8: \(\left(2x-3\right)^2-49=0\)
=>(2x-3-7)(2x-3+7)=0
=>(2x-10)(2x+4)=0
=>\(4\left(x-5\right)\left(x+2\right)=0\)
=>(x-5)(x+2)=0
=>\(\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
9: \(\left(3x-1\right)^2-\left(x+5\right)^2=0\)
=>\(\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\)
=>(2x-6)(4x+4)=0
=>\(2\left(x-3\right)\cdot4\left(x+1\right)=0\)
=>(x-3)(x+1)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
10: \(\left(2x-1\right)^2-\left(3-x\right)^2=0\)
=>\(\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\)
=>(x+2)(3x-4)=0
=>\(\left[{}\begin{matrix}x+2=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\)
Bài 18.8
\[
(2x - 3)^2 - 49 = 0
\]
\[
(2x - 3)^2 - 7^2 = 0
\]
\[
(2x - 3 - 7)(2x - 3 + 7) = 0
\]
\[
(2x - 10)(2x + 4) = 0
\]
- \( 2x - 10 = 0 \Rightarrow 2x = 10 \Rightarrow x = 5 \)
- \( 2x + 4 = 0 \Rightarrow 2x = -4 \Rightarrow x = -2 \)
Vậy \( x = 5 \) hoặc \( x = -2 \).
Bài 18.9:
\(
(3x - 1)^2 - (x + 5)^2 = 0
\)
\[
[(3x - 1) - (x + 5)] \times [(3x - 1) + (x + 5)] = 0
\]
\[
(3x - 1 - x - 5)(3x - 1 + x + 5) = 0
\]
\[
(2x - 6)(4x + 4) = 0
\]
- \( 2x - 6 = 0 \Rightarrow 2x = 6 \Rightarrow x = 3 \)
- \( 4x + 4 = 0 \Rightarrow 4x = -4 \Rightarrow x = -1 \)
Vậy \( x = 3 \) hoặc \( x = -1 \).
Bài 18.10:
\[
(2x - 1)^2 - (3 - x)^2 = 0
\]
\[
[(2x - 1) - (3 - x)] \times [(2x - 1) + (3 - x)] = 0
\]
\[
(2x - 1 - 3 + x)(2x - 1 + 3 - x) = 0
\]
\[
(3x - 4)(x + 2) = 0
\]
- \( 3x - 4 = 0 \Rightarrow 3x = 4 \Rightarrow x = \frac{4}{3} \)
- \( x + 2 = 0 \Rightarrow x = -2 \)
Vậy \( x = \frac{4}{3} \) hoặc \( x = -2 \).
