\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)

Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

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

Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).

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 \(S=x+y=-2\Rightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}\)

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 \(\Leftrightarrow x=y=\dfrac{1}{2}\)

c. Bạn kiểm tra lại đề nhé.

b. \(5x\left(2-x\right)=-5x\left(x-2\right)=-5\left(x^2-2x\right)=-5\left(x^2-2x+1-1\right)=-5\left(x-1\right)^2+5\le5\)-Dấu bằng xảy ra \(\Leftrightarrow x=1\)

a.

\(\left(80-2x\right)\left(50-2x\right)x=\dfrac{2}{3}\left(40-x\right)\left(50-2x\right)3x\le\dfrac{2}{3}\left(\dfrac{40-x+50-2x+3x}{3}\right)^3=18000\)

Dấu "=" xảy ra khi \(40-x=50-2x=3x\Leftrightarrow x=10\)

b.

\(5x\left(2-x\right)=5.x\left(2-x\right)\le\dfrac{5}{4}\left(x+2-x\right)^2=5\)

Dấu "=" xảy ra khi \(x=2-x\Rightarrow x=1\)

c.

Biểu thức này chỉ có min, ko có max

d.

\(x+y\le1\Rightarrow-\left(x+y\right)\ge-1\)

\(x+y+\dfrac{1}{x}+\dfrac{1}{y}=\left(4x+\dfrac{1}{x}\right)+\left(4y+\dfrac{1}{y}\right)-3\left(x+y\right)\ge2\sqrt{\dfrac{4x}{x}}+2\sqrt{\dfrac{4y}{y}}-3.1=5\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)

\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)

\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)

\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)

\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)