câu a ở phần mẫu của cụm đầu tiên cái \(\left(\sqrt{a+\sqrt{b}}\right)^2\rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\) giúp em với ạ ( em cảm ơn )

a

\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{a+2\sqrt{ab}+b-4\sqrt{ab}}.\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}.\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2.\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}\)

\(M=\dfrac{a\sqrt{a}-b\sqrt{b}-a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\\ M=\dfrac{a\sqrt{b}+b\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ M=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)

\(\left(1-a\right)\left(1-b\right)+2\sqrt{ab}=1\\ \Leftrightarrow1-a-b+ab+2\sqrt{ab}=1\\ \Leftrightarrow a+b-ab-2\sqrt{ab}=0\\ \Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=ab\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-\sqrt{b}=\sqrt{ab}\\\sqrt{a}-\sqrt{b}=-\sqrt{ab}\end{matrix}\right.\)

Với \(\sqrt{a}-\sqrt{b}=\sqrt{ab}\Leftrightarrow M=\dfrac{\sqrt{ab}}{\sqrt{ab}}=1\)

Với \(\sqrt{a}-\sqrt{b}=-\sqrt{ab}\Leftrightarrow M=\dfrac{\sqrt{ab}}{-\sqrt{ab}}=-1\)

\(M=\dfrac{a\sqrt{a}-b\sqrt{b}-a\left(\sqrt{a}-\sqrt{b}\right)+b\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{a\sqrt{b}+b\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)

\(\left(1-a\right)\left(1-b\right)+2\sqrt{ab}=1\)

\(\Leftrightarrow a+b-ab-2\sqrt{ab}=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=ab\Leftrightarrow\sqrt{a}-\sqrt{b}=\sqrt{ab}\)

\(M=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}}{\sqrt{ab}}=1\)

\(A=x-4-\sqrt{x^4-8x^2+16}=x-4-\sqrt{[\left(x-2\right)\left(x+2\right)]^2}\)

\(A=x-4-\left(x-2\right)\left(x+2\right)=x-4-\left(x^2-4\right)=-x^2+x\)

\(B=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)=a-b\)

a) 

\(P=\left(\dfrac{b-a}{\sqrt{b}-\sqrt{a}}-\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}\right):\dfrac{\left(\sqrt{b}-\sqrt{a}\right)^2+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)

\(=\left[\sqrt{b}+\sqrt{a}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]:\dfrac{b-\sqrt{ab}+a}{\sqrt{a}+\sqrt{b}}\)

\(=\left(\sqrt{b}+\sqrt{a}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\right).\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(=\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)\(=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}\)

b) \(P=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\)

Vì \(\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b>0;\forall a\ge0;b\ge0;a\ne b\)

\(\sqrt{ab}\ge0\)\(\forall a\ge0;b\ge0\)

\(\Rightarrow P=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\ge0\)

Vậy...

cảm ơn tất cả mọi người

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

a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)

P = \(\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)

= \(\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)

= \(\dfrac{1}{a-\sqrt{ab}+b}\)

b) có a = 16 và b = 4 (thoả mãn ĐKXĐ)

Thay a = 16, b =4 vào P có:

P = \(\dfrac{1}{16-\sqrt{16.4}+4}\)= \(\dfrac{1}{12}\)

Vậy tại a =16, b = 4 thì P = \(\dfrac{1}{12}\)

\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)

\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)

\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)

Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)

\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)

a: \(P=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)

Với `a > 0,b >= 0` có:

`Bth=[a\sqrt{b}+b]/[a-b] . \sqrt{[b(a+b-2\sqrt{ab})]/[a^2+2a\sqrt{b}+b]} . (\sqrt{a}+\sqrt{b})`

 `=[\sqrt{b}(a+\sqrt{b})]/[a-b].\sqrt{[b(\sqrt{a}-\sqrt{b})^2]/[(a+\sqrt{b})^2]}.(\sqrt{a}+\sqrt{b})`

`=[\sqrt{b}(a+\sqrt{b})|\sqrt{a}-\sqrt{b}|.\sqrt{b}.(\sqrt{a}+\sqrt{b})]/[(a-b)(a+\sqrt{b})]`

`=[b|\sqrt{a}-\sqrt{b}|]/[\sqrt{a}-\sqrt{b}]`

`={(b\text{ nếu }\sqrt{a} >= \sqrt{b}),(-b\text{ nếu }\sqrt{a} < \sqrt{b}):}`