ĐK: \(x\in N;x\ne4\)

a

Ta thấy trong 2 trường hợp \(\sqrt{x}-2>0\) và \(\sqrt{x}-2< 0\) thì Max A xảy ra trong trường hợp \(\sqrt{x}-2>0\Rightarrow\sqrt{x}-2>2\Rightarrow x>4\)

Mà \(x\in N\Rightarrow x\in\left\{5;6;7;....\right\}\Rightarrow x\ge5\Rightarrow\sqrt{x}\ge\sqrt{5}\)

\(\Rightarrow\sqrt{x}-2\ge\sqrt{5}-2\\ \Rightarrow\dfrac{3}{\sqrt{x}-2}\le\dfrac{3}{\sqrt{5}-2}\\ \Rightarrow A\le\dfrac{3}{\sqrt{5}-2}=6+3\sqrt{5}\)

Vậy Max A \(=6+3\sqrt{5}\) khi \(x=5\left(thỏa.mãn\right)\)

b

ĐK:\(x\in N;x\ne4\)

Min A xảy ra khi \(\sqrt{x}-2< 0\) \(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)

Mà \(x\in N\Rightarrow x\in\left\{0;1;2;3\right\}\)

Vậy min A \(=-6-3\sqrt{3}\) khi \(x=3\left(thỏa.mãn\right)\)

a, Ta thấy: \(\sqrt{x}\ge0\forall x\) (ĐK: \(x\ge0\))

\(\Rightarrow\sqrt{x}+10\ge10\forall x\)

\(\Rightarrow\dfrac{1}{\sqrt{x}+10}\le\dfrac{1}{10}\forall x\)

\(\Rightarrow Max_A=\dfrac{1}{10}\Leftrightarrow\dfrac{1}{\sqrt{x}+10}=\dfrac{1}{10}\)

\(\Leftrightarrow\sqrt{x}+10=10\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

b, Ta có: \(\sqrt{x}\ge0\forall x\) (ĐK: \(x\ge0;x\ne4\))

\(\Rightarrow-\sqrt{x}\le0\forall x\)

\(\Rightarrow2-\sqrt{x}\le2\forall x\)

\(\Rightarrow\dfrac{4}{2-\sqrt{x}}\ge\dfrac{4}{2}=2\)

\(\Rightarrow Min_B=2\Leftrightarrow\dfrac{4}{2-\sqrt{x}}=2\)

\(\Leftrightarrow2-\sqrt{x}=2\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy ...

#Urushi

a: ĐKXĐ: x>=0

\(\sqrt{x}+10>=10\) với mọi x thỏa mãn ĐKXĐ
=>\(A=\dfrac{1}{\sqrt{x}+10}< =\dfrac{1}{10}\) với mọi x thỏa mãn ĐKXĐ

Dấu = xảy ra khi x=0

=>Amax=1/10 khi x=0

b:Sửa đề: B nhỏ nhất

 ĐKXĐ: x>=0; x<>4

\(2-\sqrt{x}< =2\)

=>\(B=\dfrac{4}{2-\sqrt{x}}>=\dfrac{4}{2}=2\)

Dấu = xảy ra khi x=0

tính \(\frac{2012.2013-1}{2012^2+2011}\)

a)7

b)1

Làm bừa coi xem đk :b

\(M\in\Delta:y=3-x\Rightarrow M\left(x;3-x\right)\)

a/ MA+MB min

\(MA=\sqrt{\left(x_A-x_M\right)^2+\left(y_A-y_M\right)^2};MB=\sqrt{\left(x_B-x_M\right)^2+\left(y_B-y_M\right)^2}\)

\(Minkovsky:MA+MB\ge\sqrt{\left(x_M-x_A+x_M-x_B\right)^2+\left(y_M-y_A+y_M-y_B\right)^2}\)

\("="\Leftrightarrow\dfrac{x_A-x_M}{y_A-y_M}=\dfrac{x_B-x_M}{y_B-y_M}\Leftrightarrow\dfrac{1-x}{-1-3+x}=\dfrac{-x}{1-3+x}\)

\(\Leftrightarrow x=-2\Rightarrow y=5\Rightarrow M\left(-2;5\right)\)

|MA-MB| max

\(AB=\sqrt{\left(x_B-x_A\right)^2+\left(y_B-y_A\right)^2}=\sqrt{1+4}=\sqrt{5}\)

Theo bdt tam giác ta luôn có: \(\left|MA-MB\right|\le AB\)

\(\Leftrightarrow\left|\sqrt{\left(x_M-1\right)^2+\left(y_M+1\right)^2}-\sqrt{x_M^2+\left(y_M-1\right)^2}\right|\le\sqrt{5}\)

\("="\Leftrightarrow M,A,B-thang-hang\)

\(\Leftrightarrow\overrightarrow{MA}=k\overrightarrow{MB}\Leftrightarrow\left\{{}\begin{matrix}x_A-x_M=k\left(x_B-x_M\right)\\y_A-y_M=k\left(y_B-y_M\right)\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1-x}{-x}=\dfrac{-4+x}{-2+x}\Leftrightarrow x=-2\Rightarrow y=5\Rightarrow M\left(-2;5\right)\)

Câu b tương tự bạn tự làm nốt