Cho 4x.9y=64 và 81x:92y=1. Hãy tính A=x18+y12+2014
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\(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
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a) \(4x^2+12xy+9y^2\)
\(=\left(2x\right)^2+2.2x.3+\left(3y\right)^2\)
\(=\left(2x+3y\right)^2\)
b) \(81x^2-18xy+y^2\)
\(=\left(9x\right)^2-2.9x.y+y^2\)
\(=\left(9x-y\right)^2\)
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a, \(6x^3y^2.\left(2-x\right)+9x^2y^2\left(x-2\right)\)
\(=6x^3y^2.\left(2-x\right)-9x^2y^2\left(2-x\right)\)
\(=y^2.\left(2-x\right)\left(6x^3-9x^2\right)\)
\(=3x^2y^2.\left(2-x\right)\left(2x-3\right)\)
b. \(x^2-4x+4y-y^2\)
\(=\left(x^2-y^2\right)-\left(4x-4y\right)\)
\(=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-4\right)\)
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a. 25 - \(x^2\) = (5-x) (5+x)
b) -196 + 4\(x^2\) = 196 - 4\(x^2\) = (14- 2x) (14+2x)
c)\(5^4-81x^4\) = \(\left[\left(5^2\right)^2\right]-\left[\left(81x^2\right)^2\right]\) = (\(\left(5^2-81x^2\right)\left(5^2+81x^2\right)\)
\(a,25-e=\left(5-\sqrt{e}\right)\left(5+\sqrt{e}\right)\)
\(b,-196+g=-\left(196-g\right)=-\left(14-\sqrt{g}\right)\left(14+\sqrt{g}\right)\)
\(c,2^6-47^2=\left(2^3\right)^2-47^2=\left(2^3-47\right)\left(2^3+47\right)\)
\(d,5^4-81x^4=\left(5^2\right)^2-\left(9x^2\right)^2=\left(5^2-9x^2\right)\left(5^2+9x^2\right)=\left(25-9x^2\right)\left(25+9x^2\right)\)
\(i,\dfrac{25}{16}-9y^2=\left(\dfrac{5}{4}-3y\right)\left(\dfrac{5}{4}+3y\right)\)
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\(A=\dfrac{4x^2\left(x+y\right)-9y^2\left(x+y\right)}{4x^2\left(x-y\right)-9y^2\left(x-y\right)}=\dfrac{\left(4x^2-9y^2\right)\left(x+y\right)}{\left(4x^2-9y^2\right)\left(x-y\right)}=\dfrac{x+y}{x-y}\)
Với x=2014, y=14:
\(A=\dfrac{2014+14}{2014-14}=\dfrac{2028}{2000}=\dfrac{507}{500}\)
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a) 4x2+4x+1
= (2x+1)2
b) x2-16x+64
= (x-8)2
c) 4x2-9y2
= (2x+3y)(2x-3y)
d) ( x-3).(x^2+3x+9)
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Ta có:
\(\left(4x+9y+16z\right)\left(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\right)\ge\left(\sqrt{\frac{4x}{x}}+\sqrt{\frac{9y.25}{y}}+\sqrt{\frac{16z.64}{z}}\right)^2\)
\(\Leftrightarrow49\left(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\right)\ge\left(2+15+32\right)^2\)
\(\Leftrightarrow\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\ge49\)
Dấu = xảy ra tại \(x=\frac{1}{2};y=\frac{5}{3};z=2\)
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\(81x^2-6yz-9y^2-z^2\)
\(=\left(9x\right)^2-\left(6yz+9y^2+z^2\right)\)
\(=\left(9x\right)^2-\left(z+3y\right)^2\)
\(=\left(9x-z-3y\right)\left(9x+z+3y\right)\)
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\(=81x^2-\left(z+3y\right)^2\)
\(=\left(9x+z+3y\right)\left(9x-z-3y\right)\)