Question

Rút gọn biểu thức

a.\(C=\sqrt{\frac{a-2\sqrt{a-1}}{a+2\sqrt{a-1}}}:\frac{a-2}{\left(\sqrt{a-1}+1\right)^2}\)[vói \(a\ge1,a\ne2\)]

b.\(B=\left(\sqrt{x}+\sqrt{y}+1\right)\left(\sqrt{x}-\sqrt{y}-1\right)+2\sqrt{y}+1\)

c.\(C=\sqrt{a^2-2\sqrt{a^2+1}-2}\)

d.\(D=\sqrt{a+\sqrt{a+\frac{1}{2}+\sqrt{a+\frac{1}{4}}}}\)

Answer 1

a. \(C=\sqrt{\frac{a-1-2\sqrt{a-1}+1}{a-1+2\sqrt{a-1}+1}}:\frac{a-2}{\left(\sqrt{a-1}+1\right)^2}=\frac{\left(\sqrt{a-1}-1\right)^2}{\left(\sqrt{a-1}+1\right)^2}.\frac{\left(\sqrt{a-1}+1\right)^2}{a-2}=\frac{\left(\sqrt{a-1}-1\right)^2}{a-2}\)

b.\(B=\left(x-\left(\sqrt{y}-1\right)^2\right)+2\sqrt{y}+1=x-y+2\sqrt{y}-1+2\sqrt{y}+1=x+4\sqrt{y}+y\)

c.\(C=\sqrt{a^2+1-2\sqrt{a^2+1}+1}=\sqrt{a^2+1}-1\)

bn ơi mk nghĩ câu c ấy , cái chỗ -2 bn nên đổi thành 2