Question

1) Cho biểu thức \(A=\frac{\sqrt{x}+1}{x+4 \sqrt{x}+4} :\left(\frac{x}{x+2 \sqrt{x}}+\frac{x}{\sqrt{x}+2}\right)\), với x>0

a) Rút gọn A

b) Tìm tất cả các giá trị của x để \(A \geq \frac{1}{3 \sqrt{x}}\)

2) Cho biểu thức \(P=\left(1-\frac{1}{\sqrt{x}}\right) :\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\), với \(x>0\) và \(x \neq 1\)

a) Rút gọn P

b) Tim giá trị của P tại \(x=\sqrt{2022+4 \sqrt{2018}}-\sqrt{2022-4 \sqrt{2018}}\)

3) Cho biểu thức \(P=\left(\frac{x-6}{x+3 \sqrt{x}}-\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}+3}\right) : \frac{2 \sqrt{x}-6}{x+1}\), với \(x>0 ; x \neq 9\)

a) Rút gọn P

b) Tìm giá trị của x để P=1

Answer 1

https://i.imgur.com/7Gi05HK.jpg

Answer 2

https://i.imgur.com/lpCsO1V.jpg

Answer 3

\(3a.P=\left(\frac{x-6}{x+3\sqrt{x}}-\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}+3}\right):\frac{2\sqrt{x}-6}{x+1}\\ =\left(\frac{3\left(x-6\right)}{3\left(x+3\sqrt{x}\right)}-\frac{1\left(3\sqrt{x}+9\right)}{\sqrt{x}\left(3\sqrt{x}+9\right)}+\frac{1.3\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+3\right)}\right):\frac{2\sqrt{x}-6}{x+1}\\ =\left(\frac{3x-18}{3x+9\sqrt{x}}-\frac{3\sqrt{x}+9}{3x+9\sqrt{x}}+\frac{3\sqrt{x}}{3x+9\sqrt{x}}\right):\frac{2\sqrt{x}-6}{x+1}\\ =\left(\frac{3x-18-3\sqrt{x}-9+3\sqrt{x}}{3x+9\sqrt{x}}\right):\frac{2\sqrt{x}-6}{x+1}\\ =\frac{3x-27}{3x+9\sqrt{x}}:\frac{2\sqrt{x}-6}{x+1}\\ =\frac{3\left(x-9\right)}{3\left(x+3\sqrt{x}\right)}:\frac{2\sqrt{x}-6}{x+1}\\ \)

\(=\frac{x-9}{x+3\sqrt{x}}:\frac{2\sqrt{x}-6}{x+1}\\ =\frac{x-9}{x+3\sqrt{x}}.\frac{x+1}{2\sqrt{x}-6}\\ =\frac{\left(x-9\right)\left(x+1\right)}{\left(x+3\sqrt{x}\right)\left(2\sqrt{x}-6\right)}\\ =\frac{\left(x-9\right)\left(x+1\right)}{2x\sqrt{x}-18\sqrt{x}}\\ =\frac{\left(x-9\right)\left(x+1\right)}{2\sqrt{x}\left(x-9\right)}\\ =\frac{x+1}{2\sqrt{x}}\)

b) P=1

\(\frac{x+1}{2\sqrt{x}}=1\\ \Leftrightarrow2\sqrt{x}=x+1\\ \Leftrightarrow4x=x^2+2x+1\\ \Leftrightarrow-x^2+2x-1=0\\ \Leftrightarrow x^2-2x+1=0\\ \Leftrightarrow\left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)

Kết luận: P=1 thì x nhận giá trị.....

Chúc bạn học tốt , thắc mắc hỏi bên dưới nhé !

Answer 4

\(1a\\ A=\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\frac{x}{x+2\sqrt{x}}+\frac{x}{\sqrt{x}+2}\right)\\ =\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\frac{2x}{2\left(x+2\sqrt{x}\right)}+\frac{2x\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+2\right)}\right)\\ =\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\frac{2x}{2x+4\sqrt{x}}+\frac{2x\sqrt{x}}{2x+4\sqrt{x}}\right)\\ =\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\frac{2x+2x\sqrt{x}}{2x+4\sqrt{x}}\\ =\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\frac{2\sqrt{x}\left(\sqrt{x}+x\right)}{2\sqrt{x}\left(\sqrt{x}+2\right)}\\ =\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\frac{\sqrt{x}+x}{\sqrt{x}+2}\\ \)

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

b) \(A\ge\frac{1}{3\sqrt{x}}\)

\(\frac{1}{x+2\sqrt{x}}\ge\frac{1}{3\sqrt{x}}\\ \Leftrightarrow3\sqrt{x}\ge x+2\sqrt{x}\\ \Leftrightarrow\sqrt{x}-x\ge0\\ \Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)\ge0\\ \Leftrightarrow1-\sqrt{x}\ge0\\ \Leftrightarrow-\sqrt{x}\ge-1\\ \Leftrightarrow\sqrt{x}\le1\\ \Leftrightarrow x\le1\)

Kết luận.....

Chúc bạn học tốt , thắc mắc hỏi bên dưới hoặc nhắn tin nhé!