Đk: \(-1\le a\le1\)

\(P=\frac{\sqrt{1+\sqrt{1-a^2}}\left(\sqrt{\left(1+a\right)^3}-\sqrt{\left(1-a\right)^3}\right)}{2+\sqrt{1-a^2}}\)

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

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

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

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

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

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

\(A=\frac{\sqrt{6+\sqrt{12}-\sqrt{8}-\sqrt{24}}}{\sqrt{2}+\sqrt{3}+1}\)

\(A=\frac{\sqrt{1+2+3+2\sqrt{3}-2\sqrt{2}-2\sqrt{2.3}}}{\sqrt{2}+\sqrt{3}+1}\)

\(A=\frac{\sqrt{\left(1-\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{3}+1}=\frac{1-\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}+1}\)

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

\(A=\frac{\sqrt{3}-\sqrt{2}-1-\sqrt{6}+2+\sqrt{2}+3-\sqrt{6}-\sqrt{3}}{3-\left(\sqrt{2}+1\right)^2}\)

\(A=\frac{4-2\sqrt{6}}{3-3-2\sqrt{2}}=\frac{4-2\sqrt{3}}{-2\sqrt{2}}=\frac{2\left(2-\sqrt{3}\right)}{-2\sqrt{2}}=\frac{\sqrt{3}-2}{\sqrt{2}}=\frac{\sqrt{6}-2\sqrt{2}}{2}\)

Đề sai rồi. Mũ 3 mới đúng

Ta có : \(\frac{A}{B}\ge\frac{x}{4}+5\Leftrightarrow\sqrt{x}+4\ge\frac{x}{4}+5\)

\(\Leftrightarrow\frac{4\sqrt{x}+16}{4}-\frac{x}{4}-\frac{20}{4}\ge0\Leftrightarrow\frac{4\sqrt{x}-x-4}{4}\ge0\)

\(\Rightarrow-x+4\sqrt{x}-4\ge0\Leftrightarrow x-4\sqrt{x}+4\le0\)vì 4 > 0 

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\Leftrightarrow x\le4\)

Kết hợp với đk vậy \(0\le x\le4;x\ne1\)

sửa bài ĐK :  x >= 0 

\(D=x+\sqrt{x}-1=x+\sqrt{x}+\frac{1}{4}-\frac{5}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{5}{4}\ge-1\)

Dấu ''='' xảy ra khi \(x=0\)

Vậy GTNN D là -1 khi x = 0 

ĐKXĐ : \(x\ge0\)

Ta có : \(D=x+\sqrt{x}-1\)

\(\ge0+\sqrt{0}-1=-1\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy \(D_{min}=-1\Leftrightarrow x=0\)

\(C=x+2\sqrt{x}-4=x+2\sqrt{x}+1-5\)

\(=\left(\sqrt{x}+1\right)^2-5\ge-4\)

Dấu ''='' xảy ra khi \(\sqrt{x}+1\ge1\Rightarrow x=0\)

Vậy GTNN C là -4 khi x = 0 

ĐKXĐ : \(x\ge0\)

Ta có : \(C=x+2\sqrt{x}-4\)

\(\ge0+2\sqrt{0}-4=-4\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy \(C_{min}=-4\Leftrightarrow x=0\)

a) Biểu thức A :

\(A=\sqrt{28}-\sqrt{63}+\frac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{7}.\sqrt{4}-\sqrt{7}.\sqrt{9}+\frac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=\sqrt{7}.2-\sqrt{7}.3+\sqrt{7}+1-\sqrt{7}-1\)(do \(\sqrt{7};1>0\))

\(=-\sqrt{7}\)

Biểu thức B :

ĐKXĐ : \(x\ge0;x\ne9\)

Ta có : \(B=\left(\frac{1}{\sqrt{x}+3}+\frac{1}{\sqrt{x}-3}\right).\frac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\frac{\sqrt{x}-3+\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)

\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)

\(=\frac{8}{\sqrt{x}-3}\)

a, \(A=\sqrt{28}-\sqrt{63}+\frac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)

\(B=\left(\frac{1}{\sqrt{x}+3}+\frac{1}{\sqrt{x}-3}\right)\frac{4\sqrt{x}+12}{\sqrt{x}}\)ĐK : \(x>0;x\ne9\)

\(=\left(\frac{\sqrt{x}-3+\sqrt{x}+3}{x-9}\right)\frac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}=\frac{8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}=\frac{8}{\sqrt{x}-3}\)

b, Ta có : \(A>B\Rightarrow-\sqrt{7}>\frac{8}{\sqrt{x}-3}\Rightarrow-\sqrt{7}>\frac{8}{\sqrt{x}-3}\)

tự giải bft này nhé