\(\left\{{}\begin{matrix}3x+y=3\\3x-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=0\\3x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3x+y=3\\3x-y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+y+3x-y=3-3\\3x-y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x=0\\3x-y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\3.0-y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}7x=-6\\\dfrac{5\left(x+3y\right)-3\left(y-2\right)}{15}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\5x+15y-3y+6=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\12y=9-5x=9+5\cdot\dfrac{6}{7}=9+\dfrac{30}{7}=\dfrac{93}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\y=\dfrac{93}{7\cdot12}=\dfrac{93}{84}=\dfrac{31}{28}\end{matrix}\right.\)

\(\begin{cases} x-3y=-2\\2x+y=3 \end{cases} <=> \begin{cases} x-3(3-2x)=-2\\y=3-2x \end{cases} <=> \begin{cases} 7x=7\\y=3-2x \end{cases} \\<=> \begin{cases} x=1\\y=3-2.1 \end{cases} <=>\begin{cases} x=1\\y=1 \end{cases}\)

ban vao olm.vn/hoi-dap/question/709460.html nhe

good luck!!!!!!!!!

ST1+ST2=7/8*2=7/4

ST1+ST2=17/8*2=17/4

ST1*3+ST2*3=17/4

ST1+ST2=7/4

ST1*2=5/2

ST1=5/2:2=5/4

ST2=7/4-5/4=1/2

chuc ban hoc gioi nha

ko biết

Ta có: \(\left\{{}\begin{matrix}-x-y=2\\-2x-3y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\left(x+y\right)=2\\-\left(2x+3y\right)=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-2\\2x+3y=-9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2-y\\2\cdot\left(-2-y\right)+3y=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2-y\\-4-2y+3y+9=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2-y\\y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2-y\\y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2-\left(-5\right)\\y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2+5=3\\y=-5\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)

\(-15.\left(\chi+2\right)+7.\left(2\chi-3\right)=\left(-5\right).10\)

\(\Rightarrow-15\chi+\left(-30\right)+14\chi-21=-50\)

\(\Rightarrow\left(-15\chi+14\chi\right)=-50+30+21\)

\(\Rightarrow\chi=1\)

HTDT

theo đề bài, ta có:

-x/2=3y/4 = -5z/6

mà -x/2= -5x/10

=> -5x/10 = 3y/4 = -5z/6

=> -5x/10 . 1/3= 3y/4 . 1/3 = -5z/6 . 1/3

=> -5x/30 = 3y/12 = -5z/18

=> -5x/30 = y/ 4= -5z/ 18

mà y/4 = 4y/ 16

=> -5x/30 = 4y/16 = -5z/18

theo t/c của dãy tỉ số bàng nhau, ta có

- 5z-(-5x) +4y/ 18- 30 +16 = -(5z - 5x -4y)/ 4 = - 50/4 = -25/2

=> -x2 : 2= ... ( tương tự với y, z)

vậy x= ... y=... z=...

p/s bạn viết lại ra giấy cho dễ hiểu

hơi rối, mình ko viết đc ps