Question

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

a/ Rút gọn P

b/ Tìm GTNN của P

c/ Xét biểu thức Q =\(\frac{2\sqrt{x}}{p}\)

Chứng tỏ 0<Q<2

 

 

Answer 1

Chứng tỏ 0<Q<2 nha

Answer 2

\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+1=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\)

\(P+1=\frac{x^2+x+1}{x+\sqrt{x}+1}=\frac{x^2+2x+1-x}{x+\sqrt{x}+1}=\frac{\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x+\sqrt{x}+1}=x-\sqrt{x}+1\ge\frac{3}{4}\)

Answer 3

a) \(ĐKXĐ:x>1\)

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

\(=\frac{\left(\sqrt{x}\right)^4-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}.\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

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

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

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

b) Ta có: \(P=x-\sqrt{x}+1=x-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\left(\forall x>0\right)\)\(\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)\(\Leftrightarrow x=\frac{1}{4}\)

Vậy \(minP=\frac{3}{4}\Leftrightarrow x=\frac{1}{4}\)

Answer 4

               Bài làm :

 \(\text{a)}ĐKXĐ:x>1\)

Ta có :

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

\(P=\frac{\left(\sqrt{x}\right)^4-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}.\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(p=\frac{\sqrt{x}.\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)

\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)

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

\(P=x-\sqrt{x}+1\)

b) Ta có:

 \(P=x-\sqrt{x}+1\)

\(P=x-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}+\frac{3}{4}\)

\(P=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\)

 \(\text{Vì : }\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\left(\forall x>0\right)\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu " = " xảy ra khi :

 \(\sqrt{x}-\frac{1}{2}=0\)

\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{4}\)

Vậy GTNN của P = 3/4 ⇔ x=1/4