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a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
Bài 1:
a)Với x > 0;x ≠ 4 ta có:
\(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}+2\right)-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)\)
\(=\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4}{x-4}\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=b-a\)
Bài 2:
a)Với a > 0;a ≠ 1;a ≠ 2 ta có
\(P=\left(\dfrac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}\)
b)Ta có:
\(P=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2a-4}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
P nguyên khi \(2-\dfrac{8}{a+2}\) nguyên⇒\(\dfrac{8}{a+2}\) nguyên⇒\(a+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(TH1:a+2=1\Rightarrow a=-1\left(loai\right)\)
\(TH2:a+2=-1\Rightarrow a=-3\left(loai\right)\)
\(TH3:a+2=2\Rightarrow a=0\left(loai\right)\)
\(TH4:a+2=-2\Rightarrow a=-4\left(loai\right)\)
\(TH5:a+2=4\Rightarrow a=2\left(loai\right)\)
\(TH6:a+2=-4\Rightarrow a=-6\left(loai\right)\)
\(TH7:a+2=8\Rightarrow a=6\left(tm\right)\)
\(TH8:a+2=-8\Rightarrow a=-10\left(loai\right)\)
Vậy a = 6
Với `x >= 0,x ne 4` có:
`M=[(\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}(\sqrt{x}-2)-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`M=[x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`M=[3x-6\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]=[3\sqrt{x}]/[\sqrt{x}+2]`
____________
`N=(1/[\sqrt{a}-1]-1/\sqrt{a}):([\sqrt{a}+1]/[\sqrt{a}-2]-[\sqrt{a}+2]/[\sqrt{a}-1])`
- Biểu thức `N` là như vầy?
Với `a > 0,a ne 1,a ne 4` có:
`N=[\sqrt{a}-\sqrt{a}+1]/[\sqrt{a}(\sqrt{a}-1)]:[(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}+2)(\sqrt{a}-2)]/[(\sqrt{a}-2)(\sqrt{a}-1)]`
`N=1/[\sqrt{a}(\sqrt{a}-1)].[(\sqrt{a}-2)(\sqrt{a}-1)]/[a-1-a+4]`
`N=[\sqrt{a}-2]/[3\sqrt{a}]`
Với \(x\ge0;x\ne4\)
Khi đó:
\(M=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+2\sqrt{x}+\sqrt{x}+2}{x-4}+\dfrac{2x-4\sqrt{x}}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\\ =\dfrac{3x-6\sqrt{x}}{x-4}\\ =\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Với \(a>0;a\ne1;a\ne4\)
Khi đó:
\(N=(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\dfrac{\sqrt{a}}{a-\sqrt{a}}-\dfrac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\dfrac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\dfrac{1}{a-\sqrt{a}}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\\ =\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right).3}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Phần a,b,c bạn có thể tham khảo bài bên dưới.
Phần d.
ĐKXĐ: $x\geq 0; x\neq 4$
$A>5\Leftrightarrow \frac{x+9}{2\sqrt{x}}>5$ ($x> 0$)
$\Leftrightarrow x+9> 10\sqrt{x}$
$\Leftrightarrow x-10\sqrt{x}+9>0$
$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-9)>0$
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} \sqrt{x}-1>0\\ \sqrt{x}-9>0\end{matrix}\right.\\ \left\{\begin{matrix} \sqrt{x}-1<0\\ \sqrt{x}-9<0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x>1\\ x>81\end{matrix}\right.\\ \left\{\begin{matrix} 0\leq x< 1\\ 0\leq x< 81\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x>81\\ 0\leq x< 1\end{matrix}\right.\)
Kết hợp với đkxđ suy ra $x>81$ hoặc $0< x< 1$
a
Với: x \(\ge0,x\) \(\ne4\) có:
\(A=\left(\dfrac{x-\sqrt{x}+7}{x-4}+\dfrac{\sqrt{x}+2}{x-4}\right):\left(\dfrac{\left(\sqrt{x}+2\right)^2}{x-4}-\dfrac{\left(\sqrt{x}-2\right)^2}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4}{x-4}-\dfrac{x-4\sqrt{x}+4}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{2\sqrt{x}}{x-4}\right)\)
\(=\dfrac{\left(x+9\right)\left(x-4\right)}{2\sqrt{x}\left(x-4\right)}=\dfrac{x+9}{2\sqrt{x}}\)
b
Giải \(x^2-5x+4=0\)
Nhẩm nghiệm: a + b + c = 0 (1 - 5 + 4 = 0)
\(\Rightarrow x_1=1;x_2=\dfrac{c}{a}=\dfrac{4}{1}=4\)
Thay x = 1 vào A:
\(A=\dfrac{1+9}{2\sqrt{1}}=\dfrac{10}{2}=5\)
Thay x = 4 vào A:
\(A=\dfrac{4+9}{2.\sqrt{4}}=\dfrac{13}{2.2}=\dfrac{13}{4}\)
c
ĐK: x > 0
\(A=0\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}=0\)
=> \(x+9=0\Rightarrow x=-9\) (không thỏa mãn)
Vậy không xác định được giá trị x
d
ĐK: x > 0
\(A>5\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}>5\)
\(\Leftrightarrow x+9>5.2\sqrt{x}\Leftrightarrow x+9>10\sqrt{x}\)
\(\Leftrightarrow\left(x+9\right)^2>\left(10\sqrt{x}\right)^2=100x\)
<=> \(x^2+18x+81-100x>0\)
<=> \(x^2-82x+81>0\)
<=> \(x^2-81x-x+81>0\)
<=> \(x\left(x-81\right)-\left(x-81\right)>0\)
<=> \(\left(x-1\right)\left(x-81\right)>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-1>0\\x-81>0\end{matrix}\right.\\\left[{}\begin{matrix}x-1< 0\\x-81< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x>81\end{matrix}\right.\\\left[{}\begin{matrix}x< 1\\x< 81\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>81\\x< 81\end{matrix}\right.\)
Vậy để A > 5 thì x > 81 và 0 < x < 81
a/ \(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{6}=\dfrac{-1}{\sqrt{x}-2}\)
b/ \(A>0\Leftrightarrow\dfrac{-1}{\sqrt{x}-2}>0\)
Ta thấy: - 1 < 0 nên để A > 0 thì:
\(\sqrt{x}-2< 0\)\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
kết hợp với đkxđ: => \(0\le x< 4\)
\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
a: \(A=\dfrac{x\sqrt{2}}{x\sqrt{2}\left(\sqrt{x}+\sqrt{2}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{x-2}\)
\(=\dfrac{1}{\sqrt{x}+\sqrt{2}}+\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2}}=\dfrac{\sqrt{x}+1}{\sqrt{x}+\sqrt{2}}\)
b: \(M=\left(\dfrac{\sqrt{a}+a}{\sqrt{a}-2}\right)\cdot\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\left(\sqrt{a}-2\right)=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(A=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\)
\(=\dfrac{\sqrt{x}.\sqrt{2x}}{\sqrt{2x}\left(\sqrt{x}+\sqrt{2}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2}}+\dfrac{\sqrt{2}}{\sqrt{x}+\sqrt{2}}=1\)
\(M=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{\sqrt{a}+a}{\sqrt{a}-2}.\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-2}.\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)