3.a.Ta có :\(\sqrt{40^2-24^2}=\sqrt{\left(40-24\right)\left(40+24\right)}=\sqrt{16.64}=4.8=32\)
b.Ta có :\(\sqrt{52^2-48^2}=\sqrt{\left(52-48\right)\left(52+48\right)}=\sqrt{4.100}=2.10=20\)
4.a)Ta có :
\(\sqrt{4x}=8\Leftrightarrow4x=8^2\Leftrightarrow4x=64\Leftrightarrow x=16\left(tm\right)\)
Vậy x=16
b)Ta có :
\(\sqrt{0,7x}=6\Leftrightarrow0,7x=36\Leftrightarrow x=\dfrac{36}{0.7}\left(tm\right)\)
Vậy x=\(\dfrac{36}{0.7}\)
c)Ta có:
\(9-4\sqrt{x}=1\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
Vậy x=4
d)Ta có :
\(\sqrt{5x}< 6\Leftrightarrow5x< 36\Leftrightarrow x< \dfrac{36}{5}\)
vậy 0≤x<\(\dfrac{36}{5}\)
Bài 3
a) \(\sqrt{40^2-24^2}\)
\(=\sqrt{\left(40+24\right)\left(40-24\right)}\)
=\(\sqrt{64.16}=\sqrt{64}.\sqrt{16}\)
\(=8.4=24\)
b)\(\sqrt{52^2-48^2}\)
\(=\sqrt{\left(52+48\right)\left(52-48\right)}\)
\(=\sqrt{100.4}=\sqrt{100}.\sqrt{4}\)
=10.2=20
Bài 4
a)\(\sqrt{4x}=8\)
\(\Leftrightarrow2\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)(TM)
b)\(\sqrt{0,7x}=6\)
\(\Leftrightarrow\sqrt{\left(0,7x\right)^2}=6^2\)
\(\Leftrightarrow\left|0,7x\right|=36\)
\(\Leftrightarrow\left[{}\begin{matrix}0,7x=36\\0,7x=-36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{360}{7}\left(TM\right)\\x=-\dfrac{360}{7}\left(KTM\right)\end{matrix}\right.\)
c)\(9-4\sqrt{x}=1\)
\(\Leftrightarrow4\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(TM)
d)\(\sqrt{5x}< 6\)