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1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\frac{4}{9}\end{matrix}\right.\)
Ta có: \(Q=\frac{-5\sqrt{x}+4}{3\sqrt{x}-2}+\frac{6\sqrt{x}+4}{2\sqrt{x}+3}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)
\(=\frac{3\left(-5\sqrt{x}+4\right)\left(2\sqrt{x}+3\right)}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{3\left(6\sqrt{x}+4\right)\left(3\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}+\frac{29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{3\left(-10x-7\sqrt{x}+12\right)}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{3\left(18x-8\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}+\frac{29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{-30x-21\sqrt{x}+36+54x-24+29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{24x+8\sqrt{x}-16}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\left(3x+3\sqrt{x}-2\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\left(\sqrt{x}+1\right)\left(3\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}+8}{6\sqrt{x}+9}\)
2) Để \(Q>\frac{8}{3}\) thì \(Q-\frac{8}{3}>0\)
\(\Leftrightarrow\frac{8\sqrt{x}+8}{6\sqrt{x}+9}-\frac{8}{3}>0\)
\(\Leftrightarrow\frac{24\sqrt{x}+24}{3\left(6\sqrt{x}+9\right)}-\frac{8\left(6\sqrt{x}+9\right)}{3\left(6\sqrt{x}+9\right)}>0\)
\(\Leftrightarrow\frac{24\sqrt{x}+24-48\sqrt{x}-72}{9\left(2\sqrt{x}+3\right)}>0\)
mà \(9\left(2\sqrt{x}+3\right)>0\forall x\) thỏa mãn ĐKXĐ
nên \(-24\sqrt{x}-48>0\)
\(\Leftrightarrow-24\left(\sqrt{x}+2\right)>0\)
\(\Leftrightarrow\sqrt{x}+2< 0\)(Vô lý)
Vậy: Không có giá trị nào của x thỏa mãn \(Q>\frac{8}{3}\)
a)= \(\left(3+\sqrt{5}\right)\left(\sqrt{\left(3-\sqrt{5}\right)^2}\right)\)=\(\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=9-5=4\)
b)= \(\frac{2\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{\sqrt{2^2.7}}{2}-2\)=\(\frac{2\left(3-\sqrt{7}\right)}{9-7}+\sqrt{7}-2\)=1
c) =\(\frac{3}{3\left(\sqrt{7}-2\right)}-\frac{3}{3\left(\sqrt{7}+2\right)}\)=\(\frac{1}{\sqrt{7}-2}-\frac{1}{\sqrt{7}+2}=\frac{\sqrt{7}+2-\left(\sqrt{7}-2\right)}{\left(\sqrt{7}+2\right)\left(\sqrt{7}-2\right)}\)=\(\frac{4}{7-4}=\frac{4}{3}\)
d) =\(\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)^{ }\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{\left(88-44\sqrt{3}\right)}{25-3}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{22\left(4-2\sqrt{3}\right)}{22}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(1+\sqrt{3}\right)\left(\sqrt{3}-1\right)\)=3-1 = 2
e) = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{7\sqrt{x}-3}{x-9}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\)= \(\frac{x-4\sqrt{x}+3}{x-9}+\frac{7\sqrt{x}-3}{x-9}+\sqrt{x}\)= \(\frac{x+3\sqrt{x}}{x-9}+\sqrt{x}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)= \(\frac{\sqrt{x}}{\sqrt{x}-3}+\sqrt{x}=\frac{x-2\sqrt{x}}{\sqrt{x}-3}\)
4.a)\(x-2\sqrt{x}+3\)
\(=x-2\sqrt{x}+1+2\)
\(=\left(\sqrt{x}-1\right)^2+2\)
Vì \(\left(\sqrt{x}-1\right)^2\ge0,\forall x\)
\(\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow Min_{bt}=2\) khi \(\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
b)Ta có:
\(x-4\sqrt{y}+13\ge0\)
\(\Leftrightarrow x-4\sqrt{y}\ge-13\)
Dấu "=" xảy ra khi \(x-4\sqrt{y}=0\Leftrightarrow x=4\sqrt{y}\)
Vậy \(min_{bt}=0\) khi \(x=4\sqrt{y}\)
c)Ta có:
\(2x-4\sqrt{y}+6\ge0\)
\(\Leftrightarrow x-2\sqrt{y}+3\ge0\)
\(\Leftrightarrow x-2\sqrt{y}\ge-3\)
Dấu "=" xảy ra khi \(x-2\sqrt{y}=0\Leftrightarrow x=2\sqrt{y}\)
Vậy \(Min_{bt}=0\) khi \(x=2\sqrt{y}\)
d)Ta có:
\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\)
Vì \(\left(x+1\right)^2\ge0,\forall x\)
\(\Leftrightarrow\left(x+1\right)^2+4\ge4\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)^2+4}\le\frac{1}{4}\)
\(\Leftrightarrow-\frac{1}{\left(x+1\right)^2+4}\ge-\frac{1}{4}\)
\(\Leftrightarrow-\frac{4}{\left(x+1\right)^2+4}\ge-1\)
Vậy \(Min_{bt}=-1\) khi \(x+1=0\Leftrightarrow x=-1\)
M = \(\frac{2\sqrt{x}-9x}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{9-x+2x-3\sqrt{x}}{x-5\sqrt{x}+6}\)
=\(\frac{x-\sqrt{x}}{x-5\sqrt{x}+6}\)
a) \(A=\frac{-\sqrt{x}+2+4}{\sqrt{x}-2}=-1+\frac{4}{\sqrt{x}-2}\)
Để \(A\in Z\Leftrightarrow\sqrt{x}-2\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-2;0;1;3;4;6\right\}\)
Mà \(x\in Z;\sqrt{x}\ge0\Rightarrow x\in\left\{0;1;9;16;36\right\}\)
b)\(A=\frac{4\sqrt{x}-2+3}{2\sqrt{x}-1}=2+\frac{3}{2\sqrt{x}-1}\)
Để \(A\in Z\Leftrightarrow2\sqrt{x}-1\in\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow2\sqrt{x}\in\left\{-2;0;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-1;0;1;2\right\}\Leftrightarrow x\in\left\{0;1;4\right\}\)