a. (x-4)\(^2\)=x+1
⇔ x\(^2\) - 8x + 16 -x - 1 =0
⇔ x\(^2\) - 9x + 15 = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9+\sqrt{21}}{2}\\x=\frac{9-\sqrt{21}}{2}\end{matrix}\right.\)
b. 5.(x+3)+2x.(3+x)=0
⇔ (5+ 2x ) ( x + 3 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}5+2x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{2}\\x=-3\end{matrix}\right.\)
c. (x-4)\(^2\)-36=0
⇔ ( x - 4 - 6 ) ( x - 4 + 6 ) = 0
⇔ ( x - 10 ) ( x + 2 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-10=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d. (7x-4)\(^2\)-(2x+1)\(^2\)=0
⇔ ( 7x - 4 - 2x - 1 ) ( 7x - 4 + 2x + 1 ) = 0
⇔ ( 5x - 5 ) ( 9x - 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}5x-5=0\\9x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
a. (x-4)22=x+1
⇔ x22 - 8x + 16 -x - 1 =0
⇔ x22 - 9x + 15 = 0
⇔⎡⎣x=9+√212x=9−√212⇔[x=9+212x=9−212
b. 5.(x+3)+2x.(3+x)=0
⇔ (5+ 2x ) ( x + 3 ) =0
⇔[5+2x=0x+3=0⇔[x=−52x=−3⇔[5+2x=0x+3=0⇔[x=−52x=−3
c. (x-4)22-36=0
⇔ ( x - 4 - 6 ) ( x - 4 + 6 ) = 0
⇔ ( x - 10 ) ( x + 2 ) = 0
⇔[x−10=0x+2=0⇔[x=10x=−2⇔[x−10=0x+2=0⇔[x=10x=−2
d. (7x-4)22-(2x+1)22=0
⇔ ( 7x - 4 - 2x - 1 ) ( 7x - 4 + 2x + 1 ) = 0
⇔ ( 5x - 5 ) ( 9x - 3 ) = 0
⇔[5x−5=09x−3=0⇔[x=1x=13
