\(P=\left(\frac{1}{\sqrt{a}+2}+\frac{1}{\sqrt{a}-2}\right).\frac{\sqrt{a}-2}{\sqrt{a}}\)(ĐK: \(a>0\) và \(a\ne4\))

\(=\frac{\sqrt{a}-2}{\sqrt{a}\left(\sqrt{a}+2\right)}+\frac{1}{\sqrt{a}}\)

\(=\frac{\sqrt{a}-2+\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+2\right)}\)

\(=\frac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+2\right)}\)

\(=\frac{2}{\sqrt{a}+2}\)

a)Để P>1/3 thì

\(\frac{2}{\sqrt{a}+2}>\frac{1}{3}\)

\(\Leftrightarrow\sqrt{a}+2< 6\)

\(\Leftrightarrow\sqrt{a}< 4\)

\(\Leftrightarrow a< 16\)

Kết hợp với đkxđ ta được \(0< a< 16\) và \(a\ne4\) thì P>1/3

b) Ta có:

\(Q=\frac{9}{2}P=\frac{9}{2}.\frac{2}{\sqrt{a}+2}=\frac{9}{\sqrt{a}+2}\)

Để Q nguyên thì \(9⋮\left(\sqrt{a}+2\right)\)

\(\Rightarrow\sqrt{a}+2\in\left\{-9;-3;-1;1;3;9\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{-11;-5;-3;-1;1;7\right\}\)

\(\Rightarrow a\in\left\{1;49\right\}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}-2\right)}{\sqrt{x}+1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}+1}\)

\(P=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)

\(P=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(P_{Min}=-\frac{1}{4}\) khi \(\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)

b/ \(Q=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(-x+3\sqrt{x}-2\right)}=\frac{\sqrt{x}-1}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{1}{\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)}\)

\(Q\ge\frac{1}{\frac{\left(\sqrt{x}+1+2-\sqrt{x}\right)^2}{4}}=\frac{4}{3^2}=\frac{4}{9}\)

\(Q_{min}=\frac{4}{9}\) khi \(\sqrt{x}+1=2-\sqrt{x}\Leftrightarrow x=\frac{1}{4}\)

c/ \(R=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}\)

Chắc là bạn ghi nhầm đề, với \(x< 1\) biểu thức này ko có min

Nó chỉ có min khi \(x>1\)

Khi đó: \(R=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\ge2\sqrt{\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}}+3=3+2\sqrt{2}\)

\(R_{min}=3+2\sqrt{2}\) khi \(\sqrt{x}-1=\sqrt{2}\Leftrightarrow x=3+2\sqrt{2}\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{x\sqrt{x}-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)}{\sqrt{x}^3-8}-\frac{\left(x-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}^3-8}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right)\)\(:\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\frac{\sqrt{x}^3+2x+4\sqrt{x}-\sqrt{x}^3+2x+3\sqrt{x}-6-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}.\frac{\left(x+2\sqrt{x}+4\right)}{\sqrt{x}+7}\)

\(=\)\(\frac{\left(4x-16\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}=\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

Sai đề không ?

A= \(\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-\left(x-3\right)\left(\sqrt{x}-2\right)-7\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\right)\)     .  \(\frac{x+2\sqrt{x}+4}{\sqrt{x}+7}\)

= \(\frac{x\sqrt{x}+2x+4\sqrt{x}-x\sqrt{x}+3\sqrt{x}-6+2x-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4x-16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

=\(\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4\left(\sqrt{x}+2\right)}{\sqrt{x}+7}\)

= \(\frac{4\sqrt{x}+8}{\sqrt{x}+7}\)

#mã mã#

tth