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Làm hộ em nha mng cần gấp Lắm
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Adela
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`a)`
`D = RR`
`b)`
`x² – 5x – 6 ne 0`
`<=> x² + x – 6x – 6 ne 0`
`<=> (x + 1)(x – 6) ne 0`
`<=>` \(\left[ \begin{array}{l}x \ne -1\\x \ne 6\end{array} \right.\)
`=> D = RR` \ `{-1; 6}`
`c)`
`x² – 2x + 1 ne 0`
`<=> (x – 1)² ne 0`
`<=> x – 1 ne 0`
`<=> x ne 1`
`=> D = RR` \ `{1}`
`d)`
`x² – x + 1 ne 0`
`<=> x² – 2.(1)/(2).x + 1/4 + 3/4 ne 0`
`<=> (x – 1/2)² + 3/4 ne 0`
Vì `(x – 1/2)² + 3/4 > 0` với `∀ x ∈ RR`
`=> D = RR`
`e)`
\(\left\{ \begin{array}{l}x + 1 \ne 0\\1 – x ≥ 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x \ne -1\\x ≤ 1\end{array} \right.\)
`=> D = (-∞; 1]` \ `{-1}`
`f)`
\(\left\{ \begin{array}{l}x + 5 ≥ 0\\4 – 2x ≥ 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x ≥ -5\\x ≤ 2\end{array} \right.\)
`=> D = [-5; 2]`
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