Tìm GTNN của
a)\(A=\frac{3x^2-6x+17}{x^2-2x+5}\)
b)\(C=\frac{x^6+27}{x^4-3x^3+6x^2-9x+9}\)
c)\(D=\frac{x^6+512}{x^2+8}\)
Tìm max của:
C = \(\frac{x^6+27}{x^4-3x^3+6x^2-9x+9}\)
D = \(\frac{x^6+512}{x^2+8}\)
tìm GTNN của
a, \(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)
b, \(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)
c, \(C=\dfrac{x^6+27}{x^4-3x^3+6x^2-9x+9}\)
d, \(D=\dfrac{x^6+512}{x^2+8}\)
\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)
= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)
=\(\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)
= \(\dfrac{3\cdot\left(x^2-2x+5\right)}{x^2-2x+5}+\dfrac{2}{x^2-2x+5}\)
= \(3+\dfrac{2}{x^2-2x+5}\)
= \(3+\dfrac{2}{x^2-2x+1+4}\)
= \(3+\dfrac{2}{\left(x-1\right)^2+4}\)
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 ....
Ta có:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)
\(B=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)
\(B=2-\dfrac{3}{x^2-8x+16+6}\)
\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)
\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)
d)\(D=\dfrac{x^6+512}{x^2+8}\)
\(D=\dfrac{x^6+8x^4-8x^4-64x^2+64x^2+512}{x^2+8}\)
\(D=\dfrac{x^4\left(x^2+8\right)-8x^2\left(x^2+8\right)+64\left(x^2+8\right)}{x^2+8}\)
\(D=\dfrac{\left(x^2+8\right)\left(x^4-8x^2+64\right)}{x^2+8}\)
\(D=x^4-8x^2+64\)
\(D=\left(x^2-4\right)^2+48\ge48\)
\(\Rightarrow MIND=48\Leftrightarrow x=\pm2\)
tìm GTNN
A= X^6+27/(X^4-3X^3+6X^2-9X+9)
B= X^6+512/(X^2+8)
C=27-12X/(X^2+9)
D=8X+3/(4X^2+1)
Tìm GTNN của
a. C= \(\dfrac{x^6+27}{x^4-3x^3+6x^2-9x+9}\)
b. D = \(\dfrac{x^6+512}{x^2+8}\)
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
Cho biểu thức \(M=\left(1-\frac{6-2x^3}{x^6-9}\right).\frac{4}{x^5+3x^2}:\left(\frac{6x^6-24}{x^9+6x^6+9x^3}:\left(\frac{3x^2}{2}+\frac{3}{x}\right)\right)\)
a/ Rút gọn M
b/ Tìm các giá trị nguyên của x để M đạt GTLN. Tìm GTLN đó
\(A=\frac{X^6+27}{X^4-3x^3+6x^2-9x+9}\)
\(B=\frac{X^6+512}{x^2+8}\)
Mình cần gấp mọi người giúp mình nha!!
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
Bài 2: Giải phương trình sau:
a)2x – 6 = 5x – 9
b) 4x2 – 6x = 0
c)\(\frac{4+3x}{3}=\frac{x^2+1}{x}\)
d) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
a, 2x-6=5x-9
=>2x-5x=-9+6
=>-3x=-3
=>x=1
b, 4x2-6x=0
=> 2x(2x-3)=0
=>x=0 hoặc x=\(\frac{3}{2}\)
c,
\(\frac{4+3x}{3}=\frac{x^2+1}{x}\\ =>\frac{4x+3x^2-3x^2-3}{x}=0\\ =>\frac{4x-3}{x}=0\\ =>4x-3=0\\ =>x=\frac{3}{4}\)
d,
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x}{9x^2-4}\\ =>\left(3x+2\right)^2-6\left(3x-2\right)=9x\\ =>9x^2+12x+4-18x-12-9x=0\\ =>9x^2-15x+16=0\\ =>x=..\)
Q= \(\left(x^3-1-\frac{7-x^3}{3+x^3}\right).\frac{4}{x^5+3x^2}:\left(\frac{6x^4-24}{x^9+6x^6+9x^3}.\frac{2x}{3x^3+6}\right)\)
Rút gọn
a) \(\left(\frac{4}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
b) \(\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
c) \(\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)