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a) ĐKXĐ: \(\hept{\begin{cases}x-2\ne0\\x\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)
b)Ta có: P = \(\frac{x^2}{x-2}\left(\frac{x^2+4}{x}-4\right)+3\)
P = \(\frac{x^2}{x-2}\left(\frac{x^2+4-4x}{x}\right)+3\)
P = \(\frac{x^2}{x-2}\cdot\frac{\left(x-2\right)^2}{x}+3\)
P = \(\left(x-2\right).x+3\)
P = \(x^2-2x+3\)
c) Ta có: P = x2 - 2x + 3
P = (x2 - 2x + 1) + 2
P = (x - 1)2 + 2 \(\ge\)2 \(\forall\)x
Dấu "=" xảy ra <=> x - 1 = 0 <=> x = 1
Vậy x = 1 thì P đạt GTNN là 2
a) Phân thức được xác định khi \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)
ĐKXĐ: \(x\ne2;x\ne0\)
b) \(P=\frac{x^2}{x-2}\left(\frac{x^2+4}{x}-4\right)+3\)
\(P=\frac{x^4-4x^3+7x^2-6x}{x^2-2x}\)
\(P=\frac{x^3-4x^2+7x-6}{x-2}\)
\(P=\frac{\left(x-2\right)\left(x^2-2x+3\right)}{x-2}\)
\(P=x^2-2x+3\)
c) \(P=x^2-2x+3\)
\(P=x^2-2x+1+2\)
\(P=\left(x-1\right)^2+2\ge2\) vì \(\left(x-1\right)^2\ge0,\forall x\inℝ\)
\(\Rightarrow Min_P=2\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy: \(Min_p=2\Leftrightarrow x=1\)
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left(\frac{2}{x^2-4}-\frac{x+2}{x^2-4}\right)\)
\(A=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{x^2-4}\)
\(A=\frac{2x}{\left(x+2\right)^2}.\frac{x^2-4}{-x}=\frac{2\left(x-2\right)}{-\left(x+2\right)}=\frac{-2\left(x-2\right)}{x+2}\)
a) A có nghĩa\(\Leftrightarrow\hept{\begin{cases}2-x\ne0\\2+x\ne0\\x-3\ne0\end{cases}}\Rightarrow x\ne\pm2;x\ne3\)
\(A=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right):\frac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)
\(=\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{4-x^2}:\frac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)
\(=\frac{x^2+4x+4-4+4x-x^2+4x^2}{4-x^2}:\frac{x-3}{2-x}\)
\(=\frac{4x^2+8x}{4-x^2}.\frac{2-x}{x-3}\)
\(=\frac{4x\left(x+2\right)}{\left(2+x\right)\left(x-3\right)}=\frac{4x}{x-3}\)
b) \(A=1\Leftrightarrow4x=x-3\Leftrightarrow x=-1\)
c) \(A>0\Leftrightarrow\frac{4x}{x-3}>0\)
TH1: \(\hept{\begin{cases}4x>0\\x-3>0\end{cases}}\Leftrightarrow x>3\)
TH2: \(\hept{\begin{cases}4x< 0\\x-3< 0\end{cases}}\Leftrightarrow x< 0\)
Giúp mình với đúng mik tích cho :>>
Đk : \(x\ne5;x\ne0;x\ne4\)
a) ta có:
\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(KTM\right)\\x=3\left(TM\right)\end{cases}}\)
Thay x= 3 vào biểu thức A , ta được :
\(A=\frac{3-5}{3-4}=\frac{-2}{-1}=2\)
vậy ..............
b) \(B=\frac{x+5}{2x}-\frac{x-6}{5-x}-\frac{2x^2-2x-50}{2x^2-10x}\)
\(B=\frac{x+5}{2x}+\frac{6-x}{x-5}-\frac{2x^2-2x-50}{2x\left(x-5\right)}\)
\(B=\frac{\left(x-5\right)\left(x+5\right)+2x\left(6-x\right)-2x^2+2x+50}{2x\left(x-5\right)}\)
\(B=\frac{x^2-25+12x-2x^2-2x^2+2x+50}{2x\left(x-5\right)}\)
\(B=\frac{-3x^2+25+14x}{2x\left(x-5\right)}\)
c) Ta có :
\(P=A.B\)
\(P=\frac{x-5}{x-4}.\frac{-3x^2+25+14x}{2x\left(x-5\right)}\)
\(P=\frac{-3x^2+25+14x}{2x\left(x-4\right)}\)
\(P=\frac{-3x^2+25+14x}{2x^2-8x}\)
a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
Thiếu B nha bn