Question

C=\(\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)

a) Rút gọn C

b)Tìm x để C=\(\frac{1}{2}\)

c)Tìm x để \(\left|C\right|\)=\(\frac{1}{3}\)

Answer 1

a , C = \(\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x-1}\)

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

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

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

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

Answer 2

a) C= \(\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)

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

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

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

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