Question

Cho 2 biểu thức:

A = \(\frac{7}{\sqrt{x+8}}\)

B = \(\frac{\sqrt{x}}{\sqrt{x-3}}+\frac{2\sqrt{x}-24}{x-9}\) (x \(\ge\) 0 , x \(\ne\) 9)

a) Tính A khi x = 25

b) Chứng minh B = \(\frac{\sqrt{x}+8}{\sqrt{x}+3}\)

c) Tìm x để biểu thức P = A.B có giá trị nguyên

Answer 1

Khi x=25

=> A=\(\frac{7}{\sqrt{25+8}}=\frac{7}{\sqrt{\text{3}\text{3}}}\)=\(\frac{7\sqrt{33}}{33}\)

b)  B= \(\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}+\frac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

B=  \(\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\)

B=  \(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)