2:
a: x^2=9
=>x=3 hoặc x=-3
b: x^2=0,04
=>x^2=0,2^2
=>x=0,2 hoặc x=-0,2
c: x^2=7
=>\(x^2=\left(\sqrt{7}\right)^2\)
=>\(x=\pm\sqrt{7}\)
d: x^2=a
=>\(\left[{}\begin{matrix}x=\sqrt{a}\\x=-\sqrt{a}\end{matrix}\right.\)
e: x^2=4/9
=>x^2=(2/3)^2
=>x=2/3 hoặc x=-2/3
f: x^2-16/25=0
=>x^2=16/25
=>x^2=(4/5)^2
=>x=4/5 hoặc x=-4/5
Bài 1:
\(a,64=8^2=\left(-8\right)^2\\ b,0,09=0,3^2=\left(-0,3\right)^2\\ c,13=\left(\sqrt{13}\right)^2=\left(-\sqrt{13}\right)^2\\ d,x=\left(\sqrt{x}\right)^2=\left(-\sqrt{x}\right)^2\\ e,\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\\ f,\dfrac{49}{81}=\left(\dfrac{7}{9}\right)^2=\left(-\dfrac{7}{9}\right)^2\\ g,x^2=\left(-x\right)^2\\ h,m^4=\left(m^2\right)^2=\left(-m^2\right)^2\)
Vậy luôn có 2 cách viết một số dưới dạng bình phương của một số khác.
Bài 2:
\(a,x^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ b,x^2=0,04\\ \Leftrightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\end{matrix}\right.\\ c,x^2=7\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\\ d,x^2=a\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{a}\\x=-\sqrt{a}\end{matrix}\right.\\ e,x^2=\dfrac{4}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\\ f,x^2-\dfrac{16}{25}=0\\ \Leftrightarrow x^2=\dfrac{16}{25}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
