Toán Lớp 8: tìm GTNN:
a)P= $\frac{(4x+1)(4+x)}{x}$ với x>0
b)Q= ² $\frac{x2 +2x+1}{x+2}$ với x>-2
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Toán Lớp 8: tìm GTNN:
a)P= $\frac{(4x+1)(4+x)}{x}$ với x>0
b)Q= ² $\frac{x2 +2x+1}{x+2}$ với x>-2
Comments ( 2 )
a)\\
P = \dfrac{{\left( {4x + 1} \right)\left( {4 + x} \right)}}{x}\\
= \dfrac{{16x + 4{x^2} + 4 + x}}{x}\\
= \dfrac{{4{x^2} + 17x + 4}}{x}\\
= 4x + 17 + \dfrac{4}{x}\\
= 4x + \dfrac{4}{x} + 17\\
x > 0,Theo\,Co – si\\
\Leftrightarrow 4x + \dfrac{4}{x} \ge 2\sqrt {4x.\dfrac{4}{x}} = 8\\
\Leftrightarrow 4x + \dfrac{4}{x} + 17 \ge 8 + 17\\
\Leftrightarrow P \ge 25\\
\Leftrightarrow GTNN:P = 25\\
Khi:4x = \dfrac{4}{x}\\
\Leftrightarrow {x^2} = 1\\
\Leftrightarrow x = 1\\
b)Q = \dfrac{{{x^2} + 2x + 1}}{{x + 2}}\\
= \dfrac{{x\left( {x + 2} \right) + 1}}{{x + 2}}\\
= x + \dfrac{1}{{x + 2}}\\
= x + 2 + \dfrac{1}{{x + 2}} – 2\\
x > – 2 \Leftrightarrow x + 2 > 0\\
Theo\,Co – si:\\
x + 2 + \dfrac{1}{{x + 2}} \ge 2\sqrt {\left( {x + 2} \right).\dfrac{1}{{x + 2}}} = 2\\
\Leftrightarrow x + 2 + \dfrac{1}{{x + 2}} – 2 \ge 0\\
\Leftrightarrow Q \ge 0\\
\Leftrightarrow GTNN:Q = 0\\
Khi:x + 2 = \dfrac{1}{{x + 2}}\\
\Leftrightarrow x + 2 = 1\\
\Leftrightarrow x = – 1\left( {tmdk} \right)
\end{array}$