x¹⁰+x⁵+1

= x¹⁰+x⁹+x⁸-x⁹-x⁸-x⁷+x⁷+x⁶+x⁵-x⁶-x⁵-x⁴+x⁵+x⁴+x³-x³-x²-x+x²+x+1

= x⁸(x²+x+1)-x⁷(x²+x+1)+x⁵(x²+x+1)-x⁴(x²+x+1)+x³(x²+x+1)-x(x²+x+1)+(x²+x+1)

= (x²+x+1)(x⁸-x⁷+x⁵-x⁴+x³-x+1)

\(x+\sqrt{x}-12=\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

\(x+\sqrt{x}-12=\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

PTĐTTNT ??? :)) bn phân tích rồi đấy, đề là tìm x thôi 

Giải ( suỵt :), đừng ai nhìn thấy ... :v 

\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)

TH1 : \(2x-10=0\Leftrightarrow x=5\)

TH2 : \(x+10=0\Leftrightarrow x=-10\)

TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )

Vậy x = {5;-10}

sao lại "vô lí" vậy bạn 

lp 8 chưa học số vô tỉ babe nhá :)) 

Câu a:

\(x^2-y^2-x+3y-2=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\left(y^2-2.y.\frac{3}{2}+\frac{9}{4}\right)\)

\(< =>\left(x-\frac{1}{2}\right)^2-\left(y-\frac{3}{2}\right)^2\)

\(< =>\left(x-\frac{1}{2}+y-\frac{3}{2}\right)\left(x-\frac{1}{2}-y+\frac{3}{2}\right)=\left(x+y-2\right)\left(x-y+1\right)\)

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

x² - y - y² - x

= (x² - y²) - x - y

= (x - y)(x + y) - (x + y)

= (x - y - 1)(x + y)

\(=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)

x(x-4)+5(x-4)

(x+5)(x-4)