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

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

\(=-x-1\)

Với `a > 0,a \ne 1` có:

    `(1/[a-\sqrt{a}]+1/[\sqrt{a}-1]):[\sqrt{a}+1]/[a-2\sqrt{a}+1]`

`=[1+\sqrt{a}]/[\sqrt{a}(\sqrt{a}-1)].[(\sqrt{a}-1)^2]/[\sqrt{a}+1]`

`=[\sqrt{a}-1]/\sqrt{a}`

\(=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{1}{\sqrt{a}-1}\right)\times\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

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

\(=\dfrac{\sqrt{a}-1}{1}\)

Đk: \(a\ge0;a\ne1\)

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

a) Điều kiện: \(a\ge0;a\ne1\)

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

b) Để \(P< 7-4\sqrt{3}\Rightarrow P< \left(2-\sqrt{3}\right)^2\Rightarrow\left(a-1\right)^2< \left(2-\sqrt{3}\right)^2\)

\(\Rightarrow-\left(2-\sqrt{3}\right)< a-1< 2-\sqrt{3}\)

\(\Rightarrow\sqrt{3}-1< a< 3-\sqrt{3}\) (Thỏa mãn)

\(B=\sqrt{49x^2-22x+9}+\sqrt{49x^2+22x+9}\)

\(=\sqrt{\left[\left(7x\right)^2-2.7x.\dfrac{11}{7}+\dfrac{121}{49}\right]+\dfrac{320}{49}}+\sqrt{\left[\left(7x\right)^2+2.7x.\dfrac{11}{7}+\dfrac{121}{49}\right]+\dfrac{320}{49}}\)

\(=\)\(\sqrt{\left(\dfrac{11}{7}-7x\right)^2+\left(\dfrac{8\sqrt{5}}{7}\right)^2}+\sqrt{\left(7x+\dfrac{11}{7}\right)^2+\left(\dfrac{8\sqrt{5}}{7}\right)^2}\)(1)

Áp dụng BĐT Mincopxki, ta có:

\(\left(1\right)\ge\sqrt{\left(\dfrac{11}{7}-7x+7x+\dfrac{11}{7}\right)^2+\left(\dfrac{2.8\sqrt{5}}{7}\right)^2}\)

\(=\sqrt{36}=6\)

\(MinB=6\Leftrightarrow...\)

Ta có : \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\) (bđt Cauchy) (1) 

Tương tự được \(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(2\right);\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{xz}}\left(3\right)\)

Cộng (1) ; (2) ; (3) theo vế được 

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\)

\(\Leftrightarrow\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\) <=> ĐPCM

"=" khi x = y = z 

Áp dụng BĐT Cô si cho 2 số dương, ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}\ge2.\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}=\dfrac{2}{\sqrt{xy}}\)

Tương tự:

\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\)

\(\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{zx}}\)

\(\Rightarrow2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

Áp dụng BĐT \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\)

\(\Rightarrow1\ge\dfrac{1}{2}\left(x+y\right)^2\)

\(\Rightarrow2\ge\left(x+y\right)^2\)

\(\Rightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

Đầu tiên ta sẽ chứng minh BDT phụ \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

Thật vậy, ta có BDT luôn đúng: \(\left(a-b\right)^2\ge0\) \(\Leftrightarrow a^2+b^2\ge2ab\) \(\Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\) \(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

Vậy BDT phụ được chứng minh, dấu "=" xảy ra khi \(a=b\)

Áp dụng BĐT, ta được \(\left(x+y\right)^2\le2\left(x^2+y^2\right)=2\) (vì \(x^2+y^2=1\))

Từ đó \(-\sqrt{2}\le x+y\le\sqrt{2}\)

Dấu "=" xảy ra khi \(x=y\), kết hợp với giả thiết \(x^2+y^2=1\) thì ta có \(x=y=\dfrac{\sqrt{2}}{2}\)

\(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-5-x^2+6x\)

Ta có:

\(\sqrt{3x^2-18x+28}=\sqrt{3\left(x^2-6x+9\right)+1}=\sqrt{3\left(x-3\right)^2+1}\ge1\forall x\)

\(\sqrt{4x^2-24x+45}=\sqrt{4\left(x^2-6x+9\right)+9}=\sqrt{4\left(x-3\right)^3+9}\ge3x\forall\)

\(\Rightarrow VT\ge4\forall x\)\(\left(1\right)\)

Ta có:

\(VP=-5-x^2+6x=-\left(x^2-6x+9\right)+4=-\left(x-3\right)^2+4\le4\forall x\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\) Dấu "=" xảy ra \(\Leftrightarrow x=3\)

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

\(=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\sqrt{a}}=-8\)

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{9+2\sqrt{8}}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30.\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{43+30\sqrt{2}}\)

còn lại tự làm nốt