a ) ĐK : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)\(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^{^2}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

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

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

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

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

a)ĐKXĐ:x>0

P=\(\left(\frac{3}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\left(vớix>0\right)\)

=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]:\frac{1}{\sqrt{x}+1}\)

=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\frac{1}{\sqrt{x}+1}\)

= \(\left[\frac{3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\frac{1}{\sqrt{x}+1}\)

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

=\(\frac{4-\sqrt{x}}{\sqrt{x}-1}\)

b)Để P=\(\frac{5}{4}\left(vớix>0\right)\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}-\frac{5}{4}=0\)

\(\Leftrightarrow\frac{4\left(4-\sqrt{x}\right)}{4\left(\sqrt{x}-1\right)}-\frac{5\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}=0\)

\(\Rightarrow16-4\sqrt{x}-5\sqrt{x}+5=0\)

\(\Leftrightarrow21-9\sqrt{x}=0\)

\(\Leftrightarrow-9\sqrt{x}=-21\)

\(\Leftrightarrow\sqrt{x}=\frac{7}{3}\)

\(\Leftrightarrow x=\frac{21}{9}\)

Vậy:Để P=\(\frac{5}{4}\)thì x=\(\frac{21}{9}\)

c)Còn phần c thì mik chịu

...

sữa đề chút

a) đkxđ : \(x>2;x\ne3\)

b) ta có : \(A=\dfrac{\sqrt{x-1-2\sqrt{x-2}}}{\sqrt{x-2}-1}=\dfrac{\sqrt{\left(\sqrt{x-2}-1\right)^2}}{\sqrt{x-2}-1}=1\)

Mysterious Persondz Hung nguyendz Nguyễn Huy Thắngdz

0 câu trả lời

ĐKXĐ: (1-x)(2x-1)>=0

\(\Rightarrow\hept{\begin{cases}1-x>=0\\2\text{x}-1>=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x\le1\\x\ge\frac{1}{2}\end{cases}}\)

vậy 1/2<=x<=1

bé hơn hoặc bằng nha

cảm ơn nha

ko cs j chỉ là mk cx giống như bạn ban đầu cx ko hiểu thôi (mặc dù mk là dân bồi) :))

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

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

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

\(A=x-1\)

(ĐKXĐ là: \(x>0;x\ne1\))

a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)

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

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

b: A=2B

=>\(10=4\sqrt{x}-2\)

=>\(4\sqrt{x}=12\)

=>x=9(nhận)

Để \(\sqrt{x}-1\) được xác định cần:

\(\sqrt{x}\ge0\)

<=> \(x\ge0\)

Vậy ĐKXĐ của \(\sqrt{x}-1\) là \(x\ge0\)

Căn thức có nghiệm khi giá trị trong căn lớn hơn hoặc bằng 0 còn phép cộng đúng với mọi x nên x \(\ne1\) là sai nhé

ĐKXĐ: \(\orbr{\begin{cases}x>\sqrt{2}+1\\\frac{-1}{2}\le x< 1-\sqrt{2}\end{cases}}\)