Ta có : \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=\frac{z}{z+xz+xyz}+\frac{xz}{xz+xyz+xyz^2}+\frac{1}{1+z+xz}\)

\(=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{1+z+xz}=\frac{1+xz+z}{1+xz+z}=1\)

\(\Leftrightarrow6x+21x-2x-7-6x^2+5+6x+5=16\)

\(\Leftrightarrow6x+21x-2x-6x^2-6x=16+7-5-5\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)

help me

 PT: x(5-2x)+2x(x-1)=15 
<=> 5x - 2x² + 2x² - 2x = 15 
<=> 3x = 15 
<=> x = 5.

\(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)

\(\Leftrightarrow\left(2x^2+4x-5x-10\right)-2\left(x^2-x\right)-15=0\)

\(\Leftrightarrow2x^2-x-10-2x^2+2x-15=0\)

\(\Leftrightarrow x=25\)

a) \(x^4+3x^3-7x^2-27x-18\)

\(=\left(x^4+3x^3+2x^2\right)-\left(9x^2-27x-18\right)\)

\(=x^2\left(x^2+3x+2\right)-9\left(x^2+3x+2\right)=\left(x^2+x+2x+2\right)\left(x^2-9\right)\)

\(=\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)\)

b) \(x^4+5x^3-7x^2-41x-30\)

\(=\left(x^4+2x^3-15x^2\right)+\left(3x^3+6x^2-45x\right)+\left(2x^2+4x-30\right)\)

\(=x^2\left(x^2+2x-15\right)+3x\left(x^2+2x-15\right)+2\left(x^2+2x-15\right)\)

\(=\left(x^2+2x-15\right)\left(x^2+3x+2\right)=\left(x^2+5x-3x-15\right)\left(x^2+x+2x+2\right)\)

\(=\left(x+5\right)\left(x-3\right)\left(x+1\right)\left(x+2\right)\)

c) \(x^6-14x^4+49x^2-36\)

\(=\left(x^6-9x^4\right)+\left(-5x^4+45x^2\right)+\left(4x^2-36\right)\)

\(=x^4\left(x^2-9\right)-5x^2\left(x^2-9\right)+4\left(x^2-9\right)\)

\(=\left(x^2-9\right)\left(x^4-5x^2+4\right)=\left(x^2-9\right)\left(x^4-4x^2-x^2+4\right)\)

\(=\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)\)

Thế này nhé ^^

• Ta có : \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(=\left(a+b+c\right)\left[\left(a^2+2ab+b^2\right)-bc-ac+c^2-3ab\right]\)

\(=\left[\left(a+b\right)+c\right].\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b\right)-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=a^3+b^3+c^3-3abc\)

• \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\frac{\left(a+b+c\right)}{2}\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\right]=0\)

\(\Leftrightarrow\frac{\left(a+b+c\right)}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)