Question

b1  Cho 2 số a,b thỏa mãn \(\frac{a+b}{2}\) =1 

Tìm GTLN: A=\(\frac{2019}{2a^2+2b^2+2016}\)

b2  Cho Q=\(\frac{2x^2+2}{\left(x+1\right)^2}\)

Tìm GTNN của Q.

Answer 1

2/\(ĐKXĐ:x\ne-1\)

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

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

Đặt \(\frac{2}{x+1}=t\)

\(\Rightarrow Q=t^2-2t+2=\left(t-1\right)^2+1\ge1\forall t\)

\(\Rightarrow minQ=1\Leftrightarrow t=1\)

                           \(\Leftrightarrow\frac{2}{x+1}=1\)

                         \(\Leftrightarrow x=1\left(tmđkxđ\right)\)             

Answer 2

Ta có: \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}=\frac{2^2}{2}=2\)

=> \(A\le\frac{2019}{2.2+2016}=\frac{2019}{2020}\)

Dấu "=" xảy ra <=> a = b = 1

Answer 3

Câu 2)

*) Cách khác : ( Dùng delta hoặc mò để xét hiệu tìm ra GTNN và GTLN )

ĐKXĐ : \(x\ne-1\)

Ta có : \(Q-1=\frac{2x^2+2}{\left(x+1\right)^2}-1\)

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

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

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

Ta thấy : \(\left(x-1\right)^2\ge0\forall x,\left(x+1\right)^2\ge0\forall x\ne-1\)

\(\Rightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\ge0\forall x\ne-1\)

hay : \(Q-1\ge0\Rightarrow Q\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\) ( thỏa mãn ĐKXĐ )

Vậy : min \(Q=1\) tại \(x=1\)

Answer 4

Ta có HĐT :

\(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2=2^2=4\)\(\)

\(\Leftrightarrow2\left(a^2+b^2\right)+2016\ge2020\)

\(\Leftrightarrow\frac{2019}{2\left(a^2+b^2\right)+2016}\le\frac{2019}{2020}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b\\a+b=2\end{cases}\Leftrightarrow}a=b=1\)