Question

cho biểu thức B=\(\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)(x\(\ge\)o,x\(\ne\)1)

a)rút gọn B

b)tìm x để B=3

Answer 1

a) B=\(\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)

=\(\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(1-\sqrt{x}+x-\sqrt{x}\right)\)

=\(\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)

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

b) ta có : B=3 \(\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=16\)

vậy để B=3 thì x=16