Toán Tìm GTLN P = -|y+7|+49 Q = 2019-|3y-12| 06/09/2021 By Aubrey Tìm GTLN P = -|y+7|+49 Q = 2019-|3y-12|
a) Ta có: $|y+7|\geq 0\forall y$$\Leftrightarrow -|y+7|\leq 0\forall y$$\Leftrightarrow -|y+7|+49\leq 49\forall y$$\Leftrightarrow P\leq 49\forall y$Dấu “=” xảy ra khi $|y+7|=0\Leftrightarrow y=-7$Vậy $Max_{P}=49\Leftrightarrow y=-7$b) Ta có: $|3y-12|\geq 0\forall y$$\Leftrightarrow -|3y-12|\leq 0\forall y$$\Leftrightarrow 2019-|3y-12|\leq 2019\forall y$$\Leftrightarrow Q\leq 2019\forall y$Dấu “=” xảy ra khi $|3y-12|=0\Leftrightarrow y=4$Vậy $Max_{Q}=2019\Leftrightarrow y=4$ Trả lời
BẠN THAM KHẢO NHA! BÀI LÀM: $P=-|y+7|+49$ $-|y+7| \leq 0 \to -|y+7|+49 \leq 49 \to P\leq49 \to maxP=49$ Dấu “=” xảy ra khi $-|y+7|=0 \to y=-7$ $Q=2019-|3y-12|$ $-|3y-12| \leq 0 \to 2019-|3y-12| \leq 2019 \to Q\leq2019 \to maxQ=2019$ Dấu “=” xảy ra khi: $-|3y-12$=0 \to y=4$ Trả lời
a) Ta có: $|y+7|\geq 0\forall y$
$\Leftrightarrow -|y+7|\leq 0\forall y$
$\Leftrightarrow -|y+7|+49\leq 49\forall y$
$\Leftrightarrow P\leq 49\forall y$
Dấu “=” xảy ra khi $|y+7|=0\Leftrightarrow y=-7$
Vậy $Max_{P}=49\Leftrightarrow y=-7$
b) Ta có: $|3y-12|\geq 0\forall y$
$\Leftrightarrow -|3y-12|\leq 0\forall y$
$\Leftrightarrow 2019-|3y-12|\leq 2019\forall y$
$\Leftrightarrow Q\leq 2019\forall y$
Dấu “=” xảy ra khi $|3y-12|=0\Leftrightarrow y=4$
Vậy $Max_{Q}=2019\Leftrightarrow y=4$
BẠN THAM KHẢO NHA!
BÀI LÀM:
$P=-|y+7|+49$
$-|y+7| \leq 0 \to -|y+7|+49 \leq 49 \to P\leq49 \to maxP=49$
Dấu “=” xảy ra khi $-|y+7|=0 \to y=-7$
$Q=2019-|3y-12|$
$-|3y-12| \leq 0 \to 2019-|3y-12| \leq 2019 \to Q\leq2019 \to maxQ=2019$
Dấu “=” xảy ra khi: $-|3y-12$=0 \to y=4$