a) \(4\sqrt{4x-8}-2\sqrt{9x-18}+\sqrt{16x-32}=5\)
\(\rightarrow4.2\sqrt{x-2}-2.3\sqrt{x-2}+4\sqrt{x-2}=5\)
\(\rightarrow\sqrt{x-2}\left(8-6+4\right)=5\)
\(\rightarrow6\sqrt{x-2}=5\)
\(\rightarrow\sqrt{x-2}=\frac{5}{6}\)
\(\rightarrow x-2=\frac{25}{36}\)
\(\Rightarrow x=\frac{97}{36}\)
b)\(\sqrt{x^2+6x+9}-2=7\)
\(\rightarrow\sqrt{\left(x+3\right)^2}=9\)
\(\rightarrow x+3=9\)
\(\Rightarrow x=6\)
Nhớ tick mik nha
a, \(\Leftrightarrow4\sqrt{4\left(x-2\right)}-2\sqrt{9\left(x-2\right)}+\sqrt{16\left(x-2\right)}=5\)
\(\Leftrightarrow8\sqrt{x-2}-6\sqrt{x-2}+4\sqrt{x-2}=5\)
\(\Leftrightarrow\sqrt{x-2}\left(8-6+4\right)=5\)
\(\Leftrightarrow\sqrt{x-2}=-1\)
\(\Leftrightarrow x=3\)
b, ĐKXĐ: \(x^2+6x+9\ge0\)
\(\Leftrightarrow\sqrt{x^2+6x+9}=9\\ \Leftrightarrow x^2+6x+9=81\\ \Leftrightarrow x^2+6x-72=0\\ \Leftrightarrow\left(x-6\right)\left(x+12\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-12\end{matrix}\right.\left(tm\right)\)
Vậy nghiệm của phương trình đã cho là 6, -12
a) \(\Leftrightarrow\) 8\(\sqrt{x-2}\) - 6\(\sqrt{x-2}\) +\(4\sqrt{x-2}\) = 5
\(\Leftrightarrow\) \(6\sqrt{x-2}\) = 5
\(\Leftrightarrow\) \(\sqrt{x-2}\) = \(\frac{5}{6}\)
\(\Leftrightarrow\) x - 2 = \(\frac{25}{36}\)
\(\Leftrightarrow\) x = \(\frac{25}{36}\) + 2 = \(\frac{97}{36}\)
