Tìm x biết: 1) 3x(2x-5) – 4(5-2x) = 0 2) (2x+3)^2 – (5x-2)^2 = 0 06/08/2021 Bởi Maya Tìm x biết: 1) 3x(2x-5) – 4(5-2x) = 0 2) (2x+3)^2 – (5x-2)^2 = 0
Đáp án: Giải thích các bước giải: 1)3x.(2x-5) -4.(5-2x)=0 ⇔3x.(2x-5) +4.(2x-5)=0 ⇔(2x-5).(3x+4)=0 ⇔\(\left[ \begin{array}{l}2x-5=0\\3x+4=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x =5\\3x=-4\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\frac{5}{2} \\x =\frac{-4}{3} \end{array} \right.\) b)(2x+3)² -(5x-2)² =0 ⇔(2x+3+5x-2).(2x+3 -5x +2)=0 ⇔(7x +1).(-3x +5)=0 ⇔\(\left[ \begin{array}{l}7x+1=0\\-3x +5=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}7x=-1\\3x=5\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x =\frac{-1}{7}\\x=\frac{5}{3}\end{array} \right.\) Bình luận
Đáp án:
Giải thích các bước giải:
1)3x.(2x-5) -4.(5-2x)=0
⇔3x.(2x-5) +4.(2x-5)=0
⇔(2x-5).(3x+4)=0
⇔\(\left[ \begin{array}{l}2x-5=0\\3x+4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x =5\\3x=-4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{5}{2} \\x =\frac{-4}{3} \end{array} \right.\)
b)(2x+3)² -(5x-2)² =0
⇔(2x+3+5x-2).(2x+3 -5x +2)=0
⇔(7x +1).(-3x +5)=0
⇔\(\left[ \begin{array}{l}7x+1=0\\-3x +5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}7x=-1\\3x=5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x =\frac{-1}{7}\\x=\frac{5}{3}\end{array} \right.\)
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