Question

[Nguyễn Phương Anh]

Cho biểu thức:
Q= \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}\) - \(\frac{\sqrt{x}+3}{\sqrt{x}-2}\) - \(\frac{2\sqrt{x}+1}{3-\sqrt{x}}\) với x≥0 ; x≠4 ; x≠9
a) Rút gọn biểu thức Q
b) Tìm x để Q<1
c) Tìm x∈Z để Q nhận được giá trị nguyên
Mình cần gấp,mn giải chi tiết giúp mình phần a nhé!!!

Answer 1

[Lê Thị Thục Hiền]

a, Q=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)

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

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

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

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

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

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

b, Để Q<1 <=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}< 1\)

<=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\) <=> \(\frac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\) <=> \(\frac{4}{\sqrt{x}-3}< 0\)

<=> \(\sqrt{x}-3< 0\) <=> \(\sqrt{x}< 3\) <=> x<9. Kết hợp vs đk => \(0\le x< 9\) và \(x\ne2\)

Vậy Q<1 <=> \(0\le x< 9\) và \(x\ne2\)

c, Có \(Q=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để Q\(\in Z\) <=> \(\frac{4}{\sqrt{x}-3}\in Z\)

Vs \(x\in Z\) => \(\left\{{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\notin Z\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\sqrt{x}-3\in Z\\\sqrt{x}-3\notin Z\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\frac{4}{\sqrt{x}-3}\in Z\left(tm\right)\\\frac{4}{\sqrt{x}-3}\notin Z\left(ktm\right)\end{matrix}\right.\)

=> \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1,\pm2,\pm4\right\}\)

<=> \(\sqrt{x}\in\left\{4,2,1,5,-1,7\right\}\)

mà \(\sqrt{x}\ge0,\sqrt{x}\ne2\)

=> \(\sqrt{x}\in\left\{1,4,5,7\right\}\)

<=> x\(\in\left\{1,16,25,49\right\}\)

Vậy x\(\in\left\{1,16,25,49\right\}\) thì Q\(\in Z\)

Answer 2

[Nguyễn Phương Anh]

Cảm ơn nhé