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ĐKXĐ: \(x\ge0\) và \(x\ne9\)
a/ \(\frac{x\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\frac{2\sqrt{x}-6}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{x\sqrt{x}-3-\left(2\sqrt{x}-6\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}+8\sqrt{x}-3x-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}\left(x+8\right)-3\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(x+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{x+8}{\sqrt{x}+1}\)
b/ Thay \(x=14-6\sqrt{5}\) vào P ta được:
\(P=\frac{14-6\sqrt{5}+8}{\sqrt{14-6\sqrt{5}}+1}=\frac{22-6\sqrt{5}}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}\)

1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)
Đạt được khi x = 9
2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)
\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)
Không có GTLN nhé

a) \(P=\frac{\left(x\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-\left(x+4\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{x+8}{\sqrt{x}+1}\)
b) Ta có \(x=14-6\sqrt{5}=9-2.3.\sqrt{5}+5=\left(3-\sqrt{5}\right)^2\)
Vậy nên \(\sqrt{x}=3-\sqrt{5}\)
Suy ra \(P=\frac{\left(3-\sqrt{5}\right)^2+8}{3-\sqrt{5}+1}=\frac{58-2\sqrt{5}}{11}\)
c) \(P=\frac{x+8}{\sqrt{x}+1}=\frac{\left(x-1\right)+9}{\sqrt{x}+1}=\left(\sqrt{x}-1\right)+\frac{9}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}+1\right)+\frac{9}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{9}{\sqrt{x}+1}}-2=4\)
minP = 4 khi \(\sqrt{x}+1=\frac{9}{\sqrt{x}+1}\Rightarrow\sqrt{x}+1=3\Rightarrow x=4.\)

1) \(A=\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=3-2\sqrt{2}\)
\(B=\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}-1+2-\sqrt{3}=1\)
\(C=\sqrt{63}-\sqrt{28}-\sqrt{7}=3\sqrt{7}-2\sqrt{7}-\sqrt{7}=0\)
\(D=\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}=\frac{4}{2}=2\)
\(M=\left(\frac{1}{3-\sqrt{5}}-\frac{1}{3+\sqrt{5}}\right):\frac{5-\sqrt{5}}{\sqrt{5}-1}=\frac{3+\sqrt{5}-3+\sqrt{5}}{9-5}.\frac{\sqrt{5}-1}{\sqrt{5}\left(\sqrt{5}-1\right)}=\frac{2}{4}=\frac{1}{2}\)

......................?
mik ko biết
mong bn thông cảm
nha ................
chơi delta là ra mà :
\(P=\frac{\sqrt{x-5}}{\sqrt{x+3}}\ge\frac{0}{\sqrt{x+3}}=0\)