Tìm Min của
A=5x2-2x+7
C=x(x-1)(x-2)(x-3)+10
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\(VT=\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}.\left(a+b+c\right)\)
\(VT=\frac{a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2}{2}.\left(a+b+c\right)\)
\(VT=\frac{2a^2+2b^2+2c^2-2ab-2bc-2ca}{2}.\left(a+b+c\right)\)
\(VT=\frac{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}{2}.\left(a+b+c\right)\)
\(VT=\left(a^2+b^2+c^2-ab-bc-ca\right).\left(a+b+c\right)\)
\(VT=a^3+b^3+c^3-3abc=VP\left(đpcm\right)\)
Trả lời:
\(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\) \(\left(ĐK:a\ne b\ne c\right)\)
\(=\frac{1}{\left(a-b\right)\left(a-c\right)}-\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{b-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
Trả lời:
a, \(27a^2b^2-18ab+3=3\left(9a^2b^2-6ab+1\right)=3\left(3ab-1\right)^2\)
b, \(x^2+2xy+y^2-xz-yz\)
\(=\left(x^2+2xy+y^2\right)-z\left(x+y\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
c, \(a^4+a^3-a^2-a\)
\(=\left(a^4+a^3\right)-\left(a^2+a\right)\)
\(=a^3\left(a+1\right)-a\left(a+1\right)\)
\(=a\left(a+1\right)\left(a^2-1\right)\)
\(=a\left(a+1\right)\left(a-1\right)\left(a+1\right)\)
\(=a\left(a+1\right)^2\left(a-1\right)\)
d, \(a^3-b^3+2b-2a\)
\(=\left(a^3-b^3\right)-\left(2a-2b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)-2\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2-2\right)\)
a) 4x2 + 4xy + y2
= (2x + y)2
b) (2x + 1)2 - (x - 1)2
= (2x + 1 + x - 1)(2x + 1 - x + 1)
= 3x(x + 2)
c) 9 - 6x + x2 - y2
= (x2 - 6x + 9) - y2
= (x - 3)2 - y2
= (x - y - 3)(x + y - 3)
d) (-x - 2) + 3(x2 - 4)
= -(x + 2) + 3(x - 2)(x + 2)
= (x + 2)(3x - 7)
e) 5x2- 10xy2 + 5y4
= 5(x2 - 2xy2 + y4)
= 5(x - y2)2
f) \(\frac{x^4}{2}-2x^2=\frac{x^4-4x^2}{2}=\frac{x^2\left(x^2-4\right)}{2}=\frac{x^2\left(x-2\right)\left(x+2\right)}{2}\)
g) 49(x - 4)2 - 9(x + 2)2
= (7x - 28)2 - (3x + 6)2
= (10x - 22)(4x - 34)
h) (x2 + y2 - 5)2 - 2(xy + 2)2
= \(\left(x^2+y^2-5\right)^2-\left(\sqrt{2}xy+2\sqrt{2}\right)^2\)
\(=\left(x^2+y^2+2\sqrt{xy}+2\sqrt{2}-5\right)\left(x^2+y^2-\sqrt{2}xy-2\sqrt{2}-5\right)\)
\(A=5x^2-2x+7=5\left(x^2-\frac{2}{5}x+\frac{1}{25}-\frac{1}{25}\right)+7\)
\(=5\left(x-\frac{1}{5}\right)^2-\frac{1}{5}+7=5\left(x-\frac{1}{5}\right)^2+\frac{34}{5}\ge\frac{34}{5}\)
Dấu ''='' xảy ra khi x = 1/5
Vậy GTNN của A bằng 34/5 tại x = 1/5
\(C=x\left(x-1\right)\left(x-2\right)\left(x-3\right)+10\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)+10\)
Đặt \(x^2-3x=t\)
\(t\left(t+2\right)+10=t^2+2t+10=t^2+2t+1+9=\left(t+1\right)^2+9\ge9\)
Dấu ''='' xảy ra khi \(x^2-3x+1=0\Leftrightarrow x=\frac{3\pm\sqrt{5}}{2}\)
Vậy GTNN của C bằng 9 tại x = \(\frac{3\pm\sqrt{5}}{2}\)