cho A=4x^2+4x+2 b=2x^2-2x+1 c=-15-x^2+6x b,c/m c luôn âm c, tìm gtln (gtnn) của a,b,c

Quangquoctruong

2 năm trước

Loga Sinh Học lớp 12

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nhamthehao294

Đáp án:

\(\begin{array}{l}
b)dpcm\\
c)MinA = 1\\
MinB = \dfrac{1}{2}\\
MaxC = - 6
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
b)C = - 15 - {x^2} + 6x\\
= - \left( {{x^2} - 6x + 15} \right)\\
= - \left( {{x^2} - 6x + 9 + 6} \right)\\
= - {\left( {x - 3} \right)^2} - 6\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 3} \right)^2} \le 0\\
\to - {\left( {x - 3} \right)^2} - 6 < 0\\
\to dpcm\\
c)A = 4{x^2} + 4x + 2\\
= 4{x^2} + 4x + 1 + 1\\
= {\left( {2x + 1} \right)^2} + 1\\
Do:{\left( {2x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {2x + 1} \right)^2} + 1 \ge 1\\
\to MinA = 1\\
\Leftrightarrow x = - \dfrac{1}{2}\\
B = 2{x^2} - 2x + 1\\
= 2{x^2} - 2.x\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + \dfrac{1}{2}\\
= {\left( {x\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2}\\
Do:{\left( {x\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2} \ge \dfrac{1}{2}\\
\to MinB = \dfrac{1}{2}\\
\Leftrightarrow x\sqrt 2 - \dfrac{1}{{\sqrt 2 }} = 0\\
\to x = \dfrac{1}{2}\\
C = - {\left( {x - 3} \right)^2} - 6\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 3} \right)^2} \le 0\\
\to - {\left( {x - 3} \right)^2} - 6 \le - 6\\
\to MaxC = - 6\\
\Leftrightarrow x = 3
\end{array}\)

Vote (0) Phản hồi (0) 2 năm trước

lephonghao1110

Đáp án + Giải thích các bước giải:

`C=-15-x^2+6x`

`=>C=-(x^2-6x+15)`

`=>C=-(x^2-6x+9+6)`

`=>C=-(x-3)^2-6<0`

Vậy `C<0` với `AAx`

`A=4x^2+4x+2`

`=>A=4x^2+4x+1+1`

`=>A=(2x+1)^2+1\ge1`

Dấu `=` xảy ra khi: `(2x+1)^2=0`

`=>2x+1=0`

`=>2x=-1`

`=>x=-1/2`

Vậy `A_(min)=1` khi `x=-1/2`

`--`

`B=2x^2-2x+1`

`=>B=2(x^2-x+1/2)`

`=>B=2(x^2-x+1/4+1/4)`

`=>B=2(x-1/2)^2+1/2\ge1/2`

Dấu `=` xảy ra khi: `(x-1/2)^2=0`

`=>x-1/2=0`

`=>x=1/2`

Vậy `B_(min)=1/2` khi `x=1/2`

`--`

`C=-15-x^2+6x`

`=>C=-(x^2-6x+15)`

`=>C=-(x^2-6x+9+6)`

`=>C=-(x-3)^2-6\le-6`

Dấu `=` xảy ra khi: `(x-3)^2=0`

`=>x-3=0`

`=>x=3`

Vậy `C_(max)=-6` khi `x=3`

Vote (0) Phản hồi (0) 2 năm trước

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