a, ĐKXĐ: \(\hept{\begin{cases}5x+25\ne0\\x\ne0\\x^2+5x\ne0\end{cases}\Rightarrow\hept{\begin{cases}5\left(x+5\right)\ne0\\x\ne0\\x\left(x+5\right)\ne0\end{cases}\Rightarrow}}\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

b, \(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)

\(=\frac{x^3}{5x\left(x+5\right)}+\frac{5\left(2x-10\right)\left(x+5\right)}{5x\left(x+5\right)}+\frac{\left(50+5x\right).5}{5x\left(x+5\right)}\)

\(=\frac{x^3+10\left(x-5\right)\left(x+5\right)+250+25x}{5x\left(x+5\right)}\)

\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)

c, \(P=-4\Rightarrow\frac{x+5}{5}=-4\Rightarrow x+5=-20\Rightarrow x=-25\)

d, \(\frac{1}{P}\in Z\Rightarrow\frac{5}{x+5}\in Z\Rightarrow5⋮\left(x+5\right)\Rightarrow x+5\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\Rightarrow x\in\left\{-10;-6;-4;0\right\}\)

Mà x khác 0 (ĐKXĐ của P) nên \(x\in\left\{-10;-6;-4\right\}\)

a) \(ĐKXĐ:\hept{\begin{cases}5x+25\ne0\\x\ne0\\x^2+5x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

b) \(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)

\(P=\frac{x^3}{5x\left(x+5\right)}+\frac{10x^2-250}{5x\left(x+5\right)}+\frac{250+25x}{5x\left(x+5\right)}\)

\(P=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)

c) \(P=4\Leftrightarrow\frac{x+5}{5}=4\Leftrightarrow x+5=20\Leftrightarrow x=15\)

d) \(\frac{1}{P}=\frac{5}{x+5}\in Z\Leftrightarrow5⋮x+5\)

\(\Leftrightarrow x+5\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Lập bảng nhé

e) \(Q=P+\frac{x+25}{x+5}=\frac{x+30}{x+5}=1+\frac{25}{x+5}\)

\(Q_{min}\Leftrightarrow\frac{25}{x+5}_{min}\)

b)\(C=\frac{5x-19}{x-4}=\frac{5x-20+1}{x-4}=\frac{5\left(x-4\right)+1}{x-4}=5+\frac{1}{x-4}\)

Để C đạt giá trị nhỏ nhất => 1/x-5 phải đạt giá trị nhỏ nhất

=> 1/x-5=-1

=>x-5=-1

=>x=4

Giá trị nhỏ nhất của C là : 5 - 1 = 4 <=> x = 4

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

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow A=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi x=-1/2

Vậy Amin=3/4 khi x=-1/2

b,\(B=2x^2-5x-2\)

\(\Rightarrow2B=4x^2-10x-4=\left(4x^2-10x+\frac{25}{4}\right)-\frac{41}{4}=\left(2x-\frac{5}{2}\right)^2-\frac{41}{4}\)

Vì \(\left(2x-\frac{5}{2}\right)^2\ge0\Rightarrow2B=\left(2x-\frac{5}{2}\right)^2-\frac{41}{4}\ge-\frac{41}{4}\Rightarrow B\ge-\frac{41}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmin=-41/8 khi x=5/4

c,\(C=x^2+5y^2+2xy-y+3=\left(x^2+2xy+y^2\right)+\left(4y^2-y+\frac{1}{16}\right)+\frac{47}{16}=\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2+\frac{47}{16}\)

Vì\(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(2y-\frac{1}{4}\right)^2\ge0\end{cases}}\Rightarrow\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2\ge0\)

\(\Rightarrow C=\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2+\frac{47}{16}\ge\frac{47}{16}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y=0\\2y-\frac{1}{4}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{8}\\y=\frac{1}{8}\end{cases}}}\)

Vậy Cmin=47/16 khi x=-1/8,y=1/8

a) \(A=x^2-6x+11\)

\(\Rightarrow A=x^2-6x+9+2\)

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

Ta có: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\) x = 3

Vậy \(MIN\) \(A=2\Leftrightarrow x=3\)

b) \(B=2x^2+10x-1\)

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

\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{25}{2}-1\)

\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\)

Ta có: \(2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)\ge0\forall x\)

\(\Rightarrow2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\ge-\dfrac{23}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{-5}{2}\)

Vậy \(MIN\) \(B=\dfrac{-23}{2}\Leftrightarrow x=\dfrac{-5}{2}\)

c) \(C=5x-x^2\)

\(\Rightarrow C=-\left(x^2-5x\right)\)

\(\Rightarrow C=-\left(x^2-2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)

\(\Rightarrow C=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)

Ta có: \(-\left(x-\dfrac{5}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)

Vậy \(MAX\) \(C=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)

\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)

 \(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)

\(x^2+1\ge1\). dấu = xảy ra khi x²=0

=> x=0

Vậy \(B_{min}\Leftrightarrow x=0\)

ta có: \(x^2+2x-2=x^2+2x+1^2-3=\left(x+1\right)^2-3\ge-3\)

dấu = xảy ra khi \(x+1=0\)

\(\Rightarrow x=-1\)

Vậy\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)

Để A xác định 

\(\Rightarrow\hept{\begin{cases}x-1\ne0\\x^2-1\ne0\\x^2-2x+1\ne0\end{cases}}\)

\(\Rightarrow x^2-1\ne0\)

\(\Rightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

b, 

\(1,a,A=x^2-6x+25\)

\(=x^2-2.x.3+9-9+25\)

\(=\left(x-3\right)^2+16\)

Ta có :

\(\left(x-3\right)^2\ge0\)Với mọi x

\(\Rightarrow\left(x-3\right)^2+16\ge16\)

Hay \(A\ge16\)

\(\Rightarrow A_{min}=16\)

\(\Leftrightarrow x=3\)

\(b,B=4x^2+4x-2\)

\(B=4x^2+4x+1-3\)

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

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

Ta có : 

\(\left(2x+1\right)^2\ge0\)với mọi x

\(\Rightarrow\left(2x+1\right)^2-3\ge-3\)

\(\Leftrightarrow B\ge-3\)

\(\Rightarrow B_{min}=-3\)

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

A=x^2+5x+7

A=x^2+2.x.5/2+25/4+3/4

A=(x+5/2)^2+3/4>= 3/4

Vậy Min A=3/4 <=> x=-5/2

  ấ ở đây nhé !    

Mình có làm bài tìm giá trị lớn nhất trong đây rùi nhé !