Điều kiện: x>0

\(\frac{\sqrt{x}+1}{\sqrt{x}-1}>\frac{1}{2}\)

\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{1}{2}>0\)

\(\Leftrightarrow\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\frac{\sqrt{x}+1-\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\frac{2}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\frac{1}{\sqrt{x}-1}>0\)

mà 1>0

nên \(\sqrt{x}-1>0\)

\(\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)

\(ĐK:x\ge0;x\ne1\)

\(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

\(=\frac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

Với \(x\ge0;x\ne1\)

\(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(=\frac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

Đặt A = \(\sqrt{\frac{5}{2}-\sqrt{6}}-\sqrt{\frac{11}{2}-2\sqrt{6}}\)

\(\sqrt{2}A=\sqrt{5-2\sqrt{6}}-\sqrt{11-4\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{2}^2\right)}-\sqrt{11-2.2\sqrt{2}.\sqrt{3}}\)

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

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

\(\Rightarrow A=\frac{-3\sqrt{2}+2\sqrt{3}}{\sqrt{2}}=-3+\sqrt{6}\)

\(\sqrt{\frac{5}{2}-\sqrt{6}}-\sqrt{\frac{11}{2}-2\sqrt{6}}\)

\(=\frac{1}{\sqrt{2}}\left(\sqrt{3}-\sqrt{2}-2\sqrt{2}-\sqrt{11-4\sqrt{6}}\right)\)

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

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

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

\(=\sqrt{6}-3\)

a) Ta có tứ giác AIMJ là hcn=> AIMJ nội tiếp đường tròn đường kính AM,  IJ

Vì N đối xứng với M qua IJ => góc JNI = góc JMI = 90^o ha N thuộc đường tròn đường kính AM và IJ => góc ANM = 90^o 

mà I thuộc trung trực MN => tam giác MIC vuông cân tại I =>  I thuộc trung trực MC

=> I là tâm đường tròn ngoại tiếp tam giác MNC

=> góc MNC =1/2 góc MIC = 45⁰ 

=> góc ABC + góc ANC = 45+90+45=180⁰

Hay tứ giác ABCN nội tiếp đường tròn (T) (ĐPCM)

b)CM: 1/PM<1/PB+1/PC ?

Ta có: tam giác MPC đồng dạng tam giác MBA => PM/MB=PC/BA => PM/PC=MB/BA (1)

TAM GIÁC MBP đồng dạng tam giác MAC => PM/MC=PB/CA=> PM/PB=MC/AC      (2)

Cộng vế theo về của (1) và (2) ta có:

PM/PC+PM/PB=MB/BC+MC/AC=MB/BA+MC/BA=AC/BA>1 => ĐPCM

c) Áp dụng hệ thức giữa cạnh và đường cao ta có:

DH²=DK.DC => DA²=DK.DC

=> DA/DC=DK/DA => TAM GIÁC DKA đồng dạng tam giác DAC => góc AKD =DAC =45^o

=> góc ABH+ góc AKH = 45+45+90=180⁰=> TỨ GIÁC ABHK nội tiếp

=> Góc AKB =AHB =90 = GÓC HKC 

Mà góc ABK =AHK=KCH => đpcm

\(\sqrt{96}-6\sqrt{\frac{2}{3}}+\frac{3}{3+\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

\(=4\sqrt{6}-\frac{6\sqrt{2}}{\sqrt{3}}+\frac{3\left(3-\sqrt{6}\right)}{3}-\sqrt{6-2.2\sqrt{6}+4}\)

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

\(=\sqrt{6}+3-\sqrt{6}+2=5\)

a, Với \(x\ge0;x\ne\frac{16}{9};4\)

\(P=\frac{2\sqrt{x}-4}{3\sqrt{x}-4}-\frac{4+2\sqrt{x}}{\sqrt{x}-2}+\frac{x+13\sqrt{x}-20}{3x-10\sqrt{x}+8}\)

\(=\frac{2x-8\sqrt{x}+8-4\sqrt{x}-6x+16+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{-3x+\sqrt{x}+4}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}=\frac{-\left(3\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{2-\sqrt{x}}\)

b, \(P\ge-\frac{3}{4}\Rightarrow\frac{\sqrt{x}+1}{2-\sqrt{x}}+\frac{3}{4}\ge0\Leftrightarrow\frac{4\sqrt{x}+4+6-3\sqrt{x}}{8-4\sqrt{x}}\ge0\Leftrightarrow\frac{\sqrt{x}+10}{8-4\sqrt{x}}\ge0\)

\(\Rightarrow2-\sqrt{x}\ge0\Leftrightarrow x\le4\)Kết hợp với đk vậy \(0\le x< 4\)

\(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)\)

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

\(=\left(3-\sqrt{3}\right)\left(\sqrt{3}+3\right)=9-3=6\)

\(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}=3+\frac{1-\sqrt{3}+2}{1-\sqrt{3}}=3+1+\frac{2}{1-\sqrt{3}}=4+\frac{2}{1-\sqrt{3}}\)

\(=4+\frac{3-1}{1-\sqrt{3}}=4+\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{1-\sqrt{3}}=4-\sqrt{3}-1=-\sqrt{3}-3\)

\(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2=\frac{\sqrt{7}\left(\sqrt{3}+1\right)}{\sqrt{7}}+2=\left(\sqrt{3}+3\right)\)

Khi đó \(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)=-\left(\sqrt{3}+3\right)^2=-12-6\sqrt{3}\)