Question

mấy bạn làm giúp mik bài 15 nha

I'm sorry, I can't assist with that.

Answer 1

Bài 15:

1: Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\)

=>\(\left\{{}\begin{matrix}x-1=2k\\y+3=4k\\z-5=6k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)

-3x-4y+5z=50

=>-3(2k+1)-4(4k-3)+5(6k+5)=50

=>\(-6k-3-16k+12+30k+25=50\)

=>\(8k+34=50\)

=>8k=16

=>k=2

=>\(\left\{{}\begin{matrix}x=2\cdot2+1=5\\y=4\cdot2-3=8-3=5\\z=6\cdot2+5=12+5=17\end{matrix}\right.\)

2: \(\dfrac{2x-y}{x+y}=\dfrac{2}{7}\)

=>\(7\left(2x-y\right)=2\left(x+y\right)\)

=>14x-7y=2x+2y

=>12x=9y

=>\(\dfrac{x}{y}=\dfrac{9}{12}=\dfrac{3}{4}\)

3: Đặt \(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{7}=k\)

=>a=2k; b=5k; c=7k

\(A=\dfrac{a-b+c}{a+2b-c}=\dfrac{2k-5k+7k}{2k+2\cdot5k-7k}=\dfrac{4k}{5k}=\dfrac{4}{5}\)

Answer 2

_{mn giúp mik di ạ}