\(1+\left|x\right|=\orbr{\begin{cases}1+x\Leftrightarrow x\ge0\\1-x\Leftrightarrow x< 0\end{cases}}\)

Tham khảo nhé~

2x(3x-2) - 12x + 8 = 0

<=> 6x² - 4x - 12x + 8 = 0

<=> 6x² -16x + 8 = 0

<=> ( x - 2 )( 3x - 2 ) = 0

<=> x = 2 hoặc x = ²/₃

Trả lời :..............

x = 2 hoặc 2/3

Hk tốt.........................

\(x^2-25-4xy+4y^2\)

\(=\left(x^2-4xy+4y^2\right)-25\)

\(=\left[x^2-2\cdot x\cdot2y+\left(2y\right)^2\right]-25\)

\(=\left(x-2y\right)^2-5^2\)

\(=\left(x-2y-5\right)\cdot\left(x-2y+5\right)\)

\(\frac{x^2+y^2}{xy}=\frac{10}{3}\Rightarrow3x^2+3y^2-10xy=0\)

\(\Rightarrow\left(3x^2-9xy\right)-\left(xy-3y^2\right)=0\Rightarrow3x\left(x-3y\right)-y\left(x-3y\right)=0\)

\(\Rightarrow\left(x-3y\right)\left(3x-y\right)=0\Rightarrow3x-y=0\left(y>x>0\Rightarrow x-3y< 0\right)\Rightarrow3x=y\)

\(M=\frac{x-y}{x+y}=\frac{x-3x}{x+3x}=\frac{-2x}{4x}=-\frac{1}{2}\)

\(x^4+x^3y-xy^3-y^4=x^3\left(x+y\right)-y^3\left(x+y\right)\)

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

\(x^4+x^3y-xy^3-y^4\)

\(=x^3.\left(x+y\right)-y^3.\left(x+y\right)\)

\(=\left(x^3-y^3\right).\left(x+y\right)\)