## mng giúp em vs …………………………………………….

Question

mng giúp em vs …………………………………………….

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8 tháng 2020-10-29T14:50:55+00:00 1 Answers 58 views 0

## Answers ( )

1. Giải thích các bước giải:

Ta có:

$$\begin{array}{l} a,\\ 4x\left( {x + 1} \right) = 8.\left( {x + 1} \right)\\ \Leftrightarrow 4x.\left( {x + 1} \right) – 8.\left( {x + 1} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right).\left( {4x – 8} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = 0\\ 4x – 8 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = – 1\\ x = 2 \end{array} \right.\\ b,\\ x\left( {x – 1} \right) – 2\left( {1 – x} \right) = 0\\ \Leftrightarrow x\left( {x – 1} \right) + 2.\left( {x – 1} \right) = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x – 1 = 0\\ x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = – 2 \end{array} \right.\\ c,\\ 2x\left( {x – 2} \right) – {\left( {2 – x} \right)^2} = 0\\ \Leftrightarrow 2x\left( {x – 2} \right) – {\left( {x – 2} \right)^2} = 0\\ \Leftrightarrow \left( {x – 2} \right).\left[ {2x – \left( {x – 2} \right)} \right] = 0\\ \Leftrightarrow \left( {x – 2} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x – 2 = 0\\ x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = – 2 \end{array} \right.\\ d,\\ {\left( {x – 3} \right)^3} + 3 – x = 0\\ \Leftrightarrow {\left( {x – 3} \right)^3} – \left( {x – 3} \right) = 0\\ \Leftrightarrow \left( {x – 3} \right).\left[ {{{\left( {x – 3} \right)}^2} – 1} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} x – 3 = 0\\ {\left( {x – 3} \right)^2} – 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x – 3 = 0\\ x – 3 = 1\\ x – 3 = – 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 3\\ x = 4\\ x = 2 \end{array} \right.\\ e,\\ 5x.\left( {x – 2} \right) – \left( {2 – x} \right) = 0\\ \Leftrightarrow 5x.\left( {x – 2} \right) + \left( {x – 2} \right) = 0\\ \Leftrightarrow \left( {x – 2} \right).\left( {5x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x – 2 = 0\\ 5x + 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = – \frac{1}{5} \end{array} \right.\\ g,\\ 5x\left( {x – 2000} \right) – x + 2000 = 0\\ \Leftrightarrow 5x\left( {x – 2000} \right) – \left( {x – 2000} \right) = 0\\ \Leftrightarrow \left( {x – 2000} \right)\left( {5x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x – 2000 = 0\\ 5x – 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2000\\ x = \frac{1}{5} \end{array} \right.\\ h,\\ {x^2} – 4x = 0\\ \Leftrightarrow x.\left( {x – 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x – 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 4 \end{array} \right.\\ k,\\ {\left( {1 – x} \right)^2} – 1 + x = 0\\ \Leftrightarrow {\left( {1 – x} \right)^2} – \left( {1 – x} \right) = 0\\ \Leftrightarrow \left( {1 – x} \right).\left[ {\left( {1 – x} \right) – 1} \right] = 0\\ \Leftrightarrow \left( {1 – x} \right).\left( { – x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 1 – x = 0\\ – x = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 1 \end{array} \right.\\ m,\\ x + 6{x^2} = 0\\ \Leftrightarrow x.\left( {1 + 6x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 1 + 6x = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = – \frac{1}{6} \end{array} \right.\\ n,\\ \left( {x + 1} \right) = {\left( {x + 1} \right)^2}\\ \Leftrightarrow \left( {x + 1} \right) – {\left( {x + 1} \right)^2} = 0\\ \Leftrightarrow \left( {x + 1} \right).\left[ {1 – \left( {x + 1} \right)} \right] = 0\\ \Leftrightarrow \left( {x + 1} \right).\left( { – x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = 0\\ – x = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = – 1\\ x = 0 \end{array} \right. \end{array}$$