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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:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)

\(B=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)

\(B=2-\dfrac{3}{x^2-8x+16+6}\)

\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)

\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)

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

a: \(A=x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)

Dấu '=' xảy ra khi x=3/2

c: \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\)

=>\(\dfrac{3}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}< =3:\dfrac{7}{4}=\dfrac{12}{7}\)

=>C>=-12/7

Dấu '=' xảy ra khi x=1/2

10 tháng 12 2018

1.

a) \(x\left(x+4\right)+x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

b) \(x\left(x-3\right)+2x-6=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

10 tháng 12 2018

Bài 1:

a, \(x\left(x+4\right)+x+4=0\)

\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

Vậy \(x=-4\) hoặc \(x=-1\)

b, \(x\left(x-3\right)+2x-6=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x=3\) hoặc \(x=-2\)

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

1 tháng 5 2017

xin lỗi bạn nhé, mình viết nhầm đề ạ. Đúng là Tìm min bạn nhé. cảm ơn bạn !

27 tháng 11 2017

a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)

Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)

\((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)

\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)

\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4

Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4

23 tháng 8 2018

c/ đk: x khác 1; x khác -3

\(\dfrac{3x-1}{x-1}+\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)

\(\Rightarrow\left(3x+1\right)\left(x+3\right)+\left(2x+5\right)\left(x-1\right)+4=x^2+2x-3\)

\(\Leftrightarrow3x^2+10x+3+2x^2+3x-5+4=x^2+2x-3\)

\(\Leftrightarrow4x^2+11x+5=0\)

\(\Leftrightarrow\left(4x^2+2\cdot2x\cdot\dfrac{11}{4}+\dfrac{121}{16}\right)-\dfrac{41}{16}=0\)

\(\Leftrightarrow\left(2x+\dfrac{11}{4}\right)^2=\dfrac{41}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{11}{4}=\dfrac{\sqrt{41}}{4}\\2x+\dfrac{11}{4}=-\dfrac{\sqrt{41}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{41}}{8}\\x=\dfrac{-11-\sqrt{41}}{8}\end{matrix}\right.\)

Vậy.........

d/ \(\dfrac{12x+1}{6x-2}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(9x^2-1\right)}\)

đk: \(x\ne\pm\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{12x+1}{2\left(3x-1\right)}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)

\(\Rightarrow2\left(12x+1\right)\left(3x+1\right)-4\left(9x-5\right)\left(3x-1\right)=108x-36x^2-9\)

\(\Leftrightarrow72x^2+24x+6x+2-108x^2+36x-60x-20-108x+36x^2+9=0\)

\(\Leftrightarrow-102x-9=0\)

\(\Leftrightarrow-102x=9\Leftrightarrow x=-\dfrac{3}{34}\)(TM)

Vậy.........

23 tháng 8 2018

a/ \(\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\)

\(\Leftrightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\)

\(\Leftrightarrow2x\left(x^2+2x+1\right)=-24\)

\(\Leftrightarrow2x^3+4x^2+2x+24=0\)

\(\Leftrightarrow2x^3-2x^2+8x+6x^2-6x+24=0\)

\(\Leftrightarrow x\left(2x^2-2x+8\right)+3\left(2x^2-2x+8\right)=0\)

\(\Leftrightarrow\left(2x^2-2x+8\right)\left(x+3\right)=0\)

\(\Leftrightarrow2\left(x^2-x+4\right)\left(x+3\right)=0\)

Ta thấy: \(x^2-x+4=\left(x^2-2x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)

=> x+ 3 = 0 <=> x= -3

Vậy......

b/ \(2x^3+3x^2+6x+5=0\)

\(\Leftrightarrow2x^3+x^2+5x+2x^2+x+5=0\)

\(\Leftrightarrow x\left(2x^2+x+5\right)+\left(2x^2+x+5\right)=0\)

\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)

Ta thấy: \(2x^2+x+5=\left(\sqrt{2}x+2\cdot\sqrt{2}x\cdot\dfrac{\sqrt{2}}{4}+\dfrac{1}{8}\right)+\dfrac{39}{8}=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{4}\right)^2+\dfrac{39}{8}>0\)

=> x + 1 = 0 <=> x = -1

Vậy....