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a) P xác định \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Leftrightarrow x\ne\left\{-5;0\right\}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+5x^2-x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+5\right)-x\left(x+5\right)}{2x\left(x+5\right)}\)
\(P=\frac{\left(x+5\right)\left(x^2-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x-1\right)}{2x}\)
\(P=\frac{x-1}{2}\)
c) Để P = 0 thì \(x-1=0\Leftrightarrow x=1\)( thỏa mãn ĐKXĐ )
Để P = 1/4 thì \(\frac{x-1}{2}=\frac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)( thỏa mãn ĐKXĐ )
d) Để P > 0 thì \(\frac{x-1}{2}>0\)
Mà 2 > 0, do đó để P > 0 thì \(x-1>0\Leftrightarrow x>1\)
Để P < 0 thì \(\frac{x-1}{2}< 0\)
Mà 2 > 0, do đó để P < 0 thì \(x-1< 0\Leftrightarrow x< 1\)
a) Phân thức B xác định \(\Leftrightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\left\{\pm1\right\}\\x\ne-1\end{cases}\Leftrightarrow}x\ne\left\{\pm1\right\}}\)
b) \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\cdot\frac{4x^2-4}{5}\)
\(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3\cdot2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(2x\right)^2-2^2}{5}\)
\(B=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(2x-2\right)\left(2x+2\right)}{5}\)
\(B=\frac{10\cdot2\left(x-1\right)\cdot2\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)\cdot5}\)
\(B=\frac{40\left(x-1\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(B=4\)
Vậy với mọi giá trị của x thì B luôn bằng 4
Vậy giá trị của B không phụ thuộc vào biến ( đpcm )
\(Giải:\)
\(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right]=\left[\frac{x+1}{2x-2}+\frac{12}{4x^2-4}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{x+1}{2x-2}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{\left(x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(2x-2\right)}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]\)
\(=\frac{2x^2+4x+14-2x^2+2x-6x+6}{\left(2x-2\right)\left(2x+2\right)}\)
\(=\frac{6}{\left(2x-2\right)\left(2x+2\right)}\)
a) \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
( ĐKXĐ : \(x\ne0,x\ne-5\) )
\(B=\dfrac{\left(x^2+2x\right).x}{2x\left(x+5\right)}+\dfrac{\left(x-5\right).2\left(x+5\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3+2x^2+2x^2+10x-10x-50+50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3-x^2+5x^2-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^2\left(x-1\right)+5x\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{\left(x-1\right)\left(x+5\right)x}{2x\left(x+5\right)}\)
\(B=\dfrac{x-1}{2}\)
Câu b và c dễ vì đã có kết quả rút gọn rồi :)
a) P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}}\)
Vậy P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{\left(x-5\right)\left(x+5\right)2}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
Có: \(P=0\)
\(\Rightarrow P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=0\Leftrightarrow x\left(x^2+4x-5\right)=0\Leftrightarrow x^2+4x-5=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy \(P=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Bài 3:
a: DKDXĐ: x<>1
b: \(=\dfrac{x^2+2+x^2-x-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{2}{x-1}=\dfrac{x^2-2x+1}{\left(x-1\right)^2}\cdot\dfrac{2}{x^2+x+1}=\dfrac{2}{x^2+x+1}\)
c: Để C lớn nhất thì \(A=x^2+x+1_{MIN}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu = xảy ra khi x=-1/2
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
a.
ĐKXĐ: \(x\ne2\)
b.
\(P=\left(\dfrac{2x}{x-2}+\dfrac{x}{2-x}\right):\dfrac{x^2+1}{x-2}\)
\(=\left(\dfrac{2x}{x-2}-\dfrac{x}{x-2}\right)\cdot\dfrac{x-2}{x^2+1}\)
\(=\dfrac{x}{x-2}\cdot\dfrac{x-2}{x^2+1}=\dfrac{x}{x^2+1}\)
c.
\(x=-1\Rightarrow P=-\dfrac{1}{\left(-1\right)^2+1}=-\dfrac{1}{2}\)
d.
\(P=\dfrac{x}{x^2+1}\cdot\dfrac{x^2+1}{x}-\dfrac{1}{P}\ge1-\dfrac{1}{P}\)
\(\Rightarrow\dfrac{P^2+1}{P}\ge1\)
\(\Rightarrow P^2+1\ge P\) \(\Rightarrow P\left(P-1\right)\ge1\)
\(\Rightarrow P\ge2\)
Dấu "=" khi x = ...................
Bài 2:
a: \(M=\dfrac{3x+1-2x-2}{\left(3x-1\right)\left(3x+1\right)}:\dfrac{3x+1-3x}{x\left(3x+1\right)}\)
\(=\dfrac{x-1}{\left(3x-1\right)\left(3x+1\right)}\cdot\dfrac{x\left(3x+1\right)}{1}=\dfrac{x\left(x-1\right)}{3x-1}\)
b: Để M=0 thì x(x-1)=0
=>x=1(nhận) hoặc x=0(loại)
c: \(P=M\cdot\left(3x-1\right)=x\left(x-1\right)=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)
Dấu = xảy ra khi x=1/2
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(B=\left(\dfrac{2x+1}{x-1}+\dfrac{8}{x^2-1}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x^2-1}{5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{2x^2+2x+x+1+8-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{2x^2+3x+9-x^2+2x-1}{5}\)
\(=\dfrac{x^2+5x+8}{5}\)
Ta có: \(x^2+5x+8\)
\(=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{7}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\)
Ta có: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\forall x\)
\(\Leftrightarrow x^2+5x+8>0\forall x\)
\(\Leftrightarrow\dfrac{x^2+5x+8}{5}>0\forall x\) thỏa mãn ĐKXĐ(đpcm)