Đáp án:
i,$ (2x-1)²=49$
⇔\(\left[ \begin{array}{l}2x-1=7\\2x-1=-7\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x=8\\2x=-6\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\x=-3\end{array} \right.\)
k,$ (2x+7)²=9(x+2)²$
⇔\(\left[ \begin{array}{l}2x+7=3.(x+2)\\2x+7=-3.(x+2)\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x+7=3x+6\\2x+7=-3x-6\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=\frac{-13}{5}\end{array} \right.\)
j,$(5x-3)²-(4x-7)²=0$
⇔$(5x-3)²=(4x-7)²$
⇔\(\left[ \begin{array}{l}5x-3=4x-7\\5x-3=7-4x\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-4\\x=\frac{10}{9}\end{array} \right.\)
l,$4(2x+7)²=9.(x+3)²$
⇔\(\left[ \begin{array}{l}2.(x+7)=3.(x+3)\\2.(x+7)=-3.(x+3)\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x+14=3x+9\\2x+14=-3x-9\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=\frac{-23}{5}\end{array} \right.\)