Bài 38:
a.
\((x-2)^3-x(x+1)(x-1)+6x(x-3)\)
\(=x^3-3x^2.2+3.2^2x-8-x(x^2-1)+6x^2-18x\)
\(=x^3-6x^2+12x-8-x^3+x+6x^2-18x\)
\(=-5x-8\)
b.
\((x-2)(x^2-2x+4)(x+2)(x^2+2x+4)\)
\(=(x-2)(x^2+2x+4)(x+2)(x^2-2x+4)\)
\(=(x^3-2^3)(x^3+2^3)=(x^3)^2-(2^3)^2=x^6-2^6=x^6-64\)
Bài 39:
a.
\((x-3)(x^2+3x+9)+x(x+2)(2-x)=1\)
\(\Leftrightarrow x^3-3^3+x(2^2-x^2)=1\)
\(\Leftrightarrow 4x-27=1\Leftrightarrow 4x=28\Leftrightarrow x=7\)
b.
\((x+1)^3-(x-1)^3-6(x-1)^2=-10\)
\(\Leftrightarrow [(x+1)-(x-1)][(x+1)^2+(x+1)(x-1)+(x-1)^2]-6(x-1)^2=-10\)
\(\Leftrightarrow 2(x^2+2x+1+x^2-1+x^2-2x+1)-6(x-1)^2=-10\)
\(\Leftrightarrow 2(3x^2+1)-6(x^2-2x+1)=-10\)
\(\Leftrightarrow 6x^2+2-6x^2+12x-6=-10\)
\(\Leftrightarrow 12x-4=-10\Leftrightarrow 12x=-6\Leftrightarrow x=-\frac{1}{2}\)
Bài 40:
a. Đặt $b+c-a=x; a+c-b=y; a+b-c=z$ thì:
\((a+b+c)^3-(b+c-a)^3-(a+c-b)^3-(a+b-c)^3\)
\(=(x+y+z)^3-x^3-y^3-z^3=x^3+y^3+z^3+3(x+y)(y+z)(x+z)-x^3-y^3-z^3\)
\(=3(x+y)(y+z)(x+z)=3(2c)(2a)(2b)=24abc\)
b.
\((a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)\)
\(=a^3+b^3+3ab(a+b)+b^3+c^3+3bc(b+c)+c^3+a^3+3ac(c+a)-3(a+b)(b+c)(c+a)\)
\(=2(a^3+b^3+c^3)-6abc+3[ab(a+b)+bc(b+c)+ca(c+a)+2abc]-3(a+b)(b+c)(c+a)\)
\(=2(a^3+b^3+c^3)-6abc+3(a+b)(b+c)(c+a)-3(a+b)(b+c)(c+a)=2(a^3+b^3+c^3-3abc)\)
