1) \(\left(x-3\right)\left(x-5\right)+2\)
\(=x^2-8x+15+2\)
\(=\left(x^2-8x+16\right)+1\)
\(=\left(x-4\right)^2+1\)
Vì \(\left(x-4\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x-4\right)^2+1\ge1>0;\forall x\)
Vậy....
2) tương tự
\(1.\left(x-3\right)\left(x-5\right)+2\)
\(=x^2-8x+15+2\)
\(=x^2-2.4x+16+1\)
\(=\left(x-4\right)^2+1\)
Do \(\left(x-4\right)^2\ge0\)nên \(\left(x-4\right)^2+1\ge1\)
hay \(\left(x-3\right)\left(x-5\right)+2>0\)
1. Ta có: ( x-3)(x-5) + 2
= x2 - 3x - 5x + 15 + 2
= x2 - 8x + 17
= x2 - 8x + 16 + 1
= (x-4)2 + 1
Vì (x-4)2\(\ge\)0 với \(\forall x\)
=> (x-4)2 + 1 >0 với\(\forall x\)hay (x-3)(x-5)+2 >0 ( bn xem lại đề hộ mk )
2. Ta có: -x2 + 4x-5
=-(x2-4x+5)
=-(x2-4x+4+1)
=-(x-2)2-1
Vì -(x-2)2\(\le\)0 với \(\forall x\)
=> -(x-2)2-1 < 0 với \(\forall x\)
hay -x2 + 4x-5 <0 (đpcm)
chúc bn học tốt
\(-x^2+4x-5=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-4x+4+1\right)\)
\(=-\left[\left(x-2\right)^2+1\right]\)
\(=-\left(x-2\right)^2-1\le-1< 0\)
\(2.-x^2+4x-5\)
\(=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-2.2x+4+1\right)\)
\(=-\left[\left(x-2\right)^2+1\right]\)
Do (x-2)^2 >= 0 nên (x-2)^2 + 1>0
hay \(-x^2+4x-5=-\left[\left(x-2\right)^2+1\right]< 0\)
