Nguyen Sy Hoc: mình nghĩ đề đâu sai đâu nhỉ?

Có:

\(\frac{a}{1+b^2}=a.\left(1-\frac{b^2}{1+b^2}\right)\ge a\left(1-\frac{b^2}{2b}\right)=a-\frac{ab}{2}\)

Tương tự \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\); \(\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)

Cộng theo vế 3 BĐT trên ta thu được:

\(VT\ge\left(a+b+c\right)-\frac{ab+bc+ca}{2}\ge\left(a+b+c\right)-\frac{\left(a+b+c\right)^2}{6}=\frac{3}{2}^{\left(đpcm\right)}\)

Đẳng thức xảy ra khi a = b = c = 1

100% đề sai

mong bn xem lại

\(a;x^4⋮x^{2n}\Leftrightarrow4\ge2n\Leftrightarrow2\ge n\Rightarrow n=0;1;2\)

\(b;x^ny^3⋮x^2y^{n+1}\Leftrightarrow n\ge2;3\ge n+1\Leftrightarrow\hept{\begin{cases}n\ge2\\2\ge n\end{cases}\Rightarrow n=2}\)

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

\(A=x^2-x+\frac{1}{4}+\frac{3}{4}\)

\(A=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\)

\(A=\left[x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]+\frac{3}{4}\)

\(A=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

\(\text{Vì }\left(x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow A\ge\frac{3}{4}\)

\(\text{Vậy GTNN của A = }\frac{3}{4}\text{ ,dấu bằng xảy ra khi x = }\frac{1}{2}\)

\(A=x^2-x+1\) 

   \(=\left(x-0,5\right)^2+0,75\ge0,75\forall x\)

Dâu"=" xảy ra<=> \(\left(x-0,5\right)^2=0\Leftrightarrow x=0,5\)

Vậy A_min =0,75 tại x=0,5

A=\(x^2-4x+4+5\)

A= \(\left(x-2\right)^2+5\ge5\forall x\)

Amin=5 khi x=2

1) 2x(x + 1)² - (x - 1)³ - (x - 2)(x² + 2x + 4)

= 2x³ + 4x² + 2x - x³ + 3x² - 3x + 1 - x² + 8

= 7x² - x + 9

2) (3x + 5)² + (3x - 5)² - (3x + 2)(3x - 2)

= 9x² + 30x + 25 + 9x² - 30x + 25 - 9x² + 4

= 9x² + 54