Question

[k còn tên đẹp]

tìm x;y

a ) |x-1|+|x+5|=6

b)(2x-y+3)⁴+|y+2|=0

c)|3-x|+|x+5|=80 /(y-2)²+10

Giups mk với!!!

Answer 1

[Nguyệt]

\(\left|x-1\right|+\left|x+5\right|=\left|x-1\right|+\left|-x-5\right|\)

\(\Rightarrow\left|x-1\right|+\left|x+5\right|\ge\left|x-1-x-5\right|\)

\(\Rightarrow\left|x-1\right|+\left|x+5\right|\ge\left|-6\right|=6\)

dấu "=" xảy ra khi \(\left(x-1\right).\left(x+5\right)\ge0\)

\(\Rightarrow-5\le x\le1\)

Vậy x={-5,-4,-3,-2,-1,0,1}

b) \(\hept{\begin{cases}\left(2x-y+3\right)^4\ge0\\\left|y+2\right|\ge0\end{cases}}\)

mà \(\left(2x-y+3\right)^4+\left|y+2\right|=0\)

dấu "=" xảy ra khi \(\hept{\begin{cases}\left(2x-y+3\right)^4=0\\\left|y+2\right|=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{5}{2}\\y=-2\end{cases}}\)

vậy \(x=-\frac{5}{2},y=-2\)

Answer 2

[Đỗ Viết Lâm Duy]

$\left|x-1\right|+\left|x+5\right|=\left|x-1\right|+\left|-x-5\right|$

$\Rightarrow \left|x-1\right|+\left|x+5\right|\ge \left|x-1-x-5\right|$

$\Rightarrow \left|x-1\right|+\left|x+5\right|\ge \left|-6\right|=6$

dấu "=" xảy ra khi $\left(x-1\right).\left(x+5\right)\ge 0$

$\Rightarrow -5\le x\le 1$

Vậy x={-5,-4,-3,-2,-1,0,1}

b) $\text{hept}\{\begin{array}{l}{\left(2x-y+3\right)}^4\ge 0\\ \left|y+2\right|\ge 0\end{array}$

mà ${\left(2x-y+3\right)}^4+\left|y+2\right|=0$

dấu "=" xảy ra khi $\text{hept}\{\begin{array}{l}{\left(2x-y+3\right)}^4=0\\ \left|y+2\right|=0\end{array}$

$\Rightarrow \text{hept}\{\begin{array}{l}x=-\genfrac{}{}{0ex}{}{5}{2}\\ y=-2\end{array}$

vậy $x=-\genfrac{}{}{0ex}{}{5}{2},y=-2$