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P=\(5x+3y+\dfrac{12}{x}+\dfrac{16}{y}\)
=\(3x+\dfrac{12}{x}+y+\dfrac{16}{y}+2\left(x+y\right)\)
AD BĐT cô si :
Ta có \(3x+\dfrac{12}{x}\ge2\sqrt{3x.\dfrac{12}{x}}=2\sqrt{36}=12\)
\(y+\dfrac{16}{y}\ge2\sqrt{y.\dfrac{16}{y}}=2\sqrt{16}=8\)
\(2\left(x+y\right)\ge2.6=12\)
=> P\(\ge12+8+12=32\)
Dấu = xra \(\left\{{}\begin{matrix}3x=\dfrac{12}{x}\\y=\dfrac{16}{y}\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\left(x;y\right)=\left(2;4\right)\)
Vậy GTNN của P=32 khi (x;y)=(2;4)
a: \(A=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}=10\)
b: \(B=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}=-2\sqrt{y}\)
c: \(C=\dfrac{\sqrt{3}-1}{\sqrt{6}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
Bài 2. Áp dụng BĐT Cauchy dưới dạng Engel , ta có :
\(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\) ≥ \(\dfrac{\left(1+4+9\right)^2}{x+y+z}=196\)
⇒ \(P_{MIN}=196."="\) ⇔ \(x=y=z=\dfrac{1}{3}\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
Giúp mình với
Áp dụng BĐT cosi:
\(A=\left(3x+\dfrac{3}{x}\right)+\left(\dfrac{4}{9}y+\dfrac{4}{y}\right)+\left(2x+y\right)\\ A\ge2\sqrt{\dfrac{9x}{x}}+2\sqrt{\dfrac{16y}{9y}}+5\\ A\ge2\cdot3+2\cdot\dfrac{4}{3}+5=\dfrac{41}{3}\)
Vậy \(A_{min}=\dfrac{41}{3}\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{3}{x}\\\dfrac{4y}{9}=\dfrac{4}{y}\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)