bn có giải đc ko?

d. Áp dụng BĐT Caushy Schwartz ta có:

\(x+y+\dfrac{1}{x}+\dfrac{1}{y}\le x+y+\dfrac{\left(1+1\right)^2}{x+y}=x+y+\dfrac{4}{x+y}\le1+\dfrac{4}{1}=5\)

-Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

\(y=x+\sqrt[]{2\left(1-x\right)}\left(x\le1\right)\)

\(\Rightarrow y=-\left(1-x\right)+\sqrt[]{2\left(1-x\right)}+1\)

\(\Rightarrow y=-\left(1-x\right)+\sqrt[]{2\left(1-x\right)}+1\)

\(\Rightarrow y=-\left[\left(1-x\right)-\sqrt[]{2\left(1-x\right)}+\left(\dfrac{\sqrt[]{2}}{2}\right)^2\right]+1+\left(\dfrac{\sqrt[]{2}}{2}\right)^2\)

\(\Rightarrow y=-\left[\sqrt[]{1-x}-\dfrac{\sqrt[]{2}}{2}\right]^2+1+\dfrac{1}{2}\)

\(\Rightarrow y=-\left[\sqrt[]{1-x}-\dfrac{\sqrt[]{2}}{2}\right]^2+\dfrac{3}{2}\)

mà \(-\left[\sqrt[]{1-x}-\dfrac{\sqrt[]{2}}{2}\right]^2\le0,\forall x\le1\)

\(\Rightarrow y=-\left[\sqrt[]{1-x}-\dfrac{\sqrt[]{2}}{2}\right]^2+\dfrac{3}{2}\le\dfrac{3}{2}\)

Dấu "=" xảy ra khi và chỉ khi

\(\sqrt[]{1-x}-\dfrac{\sqrt[]{2}}{2}=0\)

\(\Leftrightarrow\sqrt[]{1-x}=\dfrac{\sqrt[]{2}}{2}\)

\(\Leftrightarrow1-x=\dfrac{1}{2}\)

\(\Leftrightarrow x=1-\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}\) (thỏa \(x\le1\))

\(\Rightarrow GTLN\left(y\right)=\dfrac{3}{2}\left(tạix=\dfrac{1}{2}\right)\)

\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)

\(y_{min}=-3\) khi \(x=1\)

\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)

\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)

\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)

\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)

Biết (0≤x≤1)

\(Taco:\)

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

\(\ge|x^2-x+1-x^2+x+2|=3\)

Dấu "=" xảy ra khi: \(\left(x^2-x+1\right)\left(x^2-x-2\right)\ge0\Leftrightarrow........\)

Answer:

3.

\(x^2+2y^2+2xy+7x+7y+10=0\)

\(\Rightarrow\left(x^2+2xy+y^2\right)+7x+7y+y^2+10=0\)

\(\Rightarrow\left(x+y\right)^2+7.\left(x+y\right)+y^2+10=0\)

\(\Rightarrow4S^2+28S+4y^2+40=0\)

\(\Rightarrow4S^2+28S+49+4y^2-9=0\)

\(\Rightarrow\left(2S+7\right)^2=9-4y^2\le9\left(1\right)\)

\(\Rightarrow-3\le2S+7\le3\)

\(\Rightarrow-10\le2S\le-4\)

\(\Rightarrow-5\le S\le-2\left(2\right)\)

Dấu " = " xảy ra khi: \(\left(1\right)\Rightarrow y=0\)

Vậy giá trị nhỏ nhất của \(S=x+y=-5\Rightarrow\hept{\begin{cases}y=0\\x=-5\end{cases}}\)

Vậy giá trị lớn nhất của \(S=x+y=-2\Rightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}\)