Binh nhất
Giải các hệ phương trình sau:
a) $\left\{\begin{matrix} x+y+z=12 & & & \\ ax+5y+4z=46& & & \\ 5x+ay+3z=38 & & & \end{matrix}\right.$
b) $\left\{\begin{matrix} ax+y+z=a^{2} & & & \\ x+ay+z=3a & & & \\ x+y+az=2 & & & \end{matrix}\right.$ ( a là tham số )
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Binh nhất
a.Ta có: $\left\{\begin{matrix}x+y+z=12\Leftrightarrow ax+ay+az=12a \\ ax+5y+4z=46 \\ 5x+ay+3z=38 \end{matrix}\right.$
$\Rightarrow$$\left\{\begin{matrix}x+y+z=12\Leftrightarrow ax+ay+az=12a \\ ax+5y+4z+5x+ay+3z=84\Leftrightarrow (x+y)(5+a)+7z=84\Leftrightarrow (12-z)(5+a)+7z=84\Leftrightarrow 12a-24-az+2z=0\Leftrightarrow (a-2)(12-z )=0 \end{matrix}\right.$
$\Rightarrow a=2 hoặc z=12$
$\Rightarrow đccm$
It doesn't matter if you're the slowest kid in gym class or the fastest man alive. Every one of us is running, being alive is running, running from something, running to something or someone. And no matter how fast you are. There's some things you can't outrun. Some things always manage to catch up to you.
___ THE FLASH ___
Binh nhất
b. Ta có: $\left\{\begin{matrix}ax+y+z=a^2 (1) \\ x+ay+z=3a (2) \\ x+y+az=2 (3) \end{matrix}\right.$
$\Rightarrow (1)+(2)+(3)=(x+y+z)(a+2)=a^2+3a+2=(a+1)(a+2)$
$\Rightarrow (x+y+z)(a+2)-(a+1)(a+2)=0\Rightarrow (a+2)(x+y+z-a-1)=0\Rightarrow a=-2 hoặc x+y+z=a+1$
Thay vào tính x,y,z
Bài viết đã được chỉnh sửa nội dung bởi hangnguyen2003: 19-11-2017 - 13:15
It doesn't matter if you're the slowest kid in gym class or the fastest man alive. Every one of us is running, being alive is running, running from something, running to something or someone. And no matter how fast you are. There's some things you can't outrun. Some things always manage to catch up to you.
___ THE FLASH ___
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