\(\frac{\sqrt{x-2-2\sqrt{x}-3}}{\sqrt{x-3-1}}\)

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4 tháng 9 2017

M= \(\sqrt{2}+1-\) \(\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}+1-\sqrt{2}+1=2\)

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

P= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\frac{2\sqrt{x}.\sqrt{x}}{\sqrt{x}}\) (dk \(x>0\))

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

Q= \(\sqrt{\left(\sqrt{x}+1\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\) (dk \(x\ge0\) )

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

th1 \(\sqrt{x}\ge1\Leftrightarrow x\ge1\) Q=\(\sqrt{x}+1+\sqrt{x}-1=2\sqrt{x}\)

th2 \(0\le x< 1\) Q=\(\sqrt{x}+1+1-\sqrt{x}=2\)

4 tháng 9 2017

a)  \(M=\sqrt{2}+1-\sqrt{1,5.2-2.\sqrt{2}}\)

\(=\sqrt{2}+1-\sqrt{2.\left(1,5-\sqrt{2}\right)}\)\(=\sqrt{2}+1-\sqrt{2}.\sqrt{1,5-\sqrt{2}}\)

\(=\sqrt{2}.\left(1+1,5-\sqrt{2}\right)+1=\sqrt{2}.\left(2,5-\sqrt{2}\right)+1\)

\(=\sqrt{2}.2,5-2+1=\sqrt{2}.2,5-1\)

P/s: Theo em thì em nghĩ là đúng '-' Khoảng 90% :)

23 tháng 7 2018

a)\(P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}.\)

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

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

b)P=3/2  <=>\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{3}{2}\Leftrightarrow2\sqrt{x}+1=\frac{3}{2}\sqrt{x}+\frac{3}{2}.\)

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

Với x=1 thoả nãm yêu cầu

25 tháng 7 2018

ĐKXĐ: \(x\ge0\)

\(\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}\)

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

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

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

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

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

2 tháng 3 2020

Câu 3 :

\(ĐKXĐ:x>0\)

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

\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)

\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)

b) Để P = 3

\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)

\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)

\(\Leftrightarrow x-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow\sqrt{x}-2=0\)

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

\(\Leftrightarrow x=4\)(tm)

Vậy để \(P=3\Leftrightarrow x=4\)

2 tháng 3 2020

Câu 1 : Hình như sai đề !! Mik sửa :

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)

\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)

\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)

\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

b) Để A < 2

\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)

\(\Leftrightarrow-1< 2\sqrt{x}-4\)

\(\Leftrightarrow2\sqrt{x}>3\)

\(\Leftrightarrow\sqrt{x}>1,5\)

\(\Leftrightarrow x>2,25\)

Vậy để \(A< 2\Leftrightarrow x>2,25\)

18 tháng 7 2018

mk làm luôn.

a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)

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

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

mk làm phần rút gọn xong mk bận nên bn tự làm câu b nha ^^