uses crt;

var a,m,i:integer;

s:real;

begin

clrscr;

write('Nhap a='); readln(a);

write('Nhap m='); readln(m);

s:=1;

for i:=1 to m do 

  s:=s+1/sqr(a+i);

writeln(s:4:2);

readln;

end.

Lời giải:
a. ĐKXĐ: $a\geq 0; a\neq 1$

b.

\(P=\left[\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}+1\right].\left[\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}-1\right].\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}\)

\(=(\sqrt{a}+1)(\sqrt{a}-1).\sqrt{2}=\sqrt{2}(a-1)\)

c.

\(P=\sqrt{2}(\sqrt{2+\sqrt{2}}-1)=\sqrt{4+2\sqrt{2}}-\sqrt{2}\)

a. ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ne1\\\sqrt{a}\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

b. \(P=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-1\right).\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)

\(=\left[\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right].\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right].\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\cdot\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\cdot\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)

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

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

\(a,P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1-a\right)\left(1+a\right)}}\left(-1< a< 1\right)\\ P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{2+\sqrt{\left(1-a\right)\left(1+a\right)}}\\ P=\sqrt{1-a}\\ b,a=\dfrac{24}{49}\Leftrightarrow1-a=\dfrac{25}{49}\\ \Leftrightarrow P=\sqrt{1-a}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\\ c,P=2\Leftrightarrow1-a=4\Leftrightarrow a=-3\left(ktm\right)\Leftrightarrow a\in\varnothing\)

\(a,A=\dfrac{x^2-x-2}{x^2-1}+\dfrac{1}{x-1}-\dfrac{1}{x+1}\)

\(\Rightarrow A=\dfrac{x^2-x-2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x^2-x-2x+x+1-x+1}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x^2-3x+2}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x^2-2x-x+2}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x-2}{x+1}\)

\(b,A=\dfrac{3}{4}\\ \Rightarrow\dfrac{x-2}{x+1}=\dfrac{3}{4}\\ \Rightarrow4\left(x-2\right)=3\left(x+1\right)\\ \Rightarrow4x-8=3x+3\\ \Rightarrow4x-8-3x-3=0\\ \Rightarrow x-11=0\\ \Rightarrow x=11\)

\(c,\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

Thay x=5 vào A ta có:

\(A=\dfrac{x-2}{x+1}=\dfrac{5-2}{5+1}=\dfrac{3}{6}=\dfrac{1}{2}\)

Thay x=1 vào A ta có:

\(A=\dfrac{x-2}{x+1}=\dfrac{1-2}{1+1}=\dfrac{-1}{2}\)

giúp mình với !!!!!!!!!!!!!!!!!!!!!!!!

Câu b, bài b1 chứng minh \(a=2^{2006}-1?\)