Ta có x=17 => 18 = 17 + 1

Ta có :

A(x) = x^6 - 18x^5+ 18x^4-18x^3+18x^2-18x + 2 

= 17^6-(17+1)*17^5+(17+1)*17^4-(17+1)*17^3+(17+1)*17^2-(17+1)*17+2 

= 17^6-17^6-17^5+17^5+17^4-17^4-17^3+17^3+17^2-17^2-17+2

= -17+2

=-15 

k cho mình nhé

\(a.\) \(ax^2-a^2x-x+a\)

\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)

\(=ax\left(x-a\right)-\left(x-a\right)\)

\(=\left(ax-1\right)\left(x-a\right)\)

\(b.\) \(18x^3-12x^2+2x\)

\(=2x\left(9x^2-6x+1\right)\)

\(=2x\left(3x-1\right)^2\)

\(c.\) \(x^3-5x^2-4x+20\)

\(=\left(x^3-5x^2\right)-\left(4x-20\right)\)

\(=x^2\left(x-5\right)-4\left(x-5\right)\)

\(=\left(x^2-4\right)\left(x-5\right)\)

\(=\left(x-2\right)\left(x+2\right)\left(x-5\right)\)

\(d.\) \(\left(x+7\right)\left(x+15\right)+15\)

\(=x^2+15x+7x+105+15\)

\(=x^2+22x+120\)

\(=\left(x+10\right)\left(x+12\right)\)

chắc bạn chép sai đề rồi , hai căn đầu phải 1 cộng 1 trừ chứ

Đặt

\(x=\dfrac{y+1}{2}\Rightarrow y=2x-1\)

\(\Rightarrow y=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\)

\(y^3=8+3\sqrt[3]{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=8+3y\)

\(\Rightarrow y^3-3y-8=0\\ \)

\(\Leftrightarrow8x^3-12x^2-6=0\)

\(\Rightarrow4x^3-6x^2-3=0\)

thay p vào ta có

\(P=12x^5-18x^4+4x^3-15x^2-21\)

\(=12x^5-18x^4-9x^2-4x^3-6x^2-21\)

\(=3x^2\left(4x^2-6x^2-3\right)+4x^3-6x^2-3\\ =3x^2.0+0-18\\ =-18\)

Lời giải:

a.

$=\frac{1}{2}(x^2-4y^2)=\frac{1}{2}[x^2-(2y)^2]=\frac{1}{2}(x-2y)(x+2y)$

b.

$=\frac{1}{3}x(y+3xz+3z)$

c.

$=\frac{2}{25}x(225x^2-4)=\frac{2}{25}(15x-2)(15x+2)$

d.

$=\frac{1}{5}x^2(2+25x+5y)$