Question

[le anh nhat]

cho N=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)

a. tìm ĐKXĐ và rút gọn

b.tìm x để N=5

c. tìm x\(\in\)Z để n\(\in\)Z

Answer 1

[Lê Thu Dương]

N=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)

= \(\frac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

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

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

ĐKXĐ : x ≠ 4 ; x ≠ 9

Rút gọn :

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

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

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

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

= \(\frac{x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

Để N =5 thì :

<=> \(\frac{x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\) =5

<=> x-5 = \(\left(5\sqrt{x}-10\right)\left(\sqrt{x}-3\right)\)

<=> x-5 = 5x - \(15\sqrt{x}\) - \(10\sqrt{x}\) +30

<=> x-5x-25\(\sqrt{x}\) =35

Answer 2

[Trang Hà]

a) \(\sqrt{x}\ne3;\sqrt{x}\ne2\Rightarrow x\ne4;x\ne9\)

\(N=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)

\(\Leftrightarrow N=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(\Leftrightarrow N=\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(\Rightarrow N=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) \(N=5\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}=5\)

\(\Leftrightarrow\sqrt{x}+1=5\sqrt{x}-15\Leftrightarrow4\sqrt{x}=16\)

\(\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\) (thỏa mãn)

c) \(N=\frac{\sqrt{x}+1}{\sqrt{x}-5}=\frac{\sqrt{x}-5+6}{\sqrt{x}-5}=1+\frac{6}{\sqrt{x}-5}\)

để N \(\in\) Z thì \(\left(\sqrt{x}-5\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

\(\sqrt{x}-5\)	1	-1	2	-2	3	-3	6	-6
x	36	16	49	9	64	4	121	loại