sao ko ai làm hộ tôi vậy 

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

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

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

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

b) ta có : Q=3 => \(\dfrac{-2\sqrt{x}+1}{\sqrt{x}-1}=3=>-2\sqrt{x}+1=3\sqrt{x}-3\)

=>x=16/25=0,64

vậy x=0,64 khi Q=3

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

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

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

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

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

b) ta có : Q=3 <=> \(\sqrt{x}-1=3\)

\(\Leftrightarrow\) \(\sqrt{x}=4\Leftrightarrow x=16\)

vậy để Q=3 thì x=16

cái này ms đúng

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-x+1=x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\1-x=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

b.

ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-3\end{matrix}\right.\)

Do \(\left\{{}\begin{matrix}\sqrt{x^2-3x+2}\ge0\\\sqrt{x^2+x-6}\ge0\end{matrix}\right.\) với mọi x thuộc TXĐ

\(\Rightarrow\sqrt{x^2-3x+2}+\sqrt{x^2+x-6}\ge0\)

Đẳng thức xảy ra khi:

\(\left\{{}\begin{matrix}x^2-3x+2=0\\x^2+x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=2\) (thỏa mãn ĐKXĐ)

Vậy pt có nghiệm duy nhất \(x=2\)

c.

Với \(x< 1\Rightarrow\left\{{}\begin{matrix}x-1< 0\\\sqrt{x^4-2x^2+1}\ge0\end{matrix}\right.\) phương trình vô nghiệm

Với \(x\ge1\) pt tương đương:

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

\(\Leftrightarrow\left|x^2-1\right|=x-1\)

\(\Leftrightarrow x^2-1=x-1\) (do \(x\ge1\Rightarrow x^2-1\ge0\Rightarrow\left|x^2-1\right|=x-1\))

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0< 1\left(loại\right)\\x=1\end{matrix}\right.\)

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

\(=\dfrac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\left(\text{đ}pcm\right)\)

A+B

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

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

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

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

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

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

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

\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)

a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)

c) \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}=\dfrac{7}{2\sqrt{5}-\sqrt{3}}=\dfrac{7\left(2\sqrt{5}+\sqrt{3}\right)}{\left(2\sqrt{5}+\sqrt{3}\right)\left(2\sqrt{5}-\sqrt{3}\right)}\)

\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)

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

 \(\dfrac{a+2}{\sqrt{a^2-4}}=\dfrac{\sqrt{a+2}}{\sqrt{a-2}}=\dfrac{\sqrt{a^2-4}}{a-2}\)

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

\(\dfrac{a}{\sqrt{x^2}}=\dfrac{a}{\left|x\right|}\)

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

c) \(\dfrac{2}{\sqrt{7-2\sqrt{6}}}=\dfrac{2}{\sqrt{6}-1}=\dfrac{2\left(\sqrt{6}+1\right)}{5}\)