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28 tháng 6 2017

a.ĐKXĐ;\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

b.P=\(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{x-4}\)

=\(\frac{3x-6\sqrt{x}}{x-4}=\frac{3\sqrt{x}.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)=\(\frac{3\sqrt{x}}{\sqrt{x}+2}\)

c.P=2\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}=2\Leftrightarrow3\sqrt{x}=2\sqrt{x}+\text{4}\)\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)

Vậy x=16

17 tháng 10 2018

thần đồng

12 tháng 7 2016

A>0 chứ ko phải x>0

11 tháng 8 2023

a) \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{x\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\right)\)

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

\(P=\left(\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

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

b) P = \(\dfrac{1}{2}\) khi:

\(\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\)

\(\Rightarrow2=\sqrt{x}-1\)

\(\Rightarrow\sqrt{x}=3\)

\(\Rightarrow x=9\left(tm\right)\)

a: \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{x\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\)

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

\(=\dfrac{1}{\sqrt{x}-1}\)

b: P=1/2

=>căn x-1=2

=>căn x=3

=>x=9

28 tháng 10 2023

\(P=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

\(\sqrt{x}+2>=2\)

=>\(\dfrac{3}{\sqrt{x}+2}< =\dfrac{3}{2}\)

=>\(-\dfrac{3}{\sqrt{x}+2}>=-\dfrac{3}{2}\)

=>\(P=\dfrac{-3}{\sqrt{x}+2}+1>=-\dfrac{3}{2}+1=-\dfrac{1}{2}\)

Dấu = xảy ra khi x=0

1 tháng 10 2017

\(\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}-1}\div\frac{2}{x-1}+\frac{1}{\sqrt{x}+1}.\)

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

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

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

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

\(=\frac{1+x}{\sqrt{x}}\)

a: \(P=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)

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

\(=\dfrac{x\sqrt{x}-x-4\sqrt{x}-6-2x+12\sqrt{x}-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x+8}{\sqrt{x}+1}\)

b:Đề sai rồi bạn

Vì 14-6 căn 15<0 nên x này vô nghĩa nha bạn

6 tháng 9 2019

a) \(\sqrt{x}\)\(\sqrt{2x-1}\)

x < 2x - 1

x - 2x < -1

-x < -1

x > 1

b) \(\sqrt{x}\le\sqrt{x+1}\)

< x + 1

< 1

không có x tm