\(C=\frac{x^2+5x+8}{x^2+2x+1}=\frac{x^2+2x+1+3x+3+4}{x^2+2x+1}\)

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

Đặt \(\frac{1}{x+1}=a\)\(\Rightarrow C=1+3a+4a^2\)

\(\Rightarrow C=4\left(a^2+\frac{3}{4}a+\frac{1}{4}\right)=4\left(a^2+2.\frac{3}{8}+\frac{9}{64}-\frac{9}{64}+\frac{1}{4}\right)\)

\(=4\left(a+\frac{3}{8}\right)^2+\frac{7}{16}\)

\(\Rightarrow C_{min}=\frac{7}{16}\Leftrightarrow\)\(a=-\frac{3}{8}\Leftrightarrow\frac{1}{x+1}=-\frac{3}{8}\)

\(\Rightarrow3\left(x+1\right)=-8\Rightarrow x=-\frac{11}{3}\)

Vậy \(C_{min}=\frac{16}{7}\Leftrightarrow x=-\frac{11}{3}\)

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

\(=5+3\left(3x-1\right)^2\ge5\)

\(Min=5\Leftrightarrow3x-1=0\Rightarrow x=\frac{1}{3}\)

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

Vì \(\left(2x-1\right)^2\ge0\) với mọi x

=>\(5+\left(2x-1\right)^2\ge5\)

Vậy GTNN của A là 5 khi x=1/2

ai làm được các bài nữa ko ạ. mình cần gấp lắm

giải câu b trc nha

= ((x-1)^2+2009]/x^2=(x-1)^2/x^2+2009

vậy min=2009 khi x=1

https://olm.vn//hoi-dap/question/57101.html     

Tham khảo đây nhá bạn

\(x^2-4x+1=x^2-2\cdot x\cdot2+4-4+1=\left(x-2\right)^2-4+1\)

\(=\left(x-2\right)^2-3\)    \(\forall x\in Z\)

\(\Rightarrow A_{min}=-3khix=2\)

\(a,A=x^2-4x+1=x^2-2.2.x+2^2-3=\left(x-2\right)^2-3\ge-3\)

dấu = xảy ra khi x-2=0

=> x=2

Vậy MinA=-3 khi x=2

\(b,B=5-8x-x^2=-\left(x^2+8x+5\right)=-\left(x^2+2.4.x+4^2\right)+9=-\left(x+4\right)^2+9\le9\)

dấu = xảy ra khi x+4=0

=> x=-4

Vậy MaxB=9 khi x=-4

\(c,C=5x-x^2=-\left(x^2-5x\right)=-\left(x^2-\frac{2.x.5}{2}+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

dấu = xảy ra khi \(x-\frac{5}{2}=0\)

=> x=\(\frac{5}{2}\)

Vậy Max C=\(\frac{25}{4}\)khi x=\(\frac{5}{2}\)

\(E=\frac{1}{x^2+5x+14}=\frac{1}{x^2+\frac{2.x.5}{2}+\frac{25}{4}+\frac{31}{4}}=\frac{1}{\left(x+\frac{5}{2}\right)^2+\frac{31}{4}}\)

\(\left(x+\frac{5}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\)

dấu = xảy ra khi \(x+\frac{5}{2}=0\)

=> x\(=-\frac{5}{2}\)

vì tử thức >0,mẫu thức nhỏ nhất và lớn hơn 0 => E lớnnhất khi mẫu thức nhỏ nhất 

Vậy \(MaxE=\frac{31}{4}\)khi x\(=-\frac{5}{2}\)

Tự trình bày nhé. Gợi ý thôi

\(B=5-8x-x^2\)

\(B=-\left(x^2+2.x.4+4^2\right)+21\)

\(B=-\left(x+4\right)^2+21\le21\forall x\)

\(C=5x-x^2=-\left(x^2-2.x.2,5+2,5^2\right)+6,25=-\left(x-2,5\right)^2+6,25\le6,25\forall x\)

\(D=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\)

\(D=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(D=\left(x^2+5x\right)^2-36\ge-36\forall x\)

\(A=x^2-2x+10\)

\(A=\left(x^2-2x+1\right)+9\)

\(A=\left(x-1\right)^2+9\)

Mà  \(\left(x-1\right)^2\ge0\)

\(\Rightarrow A\ge9\)

Dấu "=" xảy ra khi :

\(x-1=0\Leftrightarrow x=1\)

Vậy Min A = 9 khi x = 1

\(B=x^2-5x-7\)

\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)

\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x-\frac{5}{2}\right)^2\ge0\)

\(\Rightarrow B\ge-\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)

\(C=3x^2+3x-5\)

\(3C=9x^2+9x-15\)

\(3C=\left(9x^2+9x+\frac{9}{4}\right)-\frac{69}{4}\)

\(3C=\left(3x+\frac{3}{2}\right)^2-\frac{69}{4}\)

Mà  \(\left(3x+\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow3C\ge-\frac{69}{4}\)

\(\Leftrightarrow C\ge-\frac{23}{4}\)

Dấu "=" xảy ra khi :

\(3x+\frac{3}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy ...