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

ĐKXĐ: \(x\ne\pm3\)

\(P=\left[\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)

\(=\left[\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)

\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}=\dfrac{x+3}{x^2+9}.\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)

Ý 2 mình k hiểu ý bạn lắm

\(P=\dfrac{x+3}{x-3}=\dfrac{x-3+6}{x-3}=1+\dfrac{6}{x-3}\in Z\)

\(\Leftrightarrow\left(x-3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Kết hợp vs ĐKXĐ \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)

CẦN GẤP M.N ƠI

haaaaaaaaaaaaaa

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)

=\(\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)

= \(\dfrac{3\cdot\left(x^2-2x+5\right)}{x^2-2x+5}+\dfrac{2}{x^2-2x+5}\)

= \(3+\dfrac{2}{x^2-2x+5}\)

= \(3+\dfrac{2}{x^2-2x+1+4}\)

= \(3+\dfrac{2}{\left(x-1\right)^2+4}\)

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)\(D=\dfrac{x^6+512}{x^2+8}\)

\(D=\dfrac{x^6+8x^4-8x^4-64x^2+64x^2+512}{x^2+8}\)

\(D=\dfrac{x^4\left(x^2+8\right)-8x^2\left(x^2+8\right)+64\left(x^2+8\right)}{x^2+8}\)

\(D=\dfrac{\left(x^2+8\right)\left(x^4-8x^2+64\right)}{x^2+8}\)

\(D=x^4-8x^2+64\)

\(D=\left(x^2-4\right)^2+48\ge48\)

\(\Rightarrow MIND=48\Leftrightarrow x=\pm2\)

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

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: \(x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

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 \(\Leftrightarrow\) \(x=\pm2\)