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3 tháng 9 2021

\(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-2}\)

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

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

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

\(=\dfrac{6\sqrt{x}}{\sqrt{x}-2}\)

3 tháng 9 2021

\(C=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{\sqrt{x}-2}\) (\(x\ge0,x\ne4,x\ne9\))
\(C=\left[\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}\right].\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}.\left(\sqrt{x}+2\right)^2\)
\(C=\dfrac{2}{\sqrt{x}-2}\)
 

30 tháng 7 2016

a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\left(\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{9-x+x-4\sqrt{x}+4-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)

\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)

b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)

\(\Rightarrow9x+7\sqrt{x}-19=0\)

Mình k biết mình sai chỗ nào nữa, bạn xem giúp mình với

15 tháng 7 2021

`A=((3sqrtx+6)/(x-4)+sqrtx/(sqrtx-2)):(x-9)/(sqrtx-3)(x>=0,x ne 4,x ne 9)`

`=((3(sqrtx+2))/((sqrtx-2)(sqrtx+2))+sqrtx/(sqrtx-2)):((sqrtx-3)(sqrtx+3))/(sqrtx-3)`

`=(3/(sqrtx-2)+sqrtx/(sqrtx-2)):(sqrtx+3)`

`=(sqrtx+3)/(sqrtx-2)*1/(sqrtx+3)`

`=1/(sqrtx-2)`

15 tháng 7 2021

\(A=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\)

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

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

21 tháng 3 2017

Ủa mua Vip là giáo viên trả lời hả? :>

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

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

\(P=\left(\frac{x+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-1}{\sqrt{x}-3}\right).\left(\sqrt{x}-2\right)\)

\(P=\left(\frac{x+\sqrt{x}-4-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right).\left(\sqrt{x}-2\right)\)

\(P=\frac{x+\sqrt{x}-4-x+\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}-2\right)\)

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

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

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

21 tháng 3 2017

Mình không biết ? Đọc trong THÔNG TIN ghi là GV trả lời cho VIP's member mà bạn.

10 tháng 8 2021

Tui nhầm đề xíu, cái A kia phải là:   A=\(\sqrt{\left(1-\sqrt{5}\right)^2}-\frac{5-2\sqrt{5}}{\sqrt{5}}\)

10 tháng 8 2021

thảo nào rút gọn mãi nó chả mất căn :))

\(A=\sqrt{\left(1-\sqrt{5}\right)^2}-\frac{5-2\sqrt{5}}{\sqrt{5}}\)

\(=\sqrt{5}-1-\frac{5\sqrt{5}-10}{5}=\frac{5\sqrt{5}-5-5\sqrt{5}+10}{5}=\frac{5}{5}=1\)

Với \(x\ge0;x\ne4;9\)

\(P=\left(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}\right):\frac{x-9}{\sqrt{x}-3}\)

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

\(=\left(\frac{x+5\sqrt{x}+6}{x-4}\right):\left(\sqrt{x}+3\right)=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(x-4\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-2}\)

b, \(2P-A< 0\Rightarrow\frac{2}{\sqrt{x}-2}-1< 0\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-2}< 0\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-2}>0\)

TH1 : \(\hept{\begin{cases}\sqrt{x}-4>0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>16\\x>4\end{cases}\Leftrightarrow x>16}\)

TH2 : \(\hept{\begin{cases}\sqrt{x}-4< 0\\\sqrt{x}-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 16\\x< 4\end{cases}}\Leftrightarrow x< 4}\)

Kết hợp với đk vậy \(0\le x< 4;x>16\)