Đáp án:
Giải thích các bước giải:
$4.$
$P=(x-1)^3-4x(x+1)(x-1)+3(x-1)(x^2+x+1)$
$P=x^3-3x^2+3x-1-4x(x^2-1)+3(x^3-1)$
$P=x^3-3x^2+3x-1-4x^3+4x+3x^3-3$
$P=(x^3-4x^3+3x^3)+(3x+4x)-(1+3)-3x^2$
$P=7x-4-3x^2$
$\text{Với $x=-2$ , ta có:}$
$P=7.(-2)-4-3.(-2)^2$
$P=-14-4-3.4$
$P=-18-12=-30$
$\text{Vậy $P=-30$ khi $x=-2$}$
$9.$
$\text{Với $x=\dfrac{9}{10}$ , ta có:}$
$P=\left (\dfrac{9}{10} \right )^3+\dfrac{3}{10}.\left (\dfrac{9}{10} \right )^2+\dfrac{3}{100}.\dfrac{9}{10}+\dfrac{1}{1000}$
$P=\dfrac{729}{1000}+\dfrac{3}{10}.\dfrac{81}{100}+\dfrac{27}{1000}+\dfrac{1}{1000}$
$P=\dfrac{729+27+1}{1000}+\dfrac{243}{1000}$
$P=\dfrac{757+243}{1000}=\dfrac{1000}{1000}$
$P=1$
$\text{Vậy $P=1$ khi $x=\dfrac{9}{10}$}$
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