Question

2)cho biểu thức: \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

a)tìm các giá trị của x sao cho \(P=\frac{1}{2}\)

b) chứng minh \(P\le\frac{2}{3}\)

 

 

Answer 1

Điều kiện xác định : \(x\ge0,x\ne1\)

Rút gọn : \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)

Câu a) bạn tự tính nhé :)

b) \(P=\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{6-15\sqrt{x}}{3\left(\sqrt{x}+3\right)}=\frac{2\left(\sqrt{x}+3\right)-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}=-\frac{17\sqrt{x}}{3\left(\sqrt{x}+3\right)}+\frac{2}{3}\le\frac{2}{3}\)

Dấu "=" xảy ra khi x = 0