Toán | $x^{2}$ – 1|=2x+1 giải chi tiết nha tks 09/07/2021 By Anna | $x^{2}$ – 1|=2x+1 giải chi tiết nha tks
`|x^2-1|=2x+1\quad (x\ge -1/2)` `<=>`\(\left[ \begin{array}{l}x^2-1=2x+1\\x^2-1=-2x-1\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x^2-2x-2=0\\x^2+2x=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}(x-1)^2=3\\x(x+2)=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x-1=-\sqrt3\\x-1=\sqrt3\\x=0\\x+2=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=1-\sqrt3\ (KTM)\\x=1+\sqrt3\ (TM)\\x=0\ (TM)\\x=-2\ (KTM)\end{array} \right.\) Vậy PT có tập nghiệm `S={0;1+\sqrt3}` Trả lời
Đáp án: Giải thích các bước giải: `|x^2-1|=2x+1` `ĐK:2x+1>=0=>x>=-1/2` Ta có `|x^2-1|=2x+1` `<=>`\(\left[ \begin{array}{l}x^2-1=2x+1\\x^2-1=-2x-1\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x^2-2x-2=0\\x^2+2x=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}(x^2-2x+1)-3=0\\x(x+2)=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}(x-1)^2=3\\x(x+2)=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{array} \right.\\\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\ \end{array} \right.\) `<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{array} \right.\\\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\ \end{array} \right.\) Dựa vào điều kiện `x>=-1/2` `=>x in {\sqrt{3}+1;0}` Vậy `S= {\sqrt{3}+1;0}` Trả lời
`|x^2-1|=2x+1\quad (x\ge -1/2)`
`<=>`\(\left[ \begin{array}{l}x^2-1=2x+1\\x^2-1=-2x-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^2-2x-2=0\\x^2+2x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(x-1)^2=3\\x(x+2)=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x-1=-\sqrt3\\x-1=\sqrt3\\x=0\\x+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1-\sqrt3\ (KTM)\\x=1+\sqrt3\ (TM)\\x=0\ (TM)\\x=-2\ (KTM)\end{array} \right.\)
Vậy PT có tập nghiệm `S={0;1+\sqrt3}`
Đáp án:
Giải thích các bước giải:
`|x^2-1|=2x+1`
`ĐK:2x+1>=0=>x>=-1/2`
Ta có
`|x^2-1|=2x+1`
`<=>`\(\left[ \begin{array}{l}x^2-1=2x+1\\x^2-1=-2x-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^2-2x-2=0\\x^2+2x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(x^2-2x+1)-3=0\\x(x+2)=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(x-1)^2=3\\x(x+2)=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{array} \right.\\\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\ \end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{array} \right.\\\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\ \end{array} \right.\)
Dựa vào điều kiện `x>=-1/2`
`=>x in {\sqrt{3}+1;0}`
Vậy `S= {\sqrt{3}+1;0}`