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1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
C= x^6+27/x^4 - 3x^3 +6x^2 -9x + 9
= (x^2+3)(x^4-3x^2+9)/(x^4+3x^2)-(3x^3+9x)+(3x^2+9)
=(x^2+3)(x^4+6x^2+9-9x^2)/(x^2+3x)(x^2-3x+3)
= (x^2+3+3x)(x^2+3-3x)/x^2+3-3x =x^2+3x+3
=(x^2+3x+9/4) -9/4+3 = (x+3/2)^2 +3/4 >= 3/4
Dấu = xảy ra khi x=-3/2
Vậy Cmin = 3/4 <=> x=-3/2
Bạn ơi hai phân thức này chỉ tìm được min thôi nhé, không tìm được max đâu.Nếu tìm min thì như sau:\(C=\dfrac{x^6+27}{x^4-3x^3+6x^2-9x+9}=\dfrac{\left(x^2\right)^3+3^3}{x^4-3x^3+3x^2+3x^2-9x+9}=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^2\left(x^2-3x+3\right)+3\left(x^2-3x+3\right)}=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{\left(x^2+3\right)\left(x^2-3x+3\right)}=\dfrac{x^4-3x^2+9}{x^2-3x+3}\)\(C=\dfrac{x^4+6x^2+9-9x^2}{x^2-3x+3}=\dfrac{\left(x^2+3\right)^2-\left(3x\right)^2}{x^2-3x+3}=\dfrac{\left(x^2-3x+3\right)\left(x^2+3x+3\right)}{x^2-3x+3}=x^2+3x+3\)\(C=x^2+3x+3=x^2+2\times x\times\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{3}{4}\)
\(C=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu = xảy ra \(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=0\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy minC= 3/4 \(\Leftrightarrow\) x=-3/2
\(D=\dfrac{x^6+512}{x^2+8}=\dfrac{\left(x^2\right)^3+8^3}{x^2+8}=\dfrac{\left(x^2+8\right)\left(x^4-8x^2+64\right)}{x^2+8}\)
\(D=x^4-8x^2+64=x^4-8x^2+16+48\)
\(D=\left(x^2-4\right)^2+48\ge48\forall x\)
Dấu = xảy ra \(\Leftrightarrow\left(x^2-4\right)^2=0\Leftrightarrow x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
Vậy minD= 48 \(\Leftrightarrow\) \(x=\pm2\)
A = x2 +3x+3 min
<=>( x^2 +2x.3/2 + 9/4 ) -9/4 +3
<=> (x+3/2)^2 + 3/4 >= 3/4 ((x+3/2)^2>=0)
dấu "="xảy ra khi x=-3/2
vậy Pmin=3/4 khi x=-3/2