Tìm x a) 5(x +3) – 2x(3+x)=0 b) 4x (x-2004) – x + 2004 =0 c) (x + 1)^2 = x+1 09/08/2021 Bởi Eloise Tìm x a) 5(x +3) – 2x(3+x)=0 b) 4x (x-2004) – x + 2004 =0 c) (x + 1)^2 = x+1
a) `5(x +3) – 2x(3+x)=0` ⇔ `(5-2x)(x+3)=0` ⇔ \(\left[ \begin{array}{l}5-2x=0\\x+3=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=5/2\\x=-3\end{array} \right.\) b) `4x(x-2004)-x+2004=0` ⇔ `4x(x-2004)-(x-2004)=0` ⇔ `(4x-1)(x-2004)=0` ⇔ \(\left[ \begin{array}{l}4x-1=0\\x-2004=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=1/4\\x=2004\end{array} \right.\) c) `(x+1)^2 = x+1` ⇔ `(x+1)^2 – x-1 =0` ⇔ `(x+1)^2 -(x+1)=0` ⇔ `(x+1)(x+1-1)=0` ⇔ `(x+1)x=0` ⇔ \(\left[ \begin{array}{l}x+1=0\\x=\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-1\\x=0\end{array} \right.\) Bình luận
a) `5(x +3) – 2x(3+x)=0`
⇔ `(5-2x)(x+3)=0`
⇔ \(\left[ \begin{array}{l}5-2x=0\\x+3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=5/2\\x=-3\end{array} \right.\)
b) `4x(x-2004)-x+2004=0`
⇔ `4x(x-2004)-(x-2004)=0`
⇔ `(4x-1)(x-2004)=0`
⇔ \(\left[ \begin{array}{l}4x-1=0\\x-2004=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1/4\\x=2004\end{array} \right.\)
c) `(x+1)^2 = x+1`
⇔ `(x+1)^2 – x-1 =0`
⇔ `(x+1)^2 -(x+1)=0`
⇔ `(x+1)(x+1-1)=0`
⇔ `(x+1)x=0`
⇔ \(\left[ \begin{array}{l}x+1=0\\x=\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-1\\x=0\end{array} \right.\)
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