Question

[Hiền Thương]

1. Chứng minh biểu thức sau dương

\(M=\dfrac{1}{3}x^2+2x+10\)

2. Chứng minh biểu thức sau âm

a) \(2x-x^2-15\)

b) \(-5-\left(x-1\right)\left(x+2\right)\)

Answer 1

[Hoàng Thị Ngọc Mai]

1)

\(M=\dfrac{1}{3}x^2+2x+10\)

\(=\dfrac{1}{3}.\left(x^2+6x+30\right)\)

\(=\dfrac{1}{3}\left(x^2+2.x.3+9\right)+7\)

\(=\dfrac{1}{3}.\left(x+3\right)^2+7\) \(\ge\) 7 với \(\forall\) x

=> M luôn dương

=> đpcm

2)

a) \(2x-x^2-15\)

\(=-\left(x^2-2x+15\right)\)

\(=-\left(x^2-2x+1\right)-14\)

\(=-\left(x-1\right)^2-14\) \(\le-14\) với \(\forall\) x

=> \(2x-x^2-15\) luôn âm

=> đpcm

b) \(-5-\left(x-1\right)\left(x+2\right)\)

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

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

\(=-\left(x^2+x+3\right)\)

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

\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\) với \(\forall\) x

=> \(-5-\left(x-1\right)\left(x+2\right)\) luôn âm

=> đpcm

Answer 2

[T.Thùy Ninh]

\(M=\dfrac{1}{3}x^2+2x+10=\dfrac{1}{3}\left(x^2+6x+9\right)+7\)

\(=\dfrac{1}{3}\left(x+3\right)^2+7\)

Ta có:

\(\dfrac{1}{3}\left(x+3\right)^2\ge\forall x\Rightarrow\dfrac{1}{3}\left(x+3\right)^2+7>0\)

=>đpcm

\(2,a,2x-x^2-15\)

\(=-\left(x^2-2x+1\right)-14\)

\(=-\left(x-1\right)^2-14\)

Ta có:

\(-\left(x-1\right)^2\le0\forall x\Rightarrow-\left(x-1\right)^2-14< 0\)

=> đpcm

\(b,-5-\left(x-1\right)\left(x+2\right)\)

\(=-5-\left(x^2+x-2\right)\)

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

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

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

Ta có:

\(-\left(x+\dfrac{1}{2}\right)^2\le0\forall x\Rightarrow-\left(x+\dfrac{1}{2}\right)-\dfrac{11}{4}< 0\)=> đpcm