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Khi \(x=1,44\): \(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8,44}{1,2}=\frac{211}{30}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)(ĐK: \(x\ge0,x\ne9\))
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(S=\frac{1}{B}+A=\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{x+7}{\sqrt{x}}=\frac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+\frac{4}{\sqrt{x}}+1\)
\(\ge2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}+1=5\)
Dấu \(=\)khi \(\sqrt{x}=\frac{4}{\sqrt{x}}\Leftrightarrow x=4\)(thỏa mãn)
Ta có: \(E=1+\frac{2xy}{x^2-xy+y^2}\le1+\frac{2xy}{2xy-xy}=3\)
Dấu "=" xảy ra <=> x = y
Vậy GTLN của E là 3 tại x = y.
a) Ta có \(Q=\frac{x-9}{\sqrt{x}+3}+\frac{25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)
Áp dụng BĐT cô-si, ta có \(\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\ge10\Rightarrow Q\ge10-6=4\)
Dấu = xảy ra <=> x=4
b)Tá có \(M=x^2+4y^2+1+4xy+2x+2y+y^2-2y+1+10\)
=\(\left(x+2y+1\right)^2+\left(y-1\right)^2+10\ge10\)
dấu = xảy ra <=> y=1 và x=-3
^_^
\(P=\frac{9}{y}+\frac{18}{x}+\frac{x}{6}-\frac{5y}{12}+2018=\left(\frac{18}{x}+\frac{x}{2}\right)+\left(\frac{9}{y}+\frac{y}{4}\right)-\frac{x}{3}-\frac{2y}{3}+2018\)
Lập luận : Áp dụng BTĐ Cô si cho : \(\frac{18}{x};\frac{x}{2}>0\)(với x > 0):
\(\frac{18}{x}+\frac{x}{2}\ge2\sqrt{\frac{18}{x}.\frac{x}{2}}\Leftrightarrow\frac{18}{x}+\frac{x}{2}\ge6\)
Lập luận tương tự : Áp dụng BĐT Cô si cho : \(\frac{9}{y};\frac{y}{4}>0\)(y > 0 )
\(\frac{9}{y}+\frac{y}{4}\ge2\sqrt{\frac{9}{y}.\frac{y}{4}}\Leftrightarrow\frac{9}{y}+\frac{y}{4}\ge3\)
Và \(\frac{x}{3}-\frac{2y}{3}=\frac{x+2y}{3}\ge\frac{18}{3}\)(Do x + 2y \(\le\)18)
\(\Rightarrow P=\left(\frac{18}{x}+\frac{x}{2}\right)+\left(\frac{9}{y}+\frac{y}{4}\right)-\frac{x}{3}-\frac{2y}{3}+2018\ge6-3-\frac{18}{3}+2018=2021\)
Vậy \(P=2021\)Khi và chỉ khi \(\hept{\begin{cases}\frac{18}{x}=\frac{x}{2};\frac{9}{y}=\frac{y}{4}\\x+2y< 18;x,y>0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=6\\y=6\end{cases}}\)
2. \(P=x^2-x\sqrt{3}+1=\left(x^2-x\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=\left(x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
Dấu '=' xảy ra khi \(x=\frac{\sqrt{3}}{2}\)
Vây \(P_{min}=\frac{1}{4}\)khi \(x=\frac{\sqrt{3}}{2}\)
3. \(Y=\frac{x}{\left(x+2011\right)^2}\le\frac{x}{4x.2011}=\frac{1}{8044}\)
Dấu '=' xảy ra khi \(x=2011\)
Vây \(Y_{max}=\frac{1}{8044}\)khi \(x=2011\)
4. \(Q=\frac{1}{x-\sqrt{x}+2}=\frac{1}{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{7}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}\le\frac{4}{7}\)
Dấu '=' xảy ra khi \(x=\frac{1}{4}\)
Vậy \(Q_{max}=\frac{4}{7}\)khi \(x=\frac{1}{4}\)
\(B=\frac{x^2-2x+2018}{x^2}=\frac{2018x^2-2.2018.x+2018^2}{2018x^2}\)
\(=\frac{x^2-2.2018.x+2018^2}{2018x^2}+\frac{2017x^2}{2018x^2}\)
\(=\frac{\left(x-2018\right)^2}{x^2}+\frac{2017}{2018}\)
\(=\left(\frac{x-2018}{x}\right)^2+\frac{2017}{2018}\)
Vì : \(\left(\frac{x-2018}{x}\right)^2\ge0\forall x\)
Nên : \(B=\left(\frac{x-2018}{x}\right)^2+\frac{2017}{2018}\ge\frac{2017}{2018}\)
Vậy \(B_{min}=\frac{2017}{2018}\) khi x = 2018
\(\Leftrightarrow Bx^2-x^2+2x-2018=0\)
\(\Leftrightarrow\left(B-1\right)x^2+2x-2018=0\)
Để tồn tại x thì \(\Delta^'\ge0\)
\(\Leftrightarrow1+2018\left(B-1\right)\ge0\)
\(\Leftrightarrow B\ge\frac{2017}{2018}\)
Vậy MinB=2017/2018, dấu bằng xảy ra khi x=2018