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ĐK : a >= 0 ; a khác 36
\(K=\left[\frac{a+14\sqrt{a}+100}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}+\frac{\left(\sqrt{a}+6\right)\left(\sqrt{a}-6\right)}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}-\frac{\left(\sqrt{a}-7\right)\left(\sqrt{a}+7\right)}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\right]\div\left(\frac{\sqrt{a}-6}{\sqrt{a}-6}-\frac{\sqrt{a}-7}{\sqrt{a}-6}\right)\)
\(=\frac{a+14\sqrt{a}+100+a-36-a+49}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\div\frac{1}{\sqrt{a}-6}\)
\(=\frac{a+14\sqrt{a}+113}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\cdot\left(\sqrt{a}-6\right)=\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}\)
Để K = 2 thì \(\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}=2\Rightarrow a+14\sqrt{a}+113=2\sqrt{a}+14\Leftrightarrow a+12\sqrt{a}+99=0\)
Với a >= 0 thì \(a+12\sqrt{a}+99\ge99>0\)=> Không có giá trị x thỏa mãn K = 2
Ta có : \(K=\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}=\frac{\left(a+14\sqrt{a}+49\right)+64}{\sqrt{a}+7}=\frac{\left(\sqrt{a}+7\right)^2+64}{\sqrt{a}+7}\)
\(=\left(\sqrt{a}+7\right)+\frac{64}{\sqrt{a}+7}\ge2\sqrt{\left(\sqrt{a}+7\right)\cdot\frac{64}{\sqrt{a}+7}}=16\)( bđt AM-GM )
Dấu "=" xảy ra <=> \(\sqrt{a}+7=\frac{64}{\sqrt{a}+7}\Rightarrow a=1\left(tm\right)\). Vậy MinK = 16
Bài 1:
a. \(\sqrt{\frac{25m^2}{49}}=\frac{\sqrt{25m^2}}{\sqrt{49}}=\frac{5m}{7}\)
b. \(\frac{\sqrt{192k}}{\sqrt{3k}}=\sqrt{\frac{192k}{3k}}=\sqrt{64}=8\)
Bài 2:
a. \(\frac{a+\sqrt{a}}{\sqrt{a}}=\frac{\left(\sqrt{a}\right)^2+\sqrt{a}}{\sqrt{a}}=\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}}=\sqrt{a}+1\)
b. \(\frac{\sqrt{a}-a}{\sqrt{a}-1}=\frac{\sqrt{a}-\left(\sqrt{a}\right)^2}{\sqrt{a}-1}=\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{\sqrt{a}-1}=\frac{-\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}=-\sqrt{a}\)
c. \(\frac{a-b}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^2-\left(\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=\sqrt{a}+\sqrt{b}\)
a) \(\frac{\sqrt{110}+\sqrt{70}}{\sqrt{22}+\sqrt{14}}=\frac{\left(\sqrt{11}+\sqrt{7}\right)\sqrt{10}}{\left(\sqrt{11}+\sqrt{7}\right)\sqrt{2}}=\sqrt{5}\)
b) \(\frac{\sqrt{42}-6}{\sqrt{21}-\sqrt{18}}=\frac{\sqrt{42}-\sqrt{36}}{\sqrt{21}-\sqrt{18}}\)
\(=\frac{\left(\sqrt{7}-\sqrt{6}\right)\sqrt{6}}{\left(\sqrt{7}-\sqrt{6}\right)\sqrt{3}}=\sqrt{2}\)
c) \(\frac{\left(a-b\right)\sqrt{a^2-b^2}}{\left(a-b\right)^2}\)
\(=\frac{\sqrt{\left(a-b\right)\left(a+b\right)}}{a-b}\)