1) Phân tích đa thức thành nhân tử
a) x^3 + x^2 +x
b) 4x^2y – 5x^3y – 6xy^2
c) 15x^3y^3 – 5x^2y^3 + 20xy^2
d) 8x^3 (x-2) – x^2 + 2x
2) Tìm x
a) x^2 – 3x =0
b) 2x (x-2) – (x-2) =0
c) (x-5)^2 – x+5 = 0
1) Phân tích đa thức thành nhân tử
a) x^3 + x^2 +x
b) 4x^2y – 5x^3y – 6xy^2
c) 15x^3y^3 – 5x^2y^3 + 20xy^2
d) 8x^3 (x-2) – x^2 + 2x
2) Tìm x
a) x^2 – 3x =0
b) 2x (x-2) – (x-2) =0
c) (x-5)^2 – x+5 = 0
`1)`
`a) x^3 + x^2 +x`
`=x(x^2+x+1).`
`b) 4x^2y – 5x^3y – 6xy^2 `
`= -xy(-4x+5x^2y+6y).`
`c) 15x^3y^3 – 5x^2y^3 + 20xy^2`
`=5xy^2(3x^2y-xy+4).`
`d) 8x^3 (x-2) – x^2 + 2x`
`=x[8x^2(x-2)-x+2]`
`=x(8x^3-16x^2-x+2)`
`=x(8x^3-16x^2-x+2)`
`=x[ 8x^2(x-2)-(x-2)]`
`=x(x-2)(8x^2-1).`
`2)`
`a) x^2 – 3x =0`
`⇔x(x-3)=0`
`⇒` \(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy `x∈{0;3}.`
`b) 2x (x-2) – (x-2) =0`
`⇔(x-2)(2x-1)=0`
`⇒` \(\left[ \begin{array}{l}x-2=0\\2x-1=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `x∈{2;1/2}.`
`c) (x-5)^2 – x+5 = 0`
`⇔(x-5)^2 – (x-5 )= 0`
`⇔(x-5)(x-5-1)= 0`
`⇔(x-5)(x-6)= 0`
`⇒` \(\left[ \begin{array}{l}x-5=0\\x-6=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=5\\x=6\end{array} \right.\)
Vậy `x∈{5;6}.`
Đáp án:
Giải thích các bước giải:
$a,x^3+x^2+x=x(x^2+x+1)$
$b,4x^2y-5x^3y-6xy^2=xy(4x-5x^2-6y)$
$c,15x^3y^3-5x^2y^3+20xy^2=5xy^2(3x^2y-xy+4)$
$d,8x^3(x-2)-x^2+2x$
$=8x^3(x-2)-x(x-2)$
$=(8x^3-x)(x-2)$
2.
$a,x^2-3x=0$
$(=)x(x-3)=0$
\(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
$b,2x(x-2)-(x-2)=0$
$(=)(2x-1)(x-2)=0$
\(\left[ \begin{array}{l}2x-1=0\\x-2=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=2\end{array} \right.\)
$c,(x-5)^2-x+5=0$
$(=)x^2-10x+25-x+5=0$
$(=)x^2-11x+30=0$
$(=)(x-6)(x-5)=0$
\(\left[ \begin{array}{l}x-6=0\\x-5=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=6\\x=5\end{array} \right.\)