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Từ bđt Cauchy : \(a+b\ge2\sqrt{ab}\) ta suy ra được \(ab\le\frac{\left(a+b\right)^2}{4}\)
Áp dụng vào bài toán của bạn :
a/ \(y=\left(x+3\right)\left(5-x\right)\le\frac{\left(x+3+5-x\right)^2}{4}=...............\)
b/ Tương tự
c/ \(y=\left(x+3\right)\left(5-2x\right)=\frac{1}{2}.\left(2x+6\right)\left(5-2x\right)\le\frac{1}{2}.\frac{\left(2x+6+5-2x\right)^2}{4}=.............\)
d/ Tương tự
e/ \(y=\left(6x+3\right)\left(5-2x\right)=3\left(2x+1\right)\left(5-2x\right)\le3.\frac{\left(2x+1+5-2x\right)^2}{4}=.......\)
f/ Xét \(\frac{1}{y}=\frac{x^2+2}{x}=x+\frac{2}{x}\ge2\sqrt{x.\frac{2}{x}}=2\sqrt{2}\)
Suy ra \(y\le\frac{1}{2\sqrt{2}}\)
..........................
g/ Đặt \(t=x^2\) , \(t>0\) (Vì nếu t = 0 thì y = 0)
\(\frac{1}{y}=\frac{t^3+6t^2+12t+8}{t}=t^2+6t+\frac{8}{t}+12\)
\(=t^2+6t+\frac{8}{3t}+\frac{8}{3t}+\frac{8}{3t}+12\)
\(\ge5.\sqrt[5]{t^2.6t.\left(\frac{8}{3t}\right)^3}+12=.................\)
Từ đó đảo ngược y lại rồi đổi dấu \(\ge\) thành \(\le\)
a/ \(y=\left(x+3\right)\left(5-x\right)\le\frac{1}{4}\left(x+3+5-x\right)^2=16\)
Dấu "=" xảy ra khi \(x+3=5-x\Leftrightarrow x=1\)
b/ \(y=x\left(6-x\right)\le\frac{1}{4}\left(x+6-x\right)^2=9\)
\("="\Leftrightarrow x=3\)
c/ \(y=\frac{1}{2}\left(2x+6\right)\left(5-2x\right)\le\frac{1}{8}\left(2x+6+5-2x\right)^2=\frac{121}{8}\)
\("="\Leftrightarrow x=-\frac{1}{4}\)
d/ \(y=\frac{1}{2}\left(2x+5\right)\left(10-2x\right)\le\frac{1}{8}\left(2x+5+10-2x\right)^2=\frac{225}{8}\)
\("="\Leftrightarrow x=\frac{5}{4}\)
e/ \(y=3\left(2x+1\right)\left(5-2x\right)\le\frac{3}{4}\left(2x+1+5-2x\right)^2=27\)
\("="\Leftrightarrow x=1\)
f/ \(\frac{x}{x^2+2}\le\frac{x}{2\sqrt{x^2.2}}=\frac{1}{2\sqrt{2}}\)
\("="\Leftrightarrow x=\sqrt{2}\)
g/ \(y=\frac{x^2}{\left(x^2+\frac{3}{2}+\frac{3}{2}\right)^3}\le\frac{x^2}{\left(3\sqrt[3]{\frac{9}{4}x^2}\right)^3}=\frac{4}{243}\)
\("="\Leftrightarrow x^2=\frac{3}{2}\Leftrightarrow x=\pm\sqrt{\frac{3}{2}}\)
A = \(\frac{3x}{2}+\frac{2}{x-1}=3.\frac{x-1}{2}+\frac{2}{x-1}+\frac{3}{2}\)\(\ge2\sqrt{3}+\frac{3}{2}\)
\(\Rightarrow\)min A = \(2\sqrt{3}+\frac{3}{2}\Leftrightarrow x=\frac{2}{\sqrt{3}}+1\)(thỏa mãn)
B = \(x+\frac{3}{3x-1}=\frac{1}{3}\left(3x-1+\frac{9}{3x-1}+1\right)\)\(\ge\frac{1}{3}\left(2\sqrt{9}+1\right)=\frac{7}{3}\)
\(\Rightarrow\)min B = \(\frac{7}{3}\Leftrightarrow x=\frac{4}{3}\)
\(A\) \(=\) \(3x^2\left(8-x^2\right)\le3\frac{\left(x^2+8-x^2\right)^2}{4}=48\)
\(\Rightarrow\) maxA = 48 \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)
\(B=\) \(4x\left(8-5x\right)\)\(=\frac{4}{5}.5x\left(8-5x\right)\le\frac{4}{5}.\frac{\left(5x+8-5x\right)^2}{4}=\frac{64}{5}\)
\(\Rightarrow\)max B = \(\frac{64}{5}\Leftrightarrow x=\frac{4}{5}\)(thỏa mãn)
28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)
PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)
Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)
giai tiep
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
a/ \(0\le\sqrt{5-x^2}\le\sqrt{5}\)
Đặt \(t=\sqrt{5-x^2}\Rightarrow0\le t\le\sqrt{5}\)
\(y=-t^2-t+5\)
Ta có \(-\frac{b}{2a}=-\frac{1}{2}\notin\left[0;\sqrt{5}\right]\)
\(y\left(0\right)=5\) ; \(y\left(\sqrt{5}\right)=-\sqrt{5}\)
\(\Rightarrow y_{max}=5\) khi \(x=\pm\sqrt{5}\)
\(y_{min}=-\sqrt{5}\) khi \(x=0\)
Câu 2:
Nếu không thêm điều kiện gì thì cả min lẫn max đều ko tồn tại
Câu 3: Đề ko rõ
Câu 4: \(x>1\)
\(y=\frac{x-1}{20}+\frac{1}{2\sqrt{x-1}}+\frac{1}{2\sqrt{x-1}}+\frac{1}{20}\)
\(y\ge3\sqrt[3]{\frac{x-1}{80\left(x-1\right)}}+\frac{1}{20}=\frac{3}{2\sqrt[3]{10}}+\frac{1}{20}\)
Dấu "=" xảy ra khi \(\frac{x-1}{10}=\frac{1}{\sqrt{x-1}}\Rightarrow x=\sqrt[3]{100}+1\)