1,|x^2-4x-5|=x^2-4x-5
2,x^4-3x^2-4=0
3,|x^2-4x-5|=0
4,(x+4)(x+1)-3√x^2+5x-2=6
5,|2x-5|=|2x^2-7x+5|
Làm hộ mình với ạ
1,|x^2-4x-5|=x^2-4x-5
2,x^4-3x^2-4=0
3,|x^2-4x-5|=0
4,(x+4)(x+1)-3√x^2+5x-2=6
5,|2x-5|=|2x^2-7x+5|
Làm hộ mình với ạ
Đáp án:
1) \(\left[ \begin{array}{l}
x = 5\\
x = – 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)|{x^2} – 4x – 5| = {x^2} – 4x – 5\\
\to \left[ \begin{array}{l}
{x^2} – 4x – 5 = {x^2} – 4x – 5\left( {ld} \right)\\
{x^2} – 4x – 5 = – {x^2} + 4x + 5\left( {DK: – 1 \le x \le 5} \right)
\end{array} \right.\\
\to 2{x^2} – 8x – 10 = 0\\
\to \left[ \begin{array}{l}
x = 5\\
x = – 1
\end{array} \right.\\
2){x^4} – 3{x^2} – 4 = 0\\
\to \left[ \begin{array}{l}
{x^2} = 4\\
{x^2} = – 1\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = – 2
\end{array} \right.\\
3)|{x^2} – 4x – 5| = 0\\
\to {x^2} – 4x – 5 = 0\\
\to \left[ \begin{array}{l}
x = 5\\
x = – 1
\end{array} \right.\\
4)\left( {x + 4} \right)\left( {x + 1} \right) – 3\sqrt {{x^2} + 5x – 2} = 6\\
\to \left( {{x^2} + 5x + 4} \right) – 3\sqrt {{x^2} + 5x – 2} = 6\\
Đặt:\sqrt {{x^2} + 5x – 2} = t\left( {t \ge 0} \right)\\
\to {x^2} + 5x – 2 = {t^2}\\
\to {x^2} + 5x = {t^2} + 2\\
Pt \to \left( {{t^2} + 6} \right) – 3t = 6\\
\to t\left( {t – 3} \right) = 0\\
\to \left[ \begin{array}{l}
t = 0\\
t = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + 5x – 2 = 0\\
{x^2} + 5x – 2 = 9
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{ – 5 + \sqrt {33} }}{2}\\
x = \dfrac{{ – 5 – \sqrt {33} }}{2}\\
x = \dfrac{{ – 5 + \sqrt {69} }}{2}\\
x = \dfrac{{ – 5 – \sqrt {69} }}{2}
\end{array} \right.\\
5)|2x – 5| = |2{x^2} – 7x + 5|\\
\to \left[ \begin{array}{l}
2x – 5 = 2{x^2} – 7x + 5\left( {DK:x \ge \dfrac{5}{2}} \right)\\
– 2x + 5 = 2{x^2} – 7x + 5\left( {DK:x < \dfrac{5}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = 2\left( l \right)\\
x = 0
\end{array} \right.
\end{array}\)