Question

[kiana chan]

chung minh rang

a) (1-2x)(x-1)-5<0

b) -x^2-y^2+2x+2y-3<0

Answer 1

[Mysterious Person]

a) ta có : \(\left(1-2x\right)\left(x-1\right)-5=x-1-2x^2+2x-5\)

\(=-2x^2+3x-6=-\left(2x^2-3x+6\right)=-\left(\left(\sqrt{2}x\right)^2-2.\sqrt{2}.\dfrac{3}{2\sqrt{2}}x+\left(\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\right)\)

\(=-\left(\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\right)=-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\)

ta có : \(\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\ge0\) với mọi \(x\) \(\Rightarrow-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\le0\) với mọi \(x\)

\(-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\le\dfrac{-39}{8}< 0\) với mọi \(x\)

vậy \(\left(1-2x\right)\left(x-1\right)-5< 0\) (đpcm)

b) ta có : \(-x^2-y^2+2x+2y-3\)

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

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

ta có : \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge\forall x\\\left(y-1\right)^2\ge\forall y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-\left(x-1\right)^2\le0\forall x\\-\left(y-1\right)^2\le0\forall y\end{matrix}\right.\)

\(\Rightarrow-\left(x-1\right)^2-\left(y-1\right)^2\le0\) với mọi \(x;y\)

\(\Leftrightarrow-\left(x-1\right)^2-\left(y-1\right)^2-1\le-1< 0\) với mọi \(x;y\)

vậy \(-x^2-y^2+2x+2y-3< 0\) (đpcm)

Answer 2

[lê thị hương giang]

\(a,A=\left(1-2x\right)\left(x-1\right)-5\)

\(=x-1-2x^2+2x-5\)

\(=-2x^2+3x-6\)

\(=-\left(2x^2-3x+\dfrac{9}{8}\right)-\dfrac{39}{8}\)

\(=-\left[\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\dfrac{3}{2\sqrt{2}}+\left(\dfrac{3}{2\sqrt{2}}\right)^2\right]-\dfrac{39}{8}\)

\(=-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\)

Ta có :

\(-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\le0\) \(\Rightarrow-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\le-\dfrac{39}{8}\)

Hay A \(\le-\dfrac{39}{8}\)

Dấu = xảy ra \(\Leftrightarrow\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2=0\)

\(\Leftrightarrow\sqrt{2}x-\dfrac{3}{2\sqrt{2}}=0\) \(\Leftrightarrow\sqrt{2}x=\dfrac{3}{2\sqrt{2}}\Leftrightarrow x=\dfrac{3}{2\sqrt{2}}:\sqrt{2}\)

\(\Leftrightarrow x=\dfrac{3}{4}\)

Vậy \(Min_A=-\dfrac{39}{8}\Leftrightarrow x=\dfrac{3}{4}\)