Bài 2: 

a) Ta có: \(\left|2x-5\right|\ge0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|\le0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|+3\le3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

Đk: \(2\le x\le4\)

Áp dụng BĐT bunhiacopxki có:

\(P^2=\left(\sqrt{x-2}+3\sqrt{4-x}\right)^2\le\left(1+3^2\right)\left(x-2+4-x\right)\)

\(\Leftrightarrow P^2\le20\)\(\Leftrightarrow P\le2\sqrt{5}\)

Dấu "=" xảy ra khi \(\sqrt{x-2}=\dfrac{\sqrt{4-x}}{3}\) \(\Leftrightarrow x=\dfrac{11}{5}\) (tm đk)

Có \(P^2=8\left(4-x\right)+6\sqrt{\left(x-2\right)\left(4-x\right)}+2\ge2\)\(\Rightarrow P\ge\sqrt{2}\)

Dấu "=" xảy ra khi x=4 (tm)

Ta có: \(C=-5-\left(x+2\right)\left(x-1\right)\)

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

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

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

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

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

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

1/ Gọi B_min là GTNN của B

Ta có \(\left|3x-6\right|\ge0\)=> \(2\left|3x-6\right|\ge0\)với mọi \(x\in R\)

=> \(2\left|3x-6\right|-4\ge0\)với mọi \(x\in R\).

=> B_min = 0.

Vậy GTNN của B = 0.

2/ Gọi D_min là GTNN của D.

Ta có \(\left|x-2\right|\ge0\)với mọi \(x\in R\)

và \(\left|x-8\right|\ge0\)với mọi \(x\in R\)

=> \(\left|x-2\right|+\left|x-8\right|\ge0\)với mọi \(x\in R\)

=> D_min = 0.

=> \(\left|x-2\right|+\left|x-8\right|=0\)

=> \(\hept{\begin{cases}\left|x-2\right|=0\\\left|x-8\right|=0\end{cases}}\)=> \(\hept{\begin{cases}x-2=0\\x-8=0\end{cases}}\)=> \(\hept{\begin{cases}x=2\\x=8\end{cases}}\)(Vô lý! Không thể cùng lúc có 2 giá trị x xảy ra)

Vậy không có x thoả mãn đk khi GTNN của D = 3.

ko

tên bạn kì v

\(D=-x^2-y^2+xy+2x+2y\)

\(\Rightarrow D=-\dfrac{x^2}{2}+xy-\dfrac{y^2}{2}-\dfrac{x^2}{2}+2x-\dfrac{y^2}{2}+2y\)

\(\Rightarrow D=-\left(\dfrac{x^2}{2}-xy+\dfrac{y^2}{2}\right)-\left(\dfrac{x^2}{2}-2x\right)-\left(\dfrac{y^2}{2}-2y\right)\)

\(\Rightarrow D=-\left(\dfrac{x^2}{2}-2.\dfrac{x}{\sqrt[]{2}}.\dfrac{y}{\sqrt[]{2}}+\dfrac{y^2}{2}\right)-\left(\dfrac{x^2}{2}-2.\dfrac{x}{\sqrt[]{2}}.\sqrt[]{2}+2\right)-\left(\dfrac{y^2}{2}-2.\dfrac{y}{\sqrt[]{2}}.\sqrt[]{2}+2\right)+2+2\)

\(\Rightarrow D=-\left(\dfrac{x}{\sqrt[]{2}}-\dfrac{y}{\sqrt[]{2}}\right)^2-\left(\dfrac{x}{\sqrt[]{2}}-\sqrt[]{2}\right)^2-\left(\dfrac{y}{\sqrt[]{2}}-\sqrt[]{2}\right)^2+4\)

mà \(\left\{{}\begin{matrix}-\left(\dfrac{x}{\sqrt[]{2}}-\dfrac{y}{\sqrt[]{2}}\right)^2\le0,\forall x;y\\-\left(\dfrac{x}{\sqrt[]{2}}-\sqrt[]{2}\right)^2\le0,\forall x\\-\left(\dfrac{y}{\sqrt[]{2}}-\sqrt[]{2}\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow D=-\left(\dfrac{x}{\sqrt[]{2}}-\dfrac{y}{\sqrt[]{2}}\right)^2-\left(\dfrac{x}{\sqrt[]{2}}-\sqrt[]{2}\right)^2-\left(\dfrac{y}{\sqrt[]{2}}-\sqrt[]{2}\right)^2+4\le4\)

\(\Rightarrow GTLN\left(D\right)=4\left(tạix=y=2\right)\)