giải bài 1) thôi
1)
a) \(\left(2\sqrt{32}+3\sqrt{72}-7\sqrt{50}\right)\cdot\sqrt{2}\)
\(=\left(2\cdot4\sqrt{2}+3\cdot6\sqrt{2}-7\cdot5\sqrt{2}\right)\cdot\sqrt{2}\)
\(=\left(8\sqrt{2}+18\sqrt{2}-35\sqrt{2}\right)\cdot\sqrt{2}\)
\(=-9\sqrt{2}\cdot\sqrt{2}\)
\(=-9\cdot2\)
\(=-18\)
b) \(\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)}{\sqrt{6}}+\left|1-\sqrt{2}\right|\)
\(=\dfrac{\sqrt{2}+1}{\sqrt{2}}+\left(\sqrt{2}-1\right)\)
\(=\dfrac{\sqrt{2}\cdot\left(\sqrt{2}+1\right)}{\sqrt{2}\cdot\sqrt{2}}+\dfrac{2\left(\sqrt{2}-1\right)}{2}\)
\(=\dfrac{2+\sqrt{2}+2\sqrt{2}-2}{2}\)
\(=\dfrac{3\sqrt{2}}{2}\)
3. Giải các phương trình:
a) ĐKXĐ: \(3x-1\ge0\Leftrightarrow3x\ge1\Leftrightarrow x\ge\dfrac{1}{3}\)
\(2\sqrt{3x-1}+3=11\\ \Leftrightarrow2\sqrt{3x-1}=8\\ \Leftrightarrow4.\left|3x-1\right|=64\\ \Leftrightarrow\left|3x-1\right|=16\\ \Rightarrow\left[{}\begin{matrix}3x-1=16\\3x-1=-16\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=17\\3x=-15\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{3}\left(tm\right)\\x=-5\left(ktm\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{17}{3}\)
b) ĐKXĐ: \(x\ge0\)
Đặt \(\sqrt{x}=t\) \(\left(t\ge0\right)\)
Phương trình trở thành: \(t^2-7t=0\)
\(\Leftrightarrow t\left(t-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=0\\t-7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}t=0\left(tm\right)\\t=7\left(tm\right)\end{matrix}\right.\)
Với \(t=0\Rightarrow\sqrt{x}=0\Rightarrow x=0\) (tm)
Với t = 7 \(\Rightarrow\sqrt{x}=7\Rightarrow\left[{}\begin{matrix}x=-49\left(ktm\right)\\x=49\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;49\right\}\)
