Question

5. chứng minh rằng với mọi giá trị của x ta luôn có:

a)\(-x^2+4x-5< 0\)

b) \(x^4+3x^2+3>0\)

c) \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3>0\)

Answer 1

a ) \(-x^2+4x-5\)

\(=-\left(x^2-4x+4\right)-1\)

\(=-\left(x-2\right)^2-1\le-1< 0\forall x\left(đpcm\right)\)

b ) \(x^4+3x^2+3=x^4+3x^2+\dfrac{9}{4}+\dfrac{3}{4}=\left(x^2+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\left(đpcm\right)\)

c ) \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3\)

Đặt \(x^2+2x+3=a\) . Khi đó , ta có :

\(x\left(x+1\right)+3=x^2+x+3=x^2+x+\dfrac{1}{4}+\dfrac{11}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

\(=\left(x^2+2x+3+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

\(=\left(x^2+2x+\dfrac{7}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\left(đpcm\right)\)

Answer 2

Vội nên đánh sai :

\(a\left(a+1\right)+3=a^2+a+3=a^2+a+\dfrac{1}{4}+\dfrac{11}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)~