Question

Cho biểu thức \(P=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\) với  \(a\ge0,a\ne1\)

a) Rút gọn biểu thức P

b) Tìm giá trị nhỏ nhất cua P

 

Answer 1

chỉ làm được câu a do hơi gà 

\(P=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\)

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

\(=\frac{1-a+a^2-a^2+2}{\left(1-a\right)\left(1-a+a^2\right)}=\frac{3-a}{\left(1-a\right)\left(1-a+a^2\right)}\)

Answer 2

sửa dòng cuối :))

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

Answer 3

a) P = \(\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\)

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

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

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

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

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

b) Ta có: \(-\left(a^2+a+1\right)=-\left(a^2+a+\frac{1}{4}\right)-\frac{3}{4}=-\left(a+\frac{1}{2}\right)-\frac{3}{4}\le-\frac{3}{4}\)

=> \(\frac{1}{-\left(a^2+a+1\right)}\ge\frac{1}{-\frac{3}{4}}=-\frac{4}{3}\)

Dấu "=" xảy ra <=> x + 1/2 = 0 <=> x = -1/2

Vậy MinP = -4/3 khi x = -1/2