Question

Cho biểu thức \(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

a. Rút gọn A

b. Tính giá trị của A khi \(x=3-2\sqrt{2}\)

c. Tìm GTNN của biểu thức A

Answer 1

ĐKXĐ :x\(\ge\)0;x\(\ne\)1;x\(\ne\)3

\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{\left(\sqrt{x}-1\right)\left(x+16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x+16}{\sqrt{x}+3}\)

Answer 2

b, x =(\(\sqrt{2}-1)^2\)

Thay x =(\(\sqrt{2}-1)^2\)thỏa mãn đk vào a có:

A=\(\dfrac{\left(\sqrt{2}-1\right)^2+16}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

=\(\dfrac{2-2\sqrt{2}+1+16}{\sqrt{2}-1}\)

=\(\dfrac{19\sqrt{2}+19-4-2\sqrt{2}}{2-1}\)

=\(17\sqrt{2}+15\)

Answer 3

c, A=\(\dfrac{x+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)

=\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)

Theo BĐT cosy ta có:

\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right)\times\dfrac{25}{\sqrt{x}+3}}\)

\(\Rightarrow\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-\ge10-6\)

\(\Rightarrow A\ge4\)

\(\Rightarrow MinA=4\)

Dấu = xảy ra\(\Leftrightarrow\)\(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\)

\(\Leftrightarrow\sqrt{x}+3=5\)

\(\Leftrightarrow\sqrt{x}=2\)

\(\Leftrightarrow x=4\)