\(a,C=\frac{7}{\sqrt{x}+3}< 1\)

\(C=\frac{7}{\sqrt{x}+3}-1< 0\)

\(C=\frac{7-\sqrt{x}-3}{\sqrt{x}+3}< 0\)

\(C=\frac{4-\sqrt{x}}{\sqrt{x}+3}< 0\)

\(\sqrt{x}+3>0< =>4-\sqrt{x}< 0\)

\(\sqrt{x}>4\)

\(x>16\)

\(b,\sqrt{x}+2C=\sqrt{x}+\frac{14}{\sqrt{x}+3}\)

\(=\sqrt{x}+3+\frac{14}{\sqrt{x}+3}-3\)

\(\sqrt{x}+3+\frac{14}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{14}{\sqrt{x}+3}}=2\sqrt{14}\)

\(\sqrt{x}+2C\ge2\sqrt{14}-3\)dấu "=" xảy ra khi \(\sqrt{x}+3=\frac{14}{\sqrt{x}+3}\)

\(x+6\sqrt{x}+9=14\)

\(x+6\sqrt{x}-5=0\)

rồi bạn giải pt bậc 2 

\(< =>MIN=2\sqrt{14}-3\)

\(2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)=4\sqrt{18}-6+2\sqrt{3}\)

\(=12\sqrt{2}-6+2\sqrt{3}\)

\(11,ĐKXĐ:x\ge0;x\ne1\)

\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(B=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

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

\(B=\frac{\sqrt{x}-1}{\sqrt{x}}\)

\(b,P=9\sqrt{x}-B=9\sqrt{x}-\frac{\sqrt{x}-1}{\sqrt{x}}\)

\(P=9\sqrt{x}-1+\frac{1}{\sqrt{x}}\)

\(9\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{9\sqrt{x}.\frac{1}{\sqrt{x}}}=2\sqrt{9}=6\)( cô-si)

\(P\ge6-1=5\)dấu "=" xảy ra khi và chỉ khi \(x=\frac{1}{9}\)

\(< =>MIN:P=5\)

\(12,P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

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

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

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

\(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(b,P+\sqrt{x}=\frac{\sqrt{x}+1}{\sqrt{x}}+\sqrt{x}\)

xét\(\frac{\sqrt{x}+1}{\sqrt{x}}+\sqrt{x}-3\)

\(\frac{\sqrt{x}+1+x-3\sqrt{x}}{\sqrt{x}}\)

\(\frac{x-2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\)

mà dấu "=" xảy ra khi và chỉ khi \(x=1\left(KTM\right)\)

vậy dấu "=" ko xảy ra \(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

\(< =>P+\sqrt{x}>3\)

Bài 6.a.\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3.\sqrt{20}+9}}}\)

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)

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

b.\(\frac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\left(3+2\sqrt{2.3}+2\right)\sqrt{3-2\sqrt{2.3}+2}}{\sqrt{2}+\sqrt{3}}\)

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

Vậy ta có đpcm

a)  \(VT=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(1-\sqrt{5}\right)^2}}\)

\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

\(=\sqrt{1}\)

\(=1\left(dpcm\right)\)

\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\)

\(=\frac{\sqrt{6-2\sqrt{5}}}{2}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{2}\)

\(=\frac{\sqrt{5}-1}{2}\)

Đề sai không

\(P=\frac{\sqrt{x}+5}{3\sqrt{x}-1}=\frac{3\sqrt{x}+15}{3\sqrt{x}-1}=\frac{3\sqrt{x}-1+16}{3\sqrt{x}-1}=1+\frac{16}{3\sqrt{x}-1}\)

\(\Rightarrow3\sqrt{x}-1\inƯ\left(16\right)=\left\{\pm1;\pm2;\pm4;\pm8;\pm16\right\}\)