Đáp án:

$8)$ $\min H(x)=$`-\frac{121}{8}`

$9)$ `\max I(x)=\frac{1236}{5}`

$10)$ $\min K(x)=107$

Giải thích các bước giải:

$8)$ `H(x)=(3x-4)^2-(x-5)^2`

`=9x^2-24x+16-(x^2-10x+25)`

`=8x^2-14x-9=8(x^2-\frac{7}{4}x-\frac{9}{8})`

`=8[x^2-2·\frac{7}{8}x+(\frac{7}{8})^2-(\frac{7}{8})^2-\frac{9}{8}]`

`=8[(x-\frac{7}{8})^2-\frac{121}{64}]=8(x-\frac{7}{8})^2-\frac{121}{8}`

Vì `(x-\frac{7}{8})^2≥0∀x∈R⇔8(x-\frac{7}{8})^2≥0∀x∈R`

`⇔8(x-\frac{7}{8})^2-\frac{121}{8}≥-\frac{121}{8}∀x∈R`

`H(x)≥-\frac{121}{8}∀x∈R`

Dấu "=" xảy ra `⇔(x-\frac{7}{8})^2=0⇔x=\frac{7}{8}`

Vậy $\min H(x)=$`-\frac{121}{8}` tại `x=\frac{7}{8}`

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$8)$ `I(x)=(2x+7)^2-(3x-5)^2+55`

`=4x^2+28x+49-(9x^2-30x+25)+55`

`=-5x^2+58x+79=-5(x^2-\frac{58}{5}x-\frac{79}{5})`

`=-5[x^2-2·\frac{29}{5}x+(\frac{29}{5})^2-(\frac{29}{5})^2-\frac{79}{5}]`

`=-5[(x-\frac{29}{5})^2-\frac{1236}{25}]`

`=-5(x-\frac{29}{5})^2+\frac{1236}{5}`

Vì `(x-\frac{29}{5})^2≥0∀x∈R⇔-5(x-\frac{29}{5})^2≤0∀x∈R`

`⇔-5(x-\frac{29}{5})^2+\frac{1236}{5}≤\frac{1236}{5}∀x∈R`

`⇔I(x)≤\frac{1236}{5}∀x∈R`

Dấu "=" xảy ra `⇔(x-\frac{29}{5})^2=0⇔x=\frac{29}{5}`

Vậy `\max I(x)=\frac{1236}{5}` tại `x=\frac{29}{5}`

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$10)$ `K(x)=x^4-8x^2+123`

`=x^4-2·4x^2+4^2-4^2+123`

`=(x^2-4)^2+107`

Vì $(x^2-4)^2≥0∀x∈R⇔(x^2-4)^2+107≥107∀x∈R$

$⇔K(x)≥107∀x∈R$

Dấu "=" xảy ra $⇔(x^2-4)^2=0⇔x^2=4⇔x=±2$

Vậy $\min K(x)=107$ tại $x=±2$