a ) \(-x^2+6x-15\)
\(\Leftrightarrow-x^2+6x-9-6\)
\(\Leftrightarrow-\left(x^2-6x+9\right)-6\)
Ta có : \(\left(x-3\right)^2\ge0\)
\(\Leftrightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2-6\le-6\)
\(\RightarrowĐPCM.\)
b ) \(\left(x-3\right)\left(1-x\right)-2\)
\(\Leftrightarrow\left(x-x^2-3+3x\right)-2\)
\(\Leftrightarrow\left(-x^2+4x-3\right)-2\)
\(\Leftrightarrow-x^2+4x-3-2\)
\(\Leftrightarrow-x^2+4x-4-1\)
\(\Leftrightarrow-\left(x^2-4x+4\right)-1\)
\(\Leftrightarrow-\left(x-2\right)^2-1\)
Ta có : \(\left(x-2\right)^2\ge0\)
\(\Leftrightarrow-\left(x-2\right)^2\le0\)
\(\Leftrightarrow-\left(x-2\right)^2-1\le-1\)
\(\LeftrightarrowĐPCM.\)
c ) \(\left(x+4\right)\left(2-x\right)-10\)
\(\Leftrightarrow\left(2x-x^2+8-4x\right)-10\)
\(\Leftrightarrow\left(-x^2-2x+8\right)-10\)
\(\Leftrightarrow-x^2-2x+8-10\)
\(\Leftrightarrow-x^2-2x-2\)
\(\Leftrightarrow-x^2-2x-1-1\)
\(\Leftrightarrow-\left(x^2+2x+1\right)-1\)
\(\Leftrightarrow-\left(x+1\right)^2-1\)
Ta có : \(\left(x+1\right)^2\ge0\)
\(\Leftrightarrow-\left(x+1\right)^2\le0\)
\(\Leftrightarrow-\left(x+1\right)^2-1\le-1\)
\(\LeftrightarrowĐPCM.\)
a) \(-x^2+6x-15=-x^2+6x-9-6=-\left(x-3\right)^2-6\)
Do \(-\left(x-3\right)^2\le0\forall x\in Q\)
\(\Rightarrow......................\le0\forall x\in Q\)
Áp dụng hằng đẳng nhé mk ngại làm lắm
a) \(-x^2+6x-15=-\left(x^2-6x+15\right)=-\left(\left(x-3\right)^2+6\right)\)
= \(-\left(x-3\right)^2-6\) \(\le6< 0\forall x\) (đpcm)
b) \(\left(x-3\right).\left(1-x\right)-2=x-x^2-3+3x-2=-x^2+4x-5\)
= \(-\left(x^2-4x+5\right)\) = \(-\left(\left(x-2\right)^2+1\right)=-\left(x-2\right)^2-1\le-1< 0\forall x\) (đpcm)
c) \(\left(x+4\right)\left(2-x\right)-10=2x-x^2+8-4x-10\)
\(-x^2-2x-2=-\left(x^2+2x+2\right)=-\left(\left(x+1\right)^2+1\right)=-\left(x+1\right)^2-1\le-1< 0\forall x\)(đpcm)