Toán tìm max của biểu thức: a) A=129-5|x-7| b) B= x+1/2 -|x-2/3| 19/07/2021 By Margaret tìm max của biểu thức: a) A=129-5|x-7| b) B= x+1/2 -|x-2/3| c) C=|x+5|-|x-2| d)D=2004/2003 -|x-3/5|
Giải thích các bước giải: Ta có: \(\begin{array}{l}a,\\A = 129 – 5\left| {x – 7} \right|\\\left| {x – 7} \right| \ge 0,\forall x \Rightarrow 5\left| {x – 7} \right| \ge 0 \Rightarrow A = 129 – 5\left| {x – 7} \right| \le 129\\{A_{\max }} = 129 \Leftrightarrow x = 7\\b,\\B = x + \frac{1}{2} – \left| {x – \frac{2}{3}} \right|\\\left| a \right| \ge a,\,\,\,\,\forall a \Rightarrow \left| {x – \frac{2}{3}} \right| \ge x – \frac{2}{3}\\ \Rightarrow B = x + \frac{1}{2} – \left| {x – \frac{2}{3}} \right| \le x + \frac{1}{2} – \left( {x – \frac{2}{3}} \right) = \frac{7}{6}\\ \Rightarrow {B_{\max }} = \frac{7}{6} \Leftrightarrow x = \frac{2}{3}\\d,\\D = \frac{{2004}}{{2003}} – \left| {x – \frac{3}{5}} \right|\\\left| {x – \frac{3}{5}} \right| \ge 0 \Rightarrow D = \frac{{2004}}{{2003}} – \left| {x – \frac{3}{5}} \right| \le \frac{{2004}}{{2003}}\\ \Rightarrow {D_{\max }} = \frac{{2004}}{{2003}} \Leftrightarrow x = \frac{3}{5}\end{array}\) Trả lời
a) `5|x-7|≥0` `⇒-5|x-7|≤0` `⇒129-5|x-7|≤129` Vậy Max `A=129` đạt khi `x=7` b) `|x-2/3|≥x-2/3` `⇒-|x+=-2/3|≤-x+2/3` `⇒x+1/2-|x-2/3|≤1+1/2-x+2/3=7/6` Vậy Max `B=7/6` đạt khi `x=2/3` c) `|x+5|+|x-2|=|x+5|+|2-x|≥|x+5+2-x|=7` `⇒-|x+5|-|x-2|≤-7` Vậy Max `C=-7` đạt khi `-5≤x≤2` d) `|x-3/5|≥0` `⇒-|x-3/5|≤0` `⇒2004/2003-|x-3/5|≤2004/2003` Vậy Max `D=2004/2003` đạt khi `x=3/5` Trả lời
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = 129 – 5\left| {x – 7} \right|\\
\left| {x – 7} \right| \ge 0,\forall x \Rightarrow 5\left| {x – 7} \right| \ge 0 \Rightarrow A = 129 – 5\left| {x – 7} \right| \le 129\\
{A_{\max }} = 129 \Leftrightarrow x = 7\\
b,\\
B = x + \frac{1}{2} – \left| {x – \frac{2}{3}} \right|\\
\left| a \right| \ge a,\,\,\,\,\forall a \Rightarrow \left| {x – \frac{2}{3}} \right| \ge x – \frac{2}{3}\\
\Rightarrow B = x + \frac{1}{2} – \left| {x – \frac{2}{3}} \right| \le x + \frac{1}{2} – \left( {x – \frac{2}{3}} \right) = \frac{7}{6}\\
\Rightarrow {B_{\max }} = \frac{7}{6} \Leftrightarrow x = \frac{2}{3}\\
d,\\
D = \frac{{2004}}{{2003}} – \left| {x – \frac{3}{5}} \right|\\
\left| {x – \frac{3}{5}} \right| \ge 0 \Rightarrow D = \frac{{2004}}{{2003}} – \left| {x – \frac{3}{5}} \right| \le \frac{{2004}}{{2003}}\\
\Rightarrow {D_{\max }} = \frac{{2004}}{{2003}} \Leftrightarrow x = \frac{3}{5}
\end{array}\)
a)
`5|x-7|≥0`
`⇒-5|x-7|≤0`
`⇒129-5|x-7|≤129`
Vậy Max `A=129` đạt khi `x=7`
b)
`|x-2/3|≥x-2/3`
`⇒-|x+=-2/3|≤-x+2/3`
`⇒x+1/2-|x-2/3|≤1+1/2-x+2/3=7/6`
Vậy Max `B=7/6` đạt khi `x=2/3`
c)
`|x+5|+|x-2|=|x+5|+|2-x|≥|x+5+2-x|=7`
`⇒-|x+5|-|x-2|≤-7`
Vậy Max `C=-7` đạt khi `-5≤x≤2`
d)
`|x-3/5|≥0`
`⇒-|x-3/5|≤0`
`⇒2004/2003-|x-3/5|≤2004/2003`
Vậy Max `D=2004/2003` đạt khi `x=3/5`