GIÚP EM VS Ạ
BÀI 6 Tìm giá trị nhỏ nhất của biểu thức:
A = (x − 1)(x + 2)(x + 3)(x + 6)
B = x^2 − 2x + y^2 + 4y + 2 − 8y
C = x^2 − 4x + y^2 − 8y + 6
Bài 7. Tìm giá trị nhỏ nhất và giá trị lớn nhất của các biểu thức (nếu có):
A = x^2 – 4x + 1
B = 4x^2 + 4x + 11
C = x^2 + 4x + 8
D = 7 – 8x + x^2
E = x(x – 6)
F = (x – 3)2 + (x – 11)2
G = (x –1)(x + 3)(x + 2)(x + 6)
H = (x + 1)(x – 2)(x – 3)(x – 6)
I = 5 – 8x – x^2
J = 4x – x^2 +1
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
6,\\
a,\\
A = \left( {x – 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 6} \right)\\
= \left[ {\left( {x – 1} \right)\left( {x + 6} \right)} \right].\left[ {\left( {x + 2} \right).\left( {x + 3} \right)} \right]\\
= \left( {{x^2} + 5x – 6} \right).\left( {{x^2} + 5x + 6} \right)\\
= \left[ {\left( {{x^2} + 5x} \right) – 6} \right].\left[ {\left( {{x^2} + 5x} \right) + 6} \right]\\
= {\left( {{x^2} + 5x} \right)^2} – {6^2} = {\left( {{x^2} + 5x} \right)^2} – 36 \ge – 36,\,\,\,\forall x\\
\Rightarrow {A_{\min }} = – 36 \Leftrightarrow {\left( {{x^2} + 5x} \right)^2} = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = – 5
\end{array} \right.\\
b,\\
B = {x^2} – 2x + {y^2} + 4y + 2 – 8y\\
= {x^2} + {y^2} – 2x – 4y + 2\\
= \left( {{x^2} – 2x + 1} \right) + \left( {{y^2} – 4y + 4} \right) – 3\\
= {\left( {x – 1} \right)^2} + {\left( {y – 2} \right)^2} – 3 \ge – 3,\,\,\,\forall x,y\\
\Rightarrow {B_{\min }} = – 3 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – 1} \right)^2} = 0\\
{\left( {y – 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
c,\\
C = {x^2} – 4x + {y^2} – 8y + 6\\
= \left( {{x^2} – 4x + 4} \right) + \left( {{y^2} – 8y + 16} \right) – 14\\
= {\left( {x – 2} \right)^2} + {\left( {y – 4} \right)^2} – 14 \ge – 14,\,\,\,\forall x,y\\
\Rightarrow {C_{\min }} = – 14 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – 2} \right)^2} = 0\\
{\left( {y – 4} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 4
\end{array} \right.\\
7,\\
A = {x^2} – 4x + 1 = \left( {{x^2} – 4x + 4} \right) – 3 = {\left( {x – 2} \right)^2} – 3 \ge – 3\\
\Rightarrow {A_{\min }} = – 3 \Leftrightarrow {\left( {x – 2} \right)^2} = 0 \Leftrightarrow x = 2\\
B = 4{x^2} + 4x + 11 = \left( {4{x^2} + 4x + 1} \right) + 10 = {\left( {2x + 1} \right)^2} + 10 \ge 10\\
\Rightarrow {B_{\min }} = 10 \Leftrightarrow {\left( {2x + 1} \right)^2} = 0 \Leftrightarrow x = – \frac{1}{2}\\
C = {x^2} + 4x + 8 = \left( {{x^2} + 4x + 4} \right) + 4 = {\left( {x + 2} \right)^2} + 4 \ge 4,\,\,\,\forall x\\
\Rightarrow {C_{\min }} = 4 \Leftrightarrow {\left( {x + 2} \right)^2} = 0 \Leftrightarrow x = – 2\\
D = 7 – 8x + {x^2} = \left( {{x^2} – 8x + 16} \right) – 9 = {\left( {x – 4} \right)^2} – 9 \ge – 9,\,\,\,\forall x\\
\Rightarrow {D_{\min }} = – 9 \Leftrightarrow {\left( {x – 4} \right)^2} = 0 \Leftrightarrow x = 4\\
E = x\left( {x – 6} \right) = {x^2} – 6x = \left( {{x^2} – 6x + 9} \right) – 9 = {\left( {x – 3} \right)^2} – 9 \ge – 9,\,\,\forall x\\
\Rightarrow {E_{\min }} = – 9 \Leftrightarrow {\left( {x – 3} \right)^2} = 0 \Leftrightarrow x = 3\\
F = {\left( {x – 3} \right)^2} + {\left( {x – 11} \right)^2} = \left( {{x^2} – 6x + 9} \right) + \left( {{x^2} – 22x + 121} \right)\\
= 2{x^2} – 28x + 130 = 2.\left( {{x^2} – 14x + 49} \right) + 22 = 2.{\left( {x – 7} \right)^2} + 22 \ge 22\\
\Rightarrow {F_{\min }} = 22 \Leftrightarrow {\left( {x – 7} \right)^2} = 0 \Leftrightarrow x = 7\\
G = \left( {x – 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 6} \right)\\
= \left[ {\left( {x – 1} \right)\left( {x + 6} \right)} \right].\left[ {\left( {x + 2} \right).\left( {x + 3} \right)} \right]\\
= \left( {{x^2} + 5x – 6} \right).\left( {{x^2} + 5x + 6} \right)\\
= \left[ {\left( {{x^2} + 5x} \right) – 6} \right].\left[ {\left( {{x^2} + 5x} \right) + 6} \right]\\
= {\left( {{x^2} + 5x} \right)^2} – {6^2} = {\left( {{x^2} + 5x} \right)^2} – 36 \ge – 36,\,\,\,\forall x\\
\Rightarrow {G_{\min }} = – 36 \Leftrightarrow {\left( {{x^2} + 5x} \right)^2} = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = – 5
\end{array} \right.\\
H = \left( {x + 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x – 6} \right)\\
= \left[ {\left( {x + 1} \right)\left( {x – 6} \right)} \right].\left[ {\left( {x – 2} \right)\left( {x – 3} \right)} \right]\\
= \left( {{x^2} – 5x – 6} \right).\left( {{x^2} – 5x + 6} \right)\\
= {\left( {{x^2} – 5x} \right)^2} – 36 \ge – 36,\,\,\,\forall x\\
\Rightarrow {H_{\min }} = – 36 \Leftrightarrow {\left( {{x^2} – 5x} \right)^2} = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 5
\end{array} \right.\\
I = 5 – 8x – {x^2} = – \left( {{x^2} + 8x + 16} \right) + 21 = 21 – {\left( {x + 4} \right)^2} \le 21,\,\,\,\forall x\\
\Rightarrow {I_{\max }} = 21 \Leftrightarrow {\left( {x + 4} \right)^2} = 0 \Leftrightarrow x = – 4\\
J = 4x – {x^2} + 1 = – \left( {{x^2} – 4x + 4} \right) + 5 = 5 – {\left( {x – 2} \right)^2} \le 5,\,\,\,\forall x\\
\Rightarrow {J_{\max }} = 5 \Leftrightarrow {\left( {x – 2} \right)^2} = 0 \Leftrightarrow x = 2
\end{array}\)