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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) \(Q=P+\frac{x+25}{x+5}=\frac{x+30}{x+5}=1+\frac{25}{x+5}\)
\(Q_{min}\Leftrightarrow\frac{25}{x+5}_{min}\)
\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
\(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)
\(x^2+1\ge1\). 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\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)
Làm nốt
2/
\(T=8x^2-4x+\frac{1}{4x^2}+15\)
\(=\left(4x^2-4x+1\right)+\left(4x^2+\frac{1}{4x^2}-2\right)+16\)
\(=\left(2x-1\right)^2+\left(\frac{4x^2-1}{2x}\right)^2+16\ge16\)
\(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra <=> x = 3
Vậy MinA = 1
\(B=5x^2-10x+3=5\left(x^2-2x+1\right)-2=5\left(x-1\right)^2-2\ge-2\forall x\)
Dấu "=" xảy ra <=> x = 1
Vậy MinB = -2
\(C=2x^2+8x+y^2-10y+43=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)
Dấu "=" xảy ra <=> x = -2 ; y = 5
Vậy MinC = 10
\(A=x^2-6x+10\)
\(=\left(x^2-6x+9\right)+1\)
\(=\left(x-3\right)^2+1\ge1\forall x\)
Dấu"=" xảy ra khi \(x-3=0\Leftrightarrow x=3\)
Vậy \(Min_A=1\Leftrightarrow x=3\)
b,\(B=5x^2-10x+3\)
\(=5\left(x^2-2x+1\right)-2\)
\(=5\left(x-1\right)^2-2\ge-2\forall x\)
Dấu"=" xảy ra khi \(x-1=0\Leftrightarrow x=1\)
Vậy \(Min_B=-2\Leftrightarrow x=1\)
c,\(C=2x^3+8x+y^2-10+43\)
\(=2x^2+8x+8+y^2-10y+25+10\)
\(=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10\)
\(=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)
Dấu"=" xảy ra khi \(\orbr{\begin{cases}x+2=0\\y-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\y=5\end{cases}}}\)
Vậy \(Min_C=10\Leftrightarrow x=-2;y=5\)
Bài giải
\(B=\frac{x^2+1}{x^2-x+1}=\frac{x^2+1-x+x}{x^2-x+1}=\frac{x^2+1-x}{x^2-x+1}+\frac{x}{x^2-x+1}=1+\frac{x}{x^2-x+1}\)
\(B\) nhỏ nhất khi \(\frac{x}{x^2-x+1}\) nhỏ nhất
\(\Leftrightarrow\text{ }x\text{ nhỏ nhất}\text{ }\Rightarrow\text{ }x=0\)
Thay \(x=0\) ta có :
\(B=\frac{x^2+1}{x^2-x+1}=\frac{0^2+1}{0^2-0+1}=\frac{1}{1}=1\)
Vậy \(GTNN\) của \(B=1\)