Em làm được những bài nào rồi và cần hỗ trợ bài nào em nhỉ?
Bài 2:
a: Ta có: \(\sqrt{16x-32}-\sqrt{4x-8}+\sqrt{9x-18}=1\)
\(\Leftrightarrow4\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=1\)
\(\Leftrightarrow5\sqrt{x-2}=1\)
\(\Leftrightarrow x-2=\dfrac{1}{25}\)
hay \(x=\dfrac{51}{25}\)
b: Ta có: \(\sqrt{8x-4}-12\cdot\sqrt{\dfrac{2x-1}{9}}+\sqrt{18x-9}=3\)
\(\Leftrightarrow2\sqrt{2x-1}-12\cdot\dfrac{\sqrt{2x-1}}{3}+3\sqrt{2x-1}=3\)
\(\Leftrightarrow2x-1=9\)
\(\Leftrightarrow2x=10\)
hay x=5
c: Ta có: \(\sqrt{9x^2-12x+4}=2\)
\(\Leftrightarrow\left|3x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=2\\3x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=4\\3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=0\end{matrix}\right.\)
d: Ta có: \(\sqrt{x^2+12x+36}=\left|-7\right|\)
\(\Leftrightarrow\left|x+6\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=7\\x+6=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-13\end{matrix}\right.\)
Bài 3:
Ta có: \(H=\left(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3a+3}{a-9}\right):\left(\dfrac{2\sqrt{a}-2}{\sqrt{a}-3}-1\right)\)
\(=\dfrac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}:\dfrac{2\sqrt{a}-2-\sqrt{a}+3}{\sqrt{a}-3}\)
\(=\dfrac{-3\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-3}{\sqrt{a}+1}\)
\(=\dfrac{-3}{\sqrt{a}+3}\)
