\(K=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\\ =\left(\dfrac{\sqrt{a}}{\sqrt{â}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\\ =\left(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\\ =\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\\ =\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}.\dfrac{\sqrt{a}+1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}=\dfrac{a-1}{\sqrt{a}}\)

\(C=\dfrac{4}{\sqrt{5}-1}-\dfrac{10}{\sqrt{5}}+\dfrac{\sqrt{125}}{\sqrt{5}}+\sqrt{2}\sqrt{\dfrac{5}{2}}\\ =\dfrac{4\left(\sqrt{5}+1\right)}{5-1}-\dfrac{\sqrt{5}.\sqrt{5}.2}{\sqrt{5}}+\sqrt{\dfrac{125}{5}}+\dfrac{\sqrt{2}.\sqrt{5}}{\sqrt{2}}\\ =\dfrac{4\left(\sqrt{5}+1\right)}{4}-2\sqrt{5}+\sqrt{25}+\sqrt{5}\\ =\sqrt{5}+1-2\sqrt[]{5}+5+\sqrt{5}=6\)

\(\dfrac{ab}{\sqrt{b}}-a\sqrt{\dfrac{b}{a}}+b\sqrt{\dfrac{a}{b}}-\dfrac{ab}{\sqrt{a}}\\ =\dfrac{ab}{\sqrt{b}}-\dfrac{a\sqrt{b}}{\sqrt{a}}+\dfrac{b\sqrt{a}}{\sqrt{b}}-\dfrac{ab}{\sqrt{a}}\\ =\dfrac{ab+b\sqrt{a}}{\sqrt{b}}-\dfrac{a\sqrt{b}+ab}{\sqrt{a}}\\ =\dfrac{\sqrt{b}\left(a\sqrt{b}+\sqrt{ab}\right)}{\sqrt{b}}-\dfrac{\sqrt{a}\left(\sqrt{ab}+\sqrt{a}b\right)}{\sqrt{a}}\\ =a\sqrt{b}+\sqrt{ab}-\sqrt{ab}+\sqrt{a}b=a\sqrt{b}+b\sqrt{a}\)

Bạn ấn vào"cập nhật " rồi sửa!

Bạn ấn vào"cập nhật " rồi sửa!

Áp dụng TCDTSBN. Ta có:

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y-z}{5+4-3}=\dfrac{18}{6}=3\)

=> \(\left\{{}\begin{matrix}\dfrac{x}{5}=3\Rightarrow x=15\\\dfrac{y}{4}=3\Rightarrow y=12\\\dfrac{z}{3}=3\Rightarrow z=9\end{matrix}\right.\)

\(e)3x(x-1)+x-1=0\\<=>3x(x-1)+(x-1)=0\\<=>(x-1)(3x+1)=0\\ <=>\left[\begin{matrix} x-1=0\\ 3x+1=0\end{matrix}\right.\\<=> \left[\begin{matrix} x=1\\ x=-1/3\end{matrix}\right.\)