Tìm GTNN của:
a) \(\dfrac{x^6+27}{x^4-3x^3+6x^2-9x+9}\)
b) \(\dfrac{x^6+512}{x^2+8}\)
K
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BN
9 tháng 6 2018
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
KK
17 tháng 3 2018
\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)
= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)
=\(\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)
= \(\dfrac{3\cdot\left(x^2-2x+5\right)}{x^2-2x+5}+\dfrac{2}{x^2-2x+5}\)
= \(3+\dfrac{2}{x^2-2x+5}\)
= \(3+\dfrac{2}{x^2-2x+1+4}\)
= \(3+\dfrac{2}{\left(x-1\right)^2+4}\)
vì (x-1)2 ≥ 0 ∀ x
⇔ (x-1)2 +4 ≥ 4
⇔\(\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{1}{2}\)
⇔\(3+\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{7}{2}\)
⇔ A \(\le\dfrac{7}{2}\)
⇔ Min A =\(\dfrac{7}{2}\)
khi x-1=0
⇔ x=1
vậy ....
17 tháng 3 2018
Ta có:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)
\(B=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)
\(B=2-\dfrac{3}{x^2-8x+16+6}\)
\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)
\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)
TH
2
AL
1
1 tháng 5 2017
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 \(\Leftrightarrow\) \(x=\pm2\)
26 tháng 10 2021
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
PT
0
4 tháng 10 2021
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) \(=\left(x+3\right)^3\)