Question

p=\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)

a. rút gọn p với a ≥0 và a≠1

b. tìm a để p =0

Answer 1

a) \(P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)

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

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(a-\sqrt{a}+1-\sqrt{a}\right)\)

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

\(=\left(a-1\right)^2=a^2-2a+1\)

b) \(P=0\Rightarrow\left(a-1\right)^2=0\Rightarrow a=1\)

Answer 2

a) Ta có: \(P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)

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

\(=\left(a-1\right)^2\)

b) Để P=0 thì a-1=0

hay a=1(loại)