a: Thay x=49 vào A, ta được:

\(A=\dfrac{2\cdot7+1}{7-3}=\dfrac{14+1}{4}=\dfrac{15}{4}\)

b: \(B=\dfrac{2x+36}{x-9}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36-9\left(\sqrt{x}+3\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x+36-9\sqrt{x}-27-x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)

P>1 khi P-1>0

=>\(\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}>0\)

=>\(\sqrt{x}-2>0\)

=>\(\sqrt{x}>2\)

=>x>4

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>4\\x\ne9\end{matrix}\right.\)

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cần có điều kiện của x thì mới rút gọn được

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

\(=\frac{x+2}{4\left(x-6\right)}.\frac{\left(x-6\right).\left(x+6\right)}{x\left(x+2\right)-\left(x+2\right)}\)

\(=\frac{x+2}{4\left(x-6\right)}.\frac{\left(x-6\right).\left(x+6\right)}{\left(x+2\right).\left(x-1\right)}\)

\(=\frac{x+6}{4x-4}\)

@lê thị hương giang

\(A=\frac{x^{39}+x^{36}+x^{33}+...+x^3+1}{x^{40}+x^{38}+x^{36}+...+x^2+1}\)

Đặt \(C=x^{39}+x^{36}+x^{33}+...+x^3+1\)

\(x^3.C=x^{42}+x^{39}+x^{36}+...+x^3\)

\(\left(x^3-1\right)C=x^{42-1}\)

\(C=\frac{x^{42}-1}{x^3-1}\)

Đặt \(D=x^{40}+x^{38}+x^{36}+....+x^2+1\)

\(x^2.D=x^{42}+x^{40}+x^{38}+x^{36}+....+x^2\)

\(\left(x^2-1\right).D=x^{42}-1\)

\(D=\frac{x^{42}-1}{x^2-1}\)

Ta có :

\(C:D=\frac{x^{42}-1}{x^3-1}:\frac{x^{42}-1}{x^2-1}\)

\(C:D=\frac{x^2-1}{x^3-1}\)

\(C:D=\frac{x+1}{x^2+x+1}\)

Ta có : \(A=C:D=\frac{x+1}{x^2+x+1}\)

Vậy ...........