bình luận
Ta có: $\Large\frac{x}{4}$ = $\Large\frac{y}{3}$⇔ $x$ = $\Large\frac{4y}{3}$⇒ $x+y+z =-6$⇔ $\Large\frac{4y}{3}$$+y+z =-6$ ⇔ $\Large\frac{7y}{3}$$+z =-6$ (1).Ta lại có: $\Large\frac{y}{2}$ = $\Large\frac{z}{3}$⇔ $y$ = $\Large\frac{2z}{3}$ (2)Thay (2) vào (1) ta được:$\Large\frac{\Large\frac{14z}{3}}{3}$$+z =-6$ ⇔$\Large\frac{14z}{9}$$+z =-6$ ⇒ $z= \Large\frac{-27}{7}$ ⇒ $y= \Large\frac{-18}{7}$ ⇒ $x= \Large\frac{-24}{7}$ Bình luận
$\Large\frac{x}{4}$ = $\Large\frac{y}{3}$
<=> $x$ = $\Large\frac{4y}{3}$
=> $x+y+z =-6$
<=> $\Large\frac{4y}{3}$$+y+z =-6$
<=> $\Large\frac{7y}{3}$$+z =-6$ (1)
.
+) $\Large\frac{y}{2}$ = $\Large\frac{z}{3}$
<=> $y$ = $\Large\frac{2z}{3}$ (2)
Thay (2) vào (1) ta có:
$\Large\frac{\Large\frac{14z}{3}}{3}$$+z =-6$
$\Large\frac{14z}{9}$$+z =-6$
=> z= $\Large\frac{-27}{7}$
=> y= $\Large\frac{-18}{7}$
=> x= $\Large\frac{-24}{7}$
Ta có: $\Large\frac{x}{4}$ = $\Large\frac{y}{3}$⇔ $x$ = $\Large\frac{4y}{3}$⇒ $x+y+z =-6$
⇔ $\Large\frac{4y}{3}$$+y+z =-6$
⇔ $\Large\frac{7y}{3}$$+z =-6$ (1)
.Ta lại có: $\Large\frac{y}{2}$ = $\Large\frac{z}{3}$⇔ $y$ = $\Large\frac{2z}{3}$ (2)
Thay (2) vào (1) ta được:
$\Large\frac{\Large\frac{14z}{3}}{3}$$+z =-6$
⇔$\Large\frac{14z}{9}$$+z =-6$
⇒ $z= \Large\frac{-27}{7}$
⇒ $y= \Large\frac{-18}{7}$
⇒ $x= \Large\frac{-24}{7}$