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ta có:
\(-\frac{x}{2}=\frac{-y}{4}=\frac{6}{-8}\)
=>\(\frac{-x}{2}=\frac{6}{-8}\)
=>-8.(-x)=6.2
=>8x=12
=>x=3/2
lại có:
\(\frac{-y}{4}=\frac{6}{-8}\)
=>-8.(-y)=6.4
=>8y=24
=>y=3
Vậy x=3/2; y=3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn viết sai đề rồi nhé :)
16x8+8x24-16x11+16:0,1 đề đúng phải như vậy nhé
Ta có :
16x8+8x24-16x11+16:0.1
= 16x8+8x24-16x11+16x10
= 16x(8-11+10) + 8x24
= 16x7+8x24
= 16x7 + 16x12
= 16x(7+12)=304
bạn kiểm tra lại xem
Chúc bạn học tốt
16 . 8 + 8 . 2 . 12 - 16 . 11 + 1,6 : 0,1
= 16 . 8 + 16 . 12 - 16 . 11 + 16 . 1
= 16 ( 8 + 12 - 16 + 1 )
= 16 . 5
= 80
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=-1\\8x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x=-5\\4x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}4x+8y=-4\\4x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=-5\\x+2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}3x-6y=-12\\-3x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=-2\left(loại\right)\\-3x+6y=10\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\varnothing\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=-2\\2x+y=-2\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in R\)
\(a,\left\{{}\begin{matrix}3x-2y=-1\\4x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2\left(4x-2\right)=-1\\y=4x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-8x+4=-1\\y=4x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x=-5\\y=4x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4.1-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}x+2y=-1\\4x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-2y\\4\left(-1-2y\right)+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-2y\\-4-8y+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-2y\\-5y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-2\left(-1\right)\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}x-2y=-4\\-3x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-4\\-3\left(2y-4\right)+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-4\\-6y+12+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-4\\12=10\left(vô.lí\right)\end{matrix}\right.\)
\(d,\left\{{}\begin{matrix}2x+y=-2\\4x+2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=-2\\2x+y=-2\left(luôn.đúng\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4}{3}-\left(x-\frac{1}{5}\right)=\left|\frac{-3}{10}+\frac{1}{2}\right|-\frac{1}{6}\)
\(\frac{4}{3}-\left(x-\frac{1}{5}\right)=\frac{1}{5}-\frac{1}{6}\)
\(\frac{4}{3}-\left(x-\frac{1}{5}\right)=\frac{1}{30}\)
\(x-\frac{1}{5}=\frac{4}{3}-\frac{1}{30}\)
\(x-\frac{1}{5}=\frac{13}{10}\)
\(x=\frac{13}{10}+\frac{1}{5}\)
\(x=\frac{3}{2}\)