giai bang pp cong dai so
{x-3y=-2
{2x+y=3
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\(\left\{{}\begin{matrix}3x+y=3\\3x-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=0\\3x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3x+y=3\\3x-y=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+y+3x-y=3-3\\3x-y=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x=0\\3x-y=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\3.0-y=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x+3x=-6\\\dfrac{x+3y}{3}-\dfrac{y-2}{5}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=-6\\\dfrac{5\left(x+3y\right)-3\left(y-2\right)}{15}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\5x+15y-3y+6=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\12y=9-5x=9+5\cdot\dfrac{6}{7}=9+\dfrac{30}{7}=\dfrac{93}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\y=\dfrac{93}{7\cdot12}=\dfrac{93}{84}=\dfrac{31}{28}\end{matrix}\right.\)
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
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
\(\left\{{}\begin{matrix}x-3y=-2\\2x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-6y=-4\\2x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-7y=-7\\2x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\2x+1=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\2x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất là \(\left(1;1\right)\).
{x−3y=−22x+y=3<=>{x−3(3−2x)=−2y=3−2x<=>{7x=7y=3−2x<=>{x=1y=3−2.1<=>{x=1y=1