Tìm C min biết:
C=x2 -15x+9
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a; \(x\) - \(\dfrac{3}{4}\) = \(\dfrac{1}{7}\)
\(x\) = \(\dfrac{1}{7}\) + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{4}{28}\) + \(\dfrac{21}{28}\)
\(x\) = \(\dfrac{25}{28}\)
Vậy \(x=\dfrac{25}{28}\)
c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)
d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)
e) \(x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
f) \(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
c: \(5x^2+15x+3y+xy\)
\(=5x\left(x+3\right)+y\left(x+3\right)\)
\(=\left(x+3\right)\left(5x+y\right)\)
d: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
e: \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right)\left(x+1+y\right)\)
f: \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-9\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
a)
x 2 − 2 5 = 1 10 x 2 = 1 10 + 2 5 x 2 = 1 10 + 4 10 x 2 = 1 2 ⇒ x = 1
b)
15 x − 1 3 = 28 51 15 x = 28 51 + 1 3 15 x = 28 51 + 17 51 15 x = 15 17 ⇒ x = 17
\(C=x^2-15x+9\)
\(=\left(x\right)^2-2.x.\frac{15}{2}+\left(\frac{15}{2}\right)^2-\frac{189}{4}\)
\(=\left(x-\frac{15}{2}\right)^2-\frac{189}{4}\le-\frac{189}{4}\)
Dấu "=" xảy ra khi : \(\left(x-\frac{15}{2}\right)=0\Rightarrow x=\frac{15}{2}\)
Vậy GTNN của C là \(-\frac{182}{4}\)khi x=\(\frac{15}{2}\)