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

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

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

\(A=\frac{\sqrt{a}-2}{\sqrt{a}}\)

a/ \(P=\left(\frac{x-7\sqrt{x}+12}{x-4\sqrt{x}+3}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}.\)

\(P=\left(\frac{x-7\sqrt{x}+12}{\left(x-4\sqrt{x}+4\right)-1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-2\right)^2-1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-2-1\right)\left(\sqrt{x}-2+1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\frac{x-7\sqrt{x}+12+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\frac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

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

b/ Để P>3/4 => \(P=\frac{\sqrt{x}+3}{\sqrt{x}-1}>\frac{3}{4}\)

+/ TH1: x>1 => \(4\left(\sqrt{x}+3\right)>3\left(\sqrt{x}-1\right)\)

  <=> \(\sqrt{x}>-16\)  => x>1

+/ TH2: 0<x<1 => \(4\left(\sqrt{x}+3\right)< 3\left(\sqrt{x}-1\right)\)  => \(\sqrt{x}< -16\)=> Loại

       ĐS: x>1

c/ P=2  <=> \(P=\frac{\sqrt{x}+3}{\sqrt{x}-1}=2\)

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

<=> \(\sqrt{x}=5=>x=25\)

Mỗi biểu thức trong dấu căn có dạng:

\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\)   ( Với \(k\ge2\))

Ta có:

\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\frac{k^2\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}=\frac{k^4+2k^3+k^2+k^2+2k+1+k^2}{k^2\left(k+1\right)^2}\)

\(=\frac{k^4+2k^2\left(k+1\right)+\left(k+1\right)^2}{k^2\left(k+1\right)^2}=\frac{\left(k^2+k+1\right)^2}{\left(k\left(k+1\right)\right)^2}\)

\(\Rightarrow\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{k^2+k+1}{k^2+k}=1+\frac{1}{k\left(k+1\right)}=1+\frac{1}{k}-\frac{1}{k+1}\)

\(\Rightarrow S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2013}-\frac{1}{2014}=2014-\frac{1}{2014}\)

Mỗi biểu thức trong dấu căn có dạng:

1+1k2 +1(k+1)2    ( Với k≥2)

Ta có:

1+1k2 +1(k+1)2 =k2(k+1)2+(k+1)2+k2k2(k+1)2 =k4+2k3+k2+k2+2k+1+k2k2(k+1)2 

=k4+2k2(k+1)+(k+1)2k2(k+1)2 =(k2+k+1)2(k(k+1))2 

⇒√1+1k2 +1(k+1)2 =k2+k+1k2+k =1+1k(k+1) =1+1k −1k+1 

⇒S=1+1−12 +1+12 −13 +1+13 −14 +...+1+12013 −12014 =2014−12014 

a, dk \(x\ge0\)

ap dung bdt cosi ta co

\(\sqrt{x+3}+\frac{4x}{\sqrt{x+3}}\ge2\sqrt{4x}=4\sqrt{x}\)

dau = xay ra \(\Leftrightarrow\sqrt{x+3}=\frac{4x}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Rightarrow x=1\)(tm dk)

kl x=1 la no cua pt

1 {15;24;33;42;51;60}

1,{15,24,33,42,51}

2,{23,30,31,32,33,34,35,36,37,38,39,43}

OK bn nhé