Question

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

Answer 1

\(bpt\Leftrightarrow4x^2+4x+1+3x-3x^2\le x^2+4x+4\\ \Leftrightarrow3x\le3\\ \Leftrightarrow x\le1\)

Vậy ...........

( Xem cách làm nhưng vẫn phải biết rút ra kiến thức bạn nhé :D)

Answer 2

Ta có: \(\left(2x+1\right)^2+3x\left(1-x\right)\le\left(x+2\right)^2\)

\(\Leftrightarrow4x^2+4x+1+3x-3x^2-x^2-4x-4\le0\)

\(\Leftrightarrow3x\le3\)

hay \(x\le1\)

Answer 3

(2x+1)2+(1−x)3x≤(x+2)2

⇔ \(4x^2+2.2x.1+1+3x-3x^2\) ≤ \(x^2-2.x.2+4\)

⇔ \(4^2-3x^2-x^2\)\(+4x+3x-4x\) ≤ - 1 + 4

⇔ 3x ≤ 3

⇔ x  ≤ 1