Question

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1) Tìm x E Z biết: (x²-1).(x² - 4).(x² - 7).(x² - 10) < 0

2) Tìm Min A biết: A = |x - a| + |x - b| + |x - c| + |x - d| và a<b<c<d

Answer 1

[soyeon_Tiểubàng giải]

1) (x^2 - 1)(x^2 - 4)(x^2 - 7)(x^2 - 10) < 0

<=> [(x^2 - 1)(x^2 - 10)][(x^2 - 4)(x^2 - 7)] < 0

<=> (x^4 - x^2 - 10x^2 + 10)(x^4 - 4x^2 - 7x^2 + 28) < 0

<=> (x^4 - 11x^2 + 10)(x^4 - 11x^2 + 28) < 0

=> x^4 - 11x^2 + 10 và x^4 - 11x^2 + 28 là 2 số trái dấu

Mà x^4 - 11x^2 + 10 < x^4 - 11x^2 + 28

Nên \(\left\{\begin{matrix}x^4-11x^2+10< 0\\x^4-11x^2+28>0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}\left(x^2-\frac{11}{2}\right)^2-\frac{81}{4}< 0\\\left(x^2-\frac{11}{2}\right)^2-\frac{9}{4}>0\end{matrix}\right.\)\(\Leftrightarrow\frac{9}{4}< \left(x^2-\frac{11}{2}\right)^2< \frac{81}{4}\)

\(\Rightarrow\left[\begin{matrix}\frac{3}{2}< x^2-\frac{11}{2}< \frac{9}{2}\\-\frac{3}{2}>x^2-\frac{11}{2}>-\frac{9}{2}\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}7< x^2< 10\\4>x^2>1\end{matrix}\right.\)

do \(x\in Z\Rightarrow x^2\in N\)=> x² = 9\(\Rightarrow\left[\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy x = 3; x = -3

2) A = |x - a| + |x - b| + |x - c| + |x - d|

A = |x - a| + |x - b| + |c - x| + |d - x|\(\le\)

|x - a + x - b + c - x + d - x|= |c - a + d - b|

= c - a + d - b ( vì c - a > 0; d - b > 0)

Dấu "=" xảy ra khi \(\left\{\begin{matrix}x-a\ge0\\x-b\ge0\\x-c\le0\\x-d\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}a\le x\\b\le x\\c\ge x\\d\ge x\end{matrix}\right.\)

Vậy Min A = c - a + d - b khi \(\left\{\begin{matrix}a\le x\\b\le x\\c\ge x\\d\ge x\end{matrix}\right.\); a < b < c < d

\(\left\{\begin{matrix}a\le x\\b\le x\\c\ge x\\d\ge x\end{matrix}\right.;a< b< c< d}\)