2(x-2)^2-(x-2)(x+2)<x(x-1)
<=> 2(x^22-4x+4)-(x^2-4)<x^2-x
<=> 2x^2-8x+8 -x^2 +4<x^2-x
<=> 2x^2 -8x-x^2-x^2+x<-8-4
<=> -7x<-12
<=> x>12/7
Vậy S={x/x>12/7}
2(x−2)2−(x−2)(x+2)<x(x−1)
\(\Leftrightarrow2\left(x^2-4x+4\right)-\left(x^2-4\right)< x^2-x
\)
\(\Leftrightarrow2x^2-8x+8-x^2+4< x^2-x\)
\(\Leftrightarrow2x^2-8x-x^2-x^2+x< -8-4\)
\(\Leftrightarrow-7x< -12\)
\(\Leftrightarrow x>\dfrac{12}{7}\)
Vậy S={x/x>12/7}
easy easy thôi
\(2\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)< x\left(x-1\right)\)
\(\Leftrightarrow2\left(x^2-4x+4\right)-x^2+2x-2x+4< x^2-x\)\(\Leftrightarrow2x^2-8x+8-x^2+2x-2x+4< x^2+x\)\(\Leftrightarrow2x^2-8x+32-x^2+2x-2x+4-x^2+x< 0\)\(\Leftrightarrow-7x+12< 0\)
\(\Leftrightarrow-7x< -12\Leftrightarrow x>\dfrac{12}{7}\) ( nhớ chuyển dấu bé thành dấu lớn nha )
Vậy nghiệm bất phương trình là \(x>\dfrac{12}{7}\)