RG: B = [\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\) - \(\dfrac{\sqrt{x}}{1-x}\) ] : [ \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\) ] ; ĐKXĐ: x ≥ 0
a : \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)với a ≥ 0 x ≠ 4
b : \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
c : \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
d : \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))
\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(x+2\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x\sqrt{x}+2x+\sqrt{x}}\)
c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)
Ch.minh
a) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) = \(\dfrac{3\sqrt{x}-1}{x-1}\)
b) (\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)-\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) ) : \(\dfrac{1}{x-4}\) = -4\(\sqrt{x}\)
a.
Với \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\) có:
\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{x-1}\\ =\dfrac{x+\sqrt{x}}{x-1}-\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}\\ =\dfrac{x+\sqrt{x}}{x-1}-\dfrac{x-2\sqrt{x}+1}{x-1}\\ =\dfrac{x+\sqrt{x}-x+2\sqrt{x}-1}{x-1}\\ =\dfrac{3\sqrt{x}-1}{x-1}=VP\)
b.
Với \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\) có:
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{1}{x-4}\right)\\ =(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}).\left(\dfrac{x-4}{1}\right)\\ =(\dfrac{x-2\sqrt{x}}{x-4}-\dfrac{x+2\sqrt{x}}{x-4}).\left(x-4\right)\\ =\left(\dfrac{x-2\sqrt{x}-x-2\sqrt{x}}{x-4}\right)\left(x-4\right)\\ =\dfrac{-4\sqrt{x}\left(x-4\right)}{x-4}\\ =-4\sqrt{x}=VP\)
Cho biểu thức: B = \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{1-x}\right)\) : \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\) . Tính gtri của B biết x = \(\dfrac{2-\sqrt{3}}{2}\)
ĐKXĐ: x>=0; x<>1
\(B=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{x-1}:\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
Khi \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\left(\dfrac{\sqrt{3}-1}{2}\right)^2\) thì:
\(B=\dfrac{2\cdot\dfrac{2-\sqrt{3}}{2}+2\cdot\dfrac{\sqrt{3}-1}{2}+1}{4\cdot\dfrac{\sqrt{3}-1}{2}}\)
\(=\dfrac{2-\sqrt{3}+\sqrt{3}-1+1}{2\left(\sqrt{3}-1\right)}=\dfrac{2}{2\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{2}\)
Rút gọn các biểu thức sau
a,\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
b,\(B=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{x\sqrt{x}+1}\)
c,\(C=\left(1-\dfrac{x+3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)
d,\(D=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
e,\(E=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a) Ta có: \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) Ta có: \(B=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{x\sqrt{x}+1}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\left(\dfrac{2x\sqrt{x}-3x+3\sqrt{x}-1+3x+2\sqrt{x}-1-2x\sqrt{x}+2x-2\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}+1}{x-\sqrt{x}+1}\)
Cho biêu thức A=\(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
a. RG A
A=\(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
`=((x-2+\sqrtx)/(x+2\sqrtx).({\sqrt{x}+1}{\sqrt{x}-1})`
`=((\sqrtx-1)(\sqrtx+2))/(\sqrtx(\sqrtx+2)).({\sqrt{x}+1}{\sqrt{x}-1})`
`=(\sqrtx+1)/\sqrtx`
A=\(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
`=((x-2+\sqrtx)/(x+2\sqrtx).({\sqrt{x}+1}/{\sqrt{x}-1})`
`=((\sqrtx-1)(\sqrtx+2))/(\sqrtx(\sqrtx+2)).({\sqrt{x}+1}/{\sqrt{x}-1})`
`=(\sqrtx+1)/\sqrtx`
a) Ta có: \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)
= \(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)=\(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
1.A=\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
2.B=\(\left(\dfrac{2\sqrt{x+x+1}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
3.C=\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
Làm chi tiết hộ mình với ak mình đang cần gấp!!!
1: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\)
RG: A = \(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}\) - \(\dfrac{2\sqrt{x}}{\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\) ; ĐKXĐ: x ≥ 0
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-3\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{\left(x\sqrt{x}-x\right)+\left(16\sqrt{x}-16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{\left(\sqrt{x}-1\right)\left(x+16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x+16}{\sqrt{x}+3}\)
Rút gọn
A=\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right)\div\left(\dfrac{\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
B=\(\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\div\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\)
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(x>0,x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2x}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}:\dfrac{7-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{7-\sqrt{x}}=\dfrac{x}{\sqrt{x}-7}\)
\(B=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\left(x>0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}+1\)
\(=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}+1=-\dfrac{\sqrt{x}+1}{\sqrt{x}}+1\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}}=-\dfrac{1}{\sqrt{x}}\)