\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Lời giải:
$x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6$

$=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)$

$=(\sqrt{x}-2)(\sqrt{x}-3)$

\(x-5\sqrt{x}+6=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

x2 – x – 6

= x2 + 2x – 3x – 6

(Tách –x = 2x – 3x)

= x(x + 2) – 3(x + 2)

(có x + 2 là nhân tử chung)

= (x – 3)(x + 2)

\(x^6-y^6\\ =\left(x^3\right)^2-\left(y^3\right)^2\\ =\left(x^3-y^3\right)\left(x^3+y^3\right)\\ =\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)

x⁶-y⁶ =(x-y)(x⁵+x⁴y+x³y²+x²y³+xy⁴+y⁵)

a) \(x^2-xy+x-y\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x+1\right)\left(x-y\right)\)

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+3\right)\left(x+2\right)\)

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

x⁶y - 5x⁵ - 4x⁴y + 20x³ 

= ( x⁶y - 5x⁵ ) - ( 4x⁴y - 20x³ ) 

= x⁵( xy - 5 ) - 4x³( xy - 5 )

= ( x⁵ - 4x³ )( xy - 5 ) = x³( x² - 4 )( xy - 5 ) 

= x³( x - 2 )(x + 2 )( xy - 5 )

 = x^3.(x^3y-5x^2-4xy+20)

 = x^3.[(x^3y-5x^2)-(4xy-20)]

 = x^3.(y-5).(x^2-4) = x^3.(x-2).(x+2).(y-5)

k mk nha

Phân tích đa thức sau thành nhân tử:( x-5 )\(^2\)- 5 + x

Bài làm:

( x-5 )\(^2\)- 5 + x

= ( x - 5 )\(^2\)+ ( x - 5 )                                         Chỗ này có thể phân tích thành ( x - 5 ) ( x - 5 ) + ( x - 5 ) nha

= ( x - 5 ) \([\)( x - 5 ) + 1 \(]\)

= ( x- 5 ) ( x - 5 + 1 )

* Chúc bạn học tốt ^^

# Linh