Question

Giải bất phương trình: \(\left(x-1\right)\left(x+2\right)>\left(x-1\right)^2+3\)

Answer 1

\((x-1)(x+2)>(x-1)^2+3 \\\Leftrightarrow x^2+x-2>x^2-2x+4 \\\Leftrightarrow 3x>6 \\\Leftrightarrow x>2\)

Vậy nghiệm của BPT là \(x>2\)

Answer 2

\(\left(x-1\right)\left(x+2\right)>\left(x-1\right)^2+3\)

\(\Leftrightarrow x^2+2x-x-2>x^2-2x+1+3\)

\(\Leftrightarrow x^2+2x-x-x^2+2x>2+1+3\)

\(\Leftrightarrow3x>6\Leftrightarrow x>\dfrac{6}{3}\Leftrightarrow x>2\) vậy \(x>2\)

Answer 3

Ta có:

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

\(\Rightarrow\left(x-1\right)^2+\left(x-1\right)\ge\left(x-1\right)^2+3\)

\(\Rightarrow x-1\ge3\)

\(\Rightarrow x\ge4\)