Bài 4: Tìm x biết.
1, (x+3) (x^2 -3x+9) -x(x^2-3) = 8(5-x)
2, (2x+1)^3 + (2x +3)^3 =0
Bài5: Tìm x để A nhỏ nhất, B lớn nhất. Tìm GTLN, GTNN
1, A= x^2 -2x +2
2, A= 4x^2 +4x +5
3, A= 2x^2 +3x +3
4,A= 3x^2 +5x
5, B= 2x -x^2 -4.
6, B= -x^2 -4x
7,B= 3x -2x^2 -2
8, B= x(3-x)
9, A= (x-1) (x+1) (x+2) (x+4)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
1,\\
\left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right) – x\left( {{x^2} – 3} \right) = 8\left( {5 – x} \right)\\
\Leftrightarrow \left( {{x^3} + {3^3}} \right) – \left( {{x^3} – 3x} \right) = 40 – 8x\\
\Leftrightarrow {x^3} + 27 – {x^3} + 3x = 40 – 8x\\
\Leftrightarrow 11x = 13\\
\Leftrightarrow x = \dfrac{{13}}{{11}}\\
2,\\
{\left( {2x + 1} \right)^3} + {\left( {2x + 3} \right)^3} = 0\\
\Leftrightarrow {\left( {2x + 1} \right)^3} = – {\left( {2x + 3} \right)^3}\\
\Leftrightarrow 2x + 1 = – \left( {2x + 3} \right)\\
\Leftrightarrow 4x = – 4\\
\Leftrightarrow x = – 1\\
5,\\
1,\\
A = {x^2} – 2x + 2 = \left( {{x^2} – 2x + 1} \right) + 1 = {\left( {x – 1} \right)^2} + 1 \ge 1,\,\,\forall x\\
\Rightarrow {A_{\min }} = 1 \Leftrightarrow {\left( {x – 1} \right)^2} = 0 \Leftrightarrow x = 1\\
2,\\
A = 4{x^2} + 4x + 5 = \left( {4{x^2} + 4x + 1} \right) + 4 = {\left( {2x + 1} \right)^2} + 4 \ge 4,\,\,\,\forall x\\
\Rightarrow {A_{\min }} = 4 \Leftrightarrow {\left( {2x + 1} \right)^2} = 0 \Leftrightarrow x = – \dfrac{1}{2}\\
3,\\
A = 2{x^2} + 3x + 3 = 2.\left( {{x^2} + \dfrac{3}{2}x + \dfrac{9}{{16}}} \right) + \dfrac{{15}}{8} = 2.{\left( {x + \dfrac{3}{4}} \right)^2} + \dfrac{{15}}{8} \ge \dfrac{{15}}{8},\,\,\,\forall x\\
\Rightarrow {A_{\min }} = \dfrac{{15}}{8} \Leftrightarrow {\left( {x + \dfrac{3}{4}} \right)^2} = 0 \Leftrightarrow x = – \dfrac{3}{4}\\
4,\\
A = 3{x^2} + 5x = 3.\left( {{x^2} + \dfrac{5}{3}x + \dfrac{{25}}{{36}}} \right) – \dfrac{{25}}{{12}} = 3.{\left( {x + \dfrac{5}{6}} \right)^2} – \dfrac{{25}}{{12}} \ge – \dfrac{{25}}{{12}},\,\,\,\forall x\\
\Rightarrow {A_{\min }} = – \dfrac{{25}}{{12}} \Leftrightarrow {\left( {x + \dfrac{5}{6}} \right)^2} = 0 \Leftrightarrow x = – \dfrac{5}{6}\\
5,\\
B = 2x – {x^2} – 4 = – \left( {{x^2} – 2x + 1} \right) – 3 = – {\left( {x – 1} \right)^2} – 3 \le – 3,\,\,\,\forall x\\
\Rightarrow {B_{\max }} = – 3 \Leftrightarrow {\left( {x – 1} \right)^2} = 0 \Leftrightarrow x = 1\\
6,\\
B = – {x^2} – 4x = – \left( {{x^2} + 4x + 4} \right) + 4 = 4 – {\left( {x + 2} \right)^2} \le 4,\,\,\,\forall x\\
\Rightarrow {B_{\max }} = 4 \Leftrightarrow {\left( {x + 2} \right)^2} = 0 \Leftrightarrow x = – 2\\
7,\\
B = 3x – 2{x^2} – 2 = – 2.\left( {{x^2} – \dfrac{3}{2}x + \dfrac{9}{{16}}} \right) – \dfrac{7}{8} = – \dfrac{7}{8} – 2{\left( {x – \dfrac{3}{4}} \right)^2} \le – \dfrac{7}{8},\forall x\\
\Rightarrow {B_{\max }} = – \dfrac{7}{8} \Leftrightarrow {\left( {x – \dfrac{3}{4}} \right)^2} = 0 \Leftrightarrow x = \dfrac{3}{4}\\
8,\\
B = x.\left( {3 – x} \right) = 3x – {x^2} = – \left( {{x^2} – 3x + \dfrac{9}{4}} \right) + \dfrac{9}{4} = \dfrac{9}{4} – {\left( {x – \dfrac{3}{2}} \right)^2} \le \dfrac{9}{4},\,\,\forall x\\
\Rightarrow {B_{\max }} = \dfrac{9}{4} \Leftrightarrow {\left( {x – \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{3}{2}\\
9,\\
A = \left( {x – 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 4} \right)\\
= \left[ {\left( {x – 1} \right)\left( {x + 4} \right)} \right].\left[ {\left( {x + 1} \right)\left( {x + 2} \right)} \right]\\
= \left( {{x^2} + 3x – 4} \right)\left( {{x^2} + 3x + 2} \right)\\
= \left[ {\left( {{x^2} + 3x – 1} \right) – 3} \right].\left[ {\left( {{x^2} + 3x – 1} \right) + 3} \right]\\
= {\left( {{x^2} + 3x – 1} \right)^2} – 9 \ge – 9,\,\,\,\forall x\\
\Rightarrow {A_{\min }} = – 9 \Leftrightarrow {\left( {{x^2} + 3x – 1} \right)^2} = 0 \Leftrightarrow {x^2} + 3x – 1 = 0 \Leftrightarrow x = \dfrac{{ – 3 \pm \sqrt {13} }}{2}
\end{array}\)