Giải phương trình: a) x²-2.|x|-3=0 b) x ² – 2x + 3 – 3|x-1| =0 06/11/2021 Bởi Nevaeh Giải phương trình: a) x²-2.|x|-3=0 b) x ² – 2x + 3 – 3|x-1| =0
Đáp án + giải thích các bước giải: `x^2-2|x|-3=0` `->x^2+|x|-3|x|-3=0` `->|x|(|x|+1)-3(|x|+1)=0` `->(|x|+1)(|x|-3)=0` `->`\(\left[ \begin{array}{l}|x|+1=0\\|x|-3=0\end{array} \right.\) `->`\(\left[ \begin{array}{l}|x|=-1\\|x|=3\end{array} \right.\) `->x=3` b) `x^2-2x+3-3|x-1|=0` Với `x-1>=0` hay `x>=1`, có: `x^2-2x+3-3(x-1)=0` `->x^2-2x+3-3x+3=0` `->x^2-5x+6=0` `->x^2-2x-3x+6=0` `->x(x-2)-3(x-2)=0` `->(x-2)(x-3)=0` `->`\(\left[ \begin{array}{l}x=2\\x=3\end{array}(TM) \right.\) Với `x-1<0` hay `x<1`, có: `x^2-2x+3-3(1-x)=0` `->x^2-2x+3-3+3x=0` `->x^2+x=0` `->x(x+1)=0` `->`\(\left[ \begin{array}{l}x=0\\x=-1\end{array}(TM) \right.\) Bình luận
Đáp án: $\begin{array}{l}a){x^2} – 2.\left| x \right| – 3 = 0\\ \Rightarrow {\left( {\left| x \right|} \right)^2} – 2\left| x \right| – 3 = 0\\ \Rightarrow {\left( {\left| x \right|} \right)^2} – 3\left| x \right| + \left| x \right| – 3 = 0\\ \Rightarrow \left( {\left| x \right| – 3} \right)\left( {\left| x \right| + 1} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}\left| x \right| = 3\\\left| x \right| = – 1\left( {ktm} \right)\end{array} \right.\\ \Rightarrow x = \pm 3\\b){x^2} – 2x + 3 – 3\left| {x – 1} \right| = 0\\ \Rightarrow {\left( {x – 1} \right)^2} – 3\left| {x – 1} \right| + 2 = 0\\ \Rightarrow {\left( {\left| {x – 1} \right|} \right)^2} – 3\left| {x – 1} \right| + 2 = 0\\ \Rightarrow \left( {\left| {x – 1} \right| – 1} \right)\left( {\left| {x – 1} \right| – 2} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}\left| {x – 1} \right| = 1\\\left| {x – 1} \right| = 2\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x – 1 = 1\\x – 1 = – 1\\x – 1 = 2\\x – 1 = – 2\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = 2\\x = 0\\x = 3\\x = – 1\end{array} \right.\end{array}$ Bình luận
Đáp án + giải thích các bước giải:
`x^2-2|x|-3=0`
`->x^2+|x|-3|x|-3=0`
`->|x|(|x|+1)-3(|x|+1)=0`
`->(|x|+1)(|x|-3)=0`
`->`\(\left[ \begin{array}{l}|x|+1=0\\|x|-3=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}|x|=-1\\|x|=3\end{array} \right.\)
`->x=3`
b) `x^2-2x+3-3|x-1|=0`
Với `x-1>=0` hay `x>=1`, có:
`x^2-2x+3-3(x-1)=0`
`->x^2-2x+3-3x+3=0`
`->x^2-5x+6=0`
`->x^2-2x-3x+6=0`
`->x(x-2)-3(x-2)=0`
`->(x-2)(x-3)=0`
`->`\(\left[ \begin{array}{l}x=2\\x=3\end{array}(TM) \right.\)
Với `x-1<0` hay `x<1`, có:
`x^2-2x+3-3(1-x)=0`
`->x^2-2x+3-3+3x=0`
`->x^2+x=0`
`->x(x+1)=0`
`->`\(\left[ \begin{array}{l}x=0\\x=-1\end{array}(TM) \right.\)
Đáp án:
$\begin{array}{l}
a){x^2} – 2.\left| x \right| – 3 = 0\\
\Rightarrow {\left( {\left| x \right|} \right)^2} – 2\left| x \right| – 3 = 0\\
\Rightarrow {\left( {\left| x \right|} \right)^2} – 3\left| x \right| + \left| x \right| – 3 = 0\\
\Rightarrow \left( {\left| x \right| – 3} \right)\left( {\left| x \right| + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\left| x \right| = 3\\
\left| x \right| = – 1\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow x = \pm 3\\
b){x^2} – 2x + 3 – 3\left| {x – 1} \right| = 0\\
\Rightarrow {\left( {x – 1} \right)^2} – 3\left| {x – 1} \right| + 2 = 0\\
\Rightarrow {\left( {\left| {x – 1} \right|} \right)^2} – 3\left| {x – 1} \right| + 2 = 0\\
\Rightarrow \left( {\left| {x – 1} \right| – 1} \right)\left( {\left| {x – 1} \right| – 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\left| {x – 1} \right| = 1\\
\left| {x – 1} \right| = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x – 1 = 1\\
x – 1 = – 1\\
x – 1 = 2\\
x – 1 = – 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 0\\
x = 3\\
x = – 1
\end{array} \right.
\end{array}$