a: \(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}-2b}{a-b}\)

\(=\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)

\(=\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

a) \(\dfrac{a^2+3}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\ge2\sqrt{\sqrt{a^2+2}.\dfrac{1}{\sqrt{a^2+2}}}=2\)

Dấu = xảy ra khi \(\sqrt{a^2+2}=\dfrac{1}{\sqrt{a^2+2}}\Leftrightarrow a^2=-1\left(vn\right)\)

\(\Rightarrow\) Dấu "=" không xảy ra

Vậy \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

b)Với x,y>0,ta cm bđt phụ sau:

\(x^3+y^3\ge xy\left(x+y\right)\) (1)

Thật vậy (1)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)

\(\Leftrightarrow\cdot\left(x+y\right)\left(x^2-2xy+y^2\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) có:

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

Dấu "=" xra khi a=b

Vậy...

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

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

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

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

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

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

\(=\dfrac{2\left(\sqrt{6}-\sqrt{2}+\sqrt{3}-1+\sqrt{6}+\sqrt{2}-\sqrt{3}-1\right)}{\sqrt{3}\cdot2}\)

\(=\dfrac{2\left(2\sqrt{6}-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}-2}{\sqrt{3}}\)

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

\(\Leftrightarrow a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\)

\(\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2\)

\(\Leftrightarrow a^2+b^2-2ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

Vì (a-b)²\(\ge\)0 luôn đúng nên \(\sqrt{a^2+b^2}\ge\dfrac{a+b}{\sqrt{2}}\)

a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)

\(=a-1\)

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

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

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

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

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

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

\(a.P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Để : \(P\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\in Z\Leftrightarrow\left(\sqrt{x}+1\right)\in\left\{\pm1;\pm2\right\}\)

+) \(\sqrt{x}+1=1\Leftrightarrow x=0\left(TM\right)\)

+) \(\sqrt{x}+1=-1\Leftrightarrow vô-n^o\)

+) \(\sqrt{x}+1=2\Leftrightarrow x=1\left(KTM\right)\)

+) \(\sqrt{x}+1=-2\Leftrightarrow vô-n^o\)

KL.............

\(b.Q=\dfrac{\sqrt{a}+1}{\sqrt{a}+2}=\dfrac{\sqrt{a}+2-1}{\sqrt{a}+2}=1-\dfrac{1}{\sqrt{a}+2}\)

Để : \(Q\in Z\Leftrightarrow\dfrac{1}{\sqrt{a}+2}\in Z\Leftrightarrow\left(\sqrt{a}+2\right)\in\left\{\pm1\right\}\)

+) \(\sqrt{a}+2=1\Leftrightarrow vô-n^o\)

+) \(\sqrt{a}+2=-1\Leftrightarrow vô-n^o\)

KL............

\(c.A=\dfrac{\sqrt{a}-1}{\sqrt{a}-4}=\dfrac{\sqrt{a}-4+3}{\sqrt{a}-4}=1+\dfrac{3}{\sqrt{a}-4}\)

Để : \(A\in Z\Leftrightarrow\dfrac{3}{\sqrt{a}-4}\in Z\Leftrightarrow\left(\sqrt{a}-4\right)\in\left\{\pm1;\pm3\right\}\)

+) \(\sqrt{a}-4=1\Leftrightarrow a=25\left(TM\right)\)

+) \(\sqrt{a}-4=-1\Leftrightarrow a=9\left(TM\right)\)

+) \(\sqrt{a}-4=3\Leftrightarrow a=49\left(TM\right)\)

+) \(\sqrt{a}-4=-3\Leftrightarrow a=1\left(TM\right)\)

KL............

P/s : Mình thấy đề bài b sai nhé , mẫu phải là \(\sqrt{a}-2\) thì mới phù hợp ĐK đã cho .

\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\left(a,b>0;a\ne b\right)\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

Tick plz

Ta có: \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)

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

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

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

\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

b) Ta có: \(4x^2+x-5=0\)

\(\Leftrightarrow4x^2-4x+5x-5=0\)

\(\Leftrightarrow4x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{5}{4}\left(loại\right)\end{matrix}\right.\)

Thay x=1 vào biểu thức \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\), ta được:

\(B=\dfrac{\sqrt{1}-1}{\sqrt{1}}=0\)

Vậy: Khi \(4x^2+x-5=0\) thì B=0