Giải bpt:
a, căn(2-x) – căn(7-x) + căn(-3-2x) >0
b, (x^2+3x+2).(-x+5) >= 0
Giải bpt: a, căn(2-x) – căn(7-x) + căn(-3-2x) >0 b, (x^2+3x+2).(-x+5) >= 0
By Nevaeh
By Nevaeh
Giải bpt:
a, căn(2-x) – căn(7-x) + căn(-3-2x) >0
b, (x^2+3x+2).(-x+5) >= 0
Đáp án:
b. \(x \in \left( { – \infty ; – 2} \right] \cup \left[ {1;5} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\sqrt {2 – x} – \sqrt {7 – x} + \sqrt { – 3 – 2x} > 0\\
\to \sqrt {2 – x} + \sqrt { – 3 – 2x} > \sqrt {7 – x} \\
\to 2 – x + – 3 – 2x + 2\sqrt {\left( {2 – x} \right)\left( { – 3 – 2x} \right)} > 7 – x\left( {DK:x \le – \dfrac{3}{2}} \right)\\
\to – 1 – 3x + 2\sqrt { – 6 – 4x + 3x + 2{x^2}} > 7 – x\\
\to 2\sqrt {2{x^2} – x – 6} > 8 + 2x\\
\to \sqrt {2{x^2} – x – 6} > 4 + x\\
\to 2{x^2} – x – 6 > 16 + 8x + {x^2}\left( {x \ge – 4} \right)\\
\to {x^2} – 9x – 22 > 0\\
\to \left( {x – 11} \right)\left( {x + 2} \right) > 0\\
\to x \in \left( { – 2;11} \right)\\
KL:x \in \left( { – 2; – \dfrac{3}{2}} \right]\\
b.\left( {{x^2} + 3x + 2} \right)\left( {5 – x} \right) \ge 0\\
\to \left( {x + 1} \right)\left( {x + 2} \right)\left( {5 – x} \right) \ge 0
\end{array}\)
BXD:
x -∞ -2 -1 5 +∞
f(x) + 0 – 0 + 0 –
\(KL:x \in \left( { – \infty ; – 2} \right] \cup \left[ {1;5} \right]\)