Question

R=\(\frac{a+b}{\sqrt{a+\sqrt{b}}}:\)\(\left(\frac{a+b}{a-b}-\frac{b}{b-\sqrt{ab}}+\frac{a}{\sqrt{ab}+a}\right)-\) \(\frac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}\)

a) Rút gọn R

b) Tìm a ,b sao cho b=(a+1)² và R=-1

Answer 1

sửa lại đề đi \(\sqrt{a+\sqrt{b}}\) hay căn a+căn b

Answer 2

đk \(a>0;b>0;a\ne b\)\(R=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{a}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)-\frac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}\)

\(R=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)\)

\(-\frac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}\)

\(R=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b+\sqrt{ab}+b+a-\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)-\frac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}\)

\(R=\frac{a+b}{\sqrt{a}+\sqrt{b}}.\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{2\left(a+b\right)}-\frac{\sqrt{a}+\sqrt{b}}{2}\)

\(R=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\sqrt{a}+\sqrt{b}}{2}=\frac{-2\sqrt{b}}{2}=-\sqrt{b}\)

b) \(R=-1\Leftrightarrow-1=-\sqrt{b}\Leftrightarrow1=\sqrt{b}\Leftrightarrow b=1\)

b=(a+1)² <=> 1=(a+1)² <=> a+1=1 <=> a=0

vậy a = 0 ; b=1

Answer 3

công thức mk viết bị lỗi thông cảm