Bài 34:
a.
$x^4+2x^3+x^2=x^2(x^2+2x+1)=x^2(x+1)^2$
b.
$x^3-x+3x^2y+3xy^2+y^3-y=(x^3+y^3)-(x+y)+(3x^2y+3xy^2)$
$=(x+y)(x^2-xy+y^2)-(x+y)+3xy(x+y)$
$=(x+y)(x^2-xy+y^2-1+3xy)$
$=(x+y)[(x^2+2xy+y^2)-1]=(x+y)[(x+y)^2-1]$
$=(x+y)(x+y-1)(x+y+1)$
c.
$5x^2-10xy+5y^2-20z^2$
$=5(x^2-2xy+y^2)-20z^2=5(x-y)^2-20z^2$
$=5[(x-y)^2-(2z)^2]=5(x-y-2z)(x-y+2z)$
Bài 35:
a.
$x^2+5x-6=(x^2-x)+(6x-6)=x(x-1)+6(x-1)=(x-1)(x+6)$
b.
$5x^2+5xy-x-y=5x(x+y)-(x+y)=(x+y)(5x-1)$
c.
$7x-6x^2-2=(-6x^2+3x)+(4x-2)$
$=-3x(2x-1)+2(2x-1)=(2x-1)(2-3x)$
38
a+b+c=0
=>(a+b+c)(a^2 +b^2 +c^2 -ab -bc-ca)=0
=>a^3 +b^3 +c^3 -3abc=0
=>a^3 +b^3 +c^3=3abc
Bài 36:
a.
$x^2+4x+3=(x^2+x)+(3x+3)=x(x+1)+3(x+1)=(x+1)(x+3)$
b.
$2x^2+3x-5=(2x^2-2x)+(5x-5)=2x(x-1)+5(x-1)$
$=(x-1)(2x+5)$
c.
$16x-5x^2-3=(-5x^2+15x)+(x-3)$
$=-5x(x-3)+(x-3)=(x-3)(1-5x)$
\(37,\\ a,5x\left(x-1\right)=x-1\\ \Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\\ b,2\left(x+5\right)-x^2-5x=0\\ \Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Bài 37:
a.
$5x(x-1)=x-1$
$\Leftrightarrow 5(x-1)-(x-1)=0$
$\Leftrightarrow (x-1)(5-x)=0$
$\Leftrightarrow x-1=0$ hoặc $5-x=0$
$\Leftrightarrow x=1$ hoặc $x=5$
b.
$2(x+5)-x^2-5x=0$
$\Leftrightarrow 2(x+5)-(x^2+5x)=0$
$\Leftrightarrow 2(x+5)-x(x+5)=0$
$\Leftrightarrow (x+5)(2-x)=0$
$\Leftrightarrow x+5=0$ hoặc $2-x=0$
$\Leftrightarrow x=-5$ hoặc $x=2$
Bài 38:
$a+b+c=0\Rightarrow a+b=-c$
Áp dụng HĐT đáng nhớ:
$a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3$
$=(-c)^3-3ab(-c)+c^3=-c^3+3abc+c^3=3abc$
Ta có đpcm.
37
a)5x(x-1)=x-1 b)2x+10-x^2-5x=0
5x=0 -x^2-3x+10=0
x=0 -(x+2x3/2 +9/4)+49/4=0
-(x+3/2)^2 +49/4=0
=>x-3/2=-49/4
=>x=43/4
