Question

1.

\(A=\left(\frac{1+x\sqrt{x}}{x-1}+1\right).\left(1-\frac{1}{\sqrt{x}}\right)\)

a. Tìm ĐKXĐ

b. Rút gọn biểu thức A

c. Tính GTBT A khi \(x=4-2\sqrt{3}\)

Answer 1

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

ĐKXĐ : x > 1

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

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

= \(\sqrt{x}\)

Ta có :

x = 4 \(-2\sqrt{3}\)

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

<=> x = ( \(\sqrt{3}-1\) )²

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

= \(\sqrt{\left(\sqrt{3}-1\right)^2}\)

= \(\left|\sqrt{3}-1\right|\)