Question

3. tính

a)\(\sqrt{40^2-24^2}\) b)\(\sqrt{52^2-48^2}\)

4 tìm số x ko âm ,biết

a)\(\sqrt{4x}=8\) b)\(\sqrt{0,7x}=6\) c)\(9-4\sqrt{x}=1\) d)\(\sqrt{5x}< 6\)

giup mik voi cac ban oi mai minh hoc roi

Answer 1

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<\(\dfrac{36}{5}\)

Answer 2

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)\(\sqrt{5x}< 6\)