Bài 1:
a) cứ kiểu bị sai đề
b) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{y}+\sqrt{x}\right)}=\frac{\sqrt{x}}{\sqrt{y}}\)
c) \(\sqrt{a^2\left(a-2\right)^2}=a\left(a-2\right)\)
Bài 2:
a)\(\sqrt{19x}=15\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow19x=225\)
\(\Leftrightarrow x=\frac{225}{19}\)
b)\(\sqrt{4x^2}=8\)
\(\Leftrightarrow\left|2x\right|=8\) (1)
+)TH1: \(x\ge0\) thì pt(1) trở thành
\(\Leftrightarrow2x=8\Leftrightarrow x=4\) (ym)
+)TH2: \(x\le0\) thì pt(1) trở thành
\(\Leftrightarrow-2x=8\Leftrightarrow x=-4\) (tm)
Vậy x={-4;4}
c) \(\sqrt{4\left(x+1\right)}=\sqrt{8}\left(ĐK:x\ge-1\right)\)
\(\Leftrightarrow4\left(x+1\right)=8\)
\(\Leftrightarrow x+1=2\Leftrightarrow x=1\)
d) \(\sqrt{9\left(2-3x\right)^2}=6\)
\(\Leftrightarrow\left|3\left(2-3x\right)\right|=6\)
+)TH1:\(3\left(2-3x\right)\ge0\Leftrightarrow2-3x\ge0\Leftrightarrow-3x\ge-2\Leftrightarrow x\le\frac{2}{3}\) thì pt (2) trở thành
\(3\left(2-3x\right)=6\)
\(\Leftrightarrow2-3x=2\)
\(\Leftrightarrow x=0\)
\(\Leftrightarrow x=0\) (TM)
+)Th2:\(3\left(2-3x\right)\le0\Leftrightarrow2-3x\le0\Leftrightarrow x\ge\frac{2}{3}\) thì pt(2) trở thành
\(-3\left(2-3x\right)=6\)
\(\Leftrightarrow2-3x=-2\)
\(\Leftrightarrow-3x=-4\)
\(\Leftrightarrow x=\frac{4}{3}\) (tm)
Vậy \(x=\left\{0;\frac{4}{3}\right\}\)
