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dangthihuonglan618

2 năm trước

Loga Sinh Học lớp 12

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a) `P= -5| x+2| +7`

Với mọi `x` ta luôn có:`|x+2| ge 0`

`=> 5| x+2| ge 0`

`=> -5|x+2| le 0`

`=> -5|x+2| +7 le 7`

Dấu bằng xảy ra khi:

`|x+2|=0`

`=> x+2=0`

`=>x= -2`

Vậy giá trị lớn nhất của `P` là `7` khi `x= -2`

b) `Q = 4 -3 |2x-1|`

Với mọi `x` ta luôn có: `| 2x-1| ge 0`

`=> 3|2x-1| ge 0`

`=> -3|2x-1| le 0`

`=> 4 - 3|2x-1| le 4`

Dấu bằng xảy ra khi: `|2x-1|=0`

`=> 2x-1=0`

`=> 2x=1`

`=> x= 1/2`

Vậy giá trị lớn nhất của `Q` là `4` khi `x = 1/2`

c) `R= 14/(|x+2| +7 ) - 5`

Với mọi `x` ta luôn có: `|x+2| ge 0`

`=> |x+2| + 7 ge 7`

`=> 14/(|x+2| +7) le 2`

`=> 14/(|x+2| +7) - 5 le -3`

Dấu bằng xảy ra khi:

`|x+2|=0`

`=> x +2=0`

`=> x= -2`

Vậy giá trị lớn nhất của `R` là `-3` khi `x= -2`

Vote (0) Phản hồi (0) 2 năm trước

tien0310

Đáp án + Giải thích các bước giải:

Với `AAx` ta có: `|x+2|\ge0`

`=>-5|x+2|\le0`

`=>P=-5|x+2|+7\le7`

Dấu `=` xảy ra khi: `|x+2|=0`

`=>x+2=0`

`=>x=-2`

Vậy `P_(max)=7` khi `x=-2`

Với `AAx` ta có: `|2x-1|\ge0`

`=>-3|2x-1|\le0`

`=>Q=4-3|2x-1|\le4`

Dấu `=` xảy ra khi: `|2x-1|=0`

`=>2x-1=0`

`=>2x=1`

`=>x=1/2`

Vậy `Q_(max)=4` khi `x=1/2`

Với `AAx` ta có: `|x+2|\ge0`

`=>|x+2|+7\ge7`

`=>14/(|x+2|+7)\le2`

`=>R=14/(|x+2|+7)-5\le-3`

Dấu `=` xảy ra khi: `|x+2|=0`

`=>x+2=0`

`=>x=-2`

Vậy `R_(max)=-3` khi `x=-2`

Vote (0) Phản hồi (0) 2 năm trước

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