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\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
2)
a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:
\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)
b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=5+\sqrt{2}-4-\sqrt{2}\)
=1
Thay x=1 vào P, ta được:
\(P=\dfrac{1+2}{1+3}=\dfrac{3}{4}\)
a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:
\(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)
b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)
\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)
c: Để B là số tự nhiên thì \(\sqrt{x}+4⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;3;6\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{3;4;5;8\right\}\)
hay \(x\in\left\{16;25;64\right\}\)
a: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
Khi x=25 thì \(A=\dfrac{5+2}{5+3}=\dfrac{7}{8}\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}+2}+\dfrac{x+4}{4-x}\)
\(=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-x-4}{x-4}\)
\(=\dfrac{5\sqrt{x}-10}{x-4}=\dfrac{5}{\sqrt{x}+2}\)
c: \(A\cdot B=\dfrac{5}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{5}{\sqrt{x}+3}\)
Để A*B>1 thì \(\dfrac{5}{\sqrt{x}+3}-1>0\)
=>\(\dfrac{5-\sqrt{x}-3}{\sqrt{x}+3}>0\)
=>\(2-\sqrt{x}>0\)
=>căn x<2
=>0<=x<4
b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{x^2-9}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
