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a, điều kiện xác định là \(x\ne2;x\ne-2;x\ne0\)
\(b,\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\cdot\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=-\frac{6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}\)
\(=-\frac{1}{x-2}=\frac{1}{2-x}\)
c, Để A>0
mình làm hơi tắt nên chịu khó hiểu
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
\(a,x\ne2;x\ne-2;x\ne0\)
\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(=\frac{1}{2-x}\)
\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.
a) Để phân thức trên xác định \(\Leftrightarrow x^3-8\ne0\Leftrightarrow x\ne2\)
b) \(\frac{3x^2+6x+12}{x^3-8}\)
\(=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\frac{3}{x-2}\)
a, \(ĐKXĐ\hept{\begin{cases}2-x\ne0\\2+x\ne0\end{cases}\Leftrightarrow x\ne\pm2}\)
b, Ta có: \(A=\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\)
\(=\frac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x}{x-2}\)
a) ĐKXĐ: \(\hept{\begin{cases}2-x\ne0\\x^2-4\ne0\\2+x\ne0\end{cases}}\)<=>\(\hept{\begin{cases}2-x\ne0\\2+x\ne0\\\left(x-2\right)\left(x+2\right)\ne0\end{cases}}\)<=>\(x\ne\pm2\)
b)\(A=\frac{2+x}{2-x}-\frac{4x}{x^2-4}-\frac{2-x}{2+x}\)
\(\Leftrightarrow A=\frac{2+x}{2-x}+\frac{4x}{4-x^2}-\frac{2-x}{2+x}\)
\(\Leftrightarrow A=\frac{\left(2+x\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\frac{4x}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)\left(2-x\right)}{\left(2+x\right)\left(2-x\right)}\)
\(\Leftrightarrow A=\frac{x^2+4x+4+4x-x^2+4x-4}{\left(2+x\right)\left(2-x\right)}\)
\(\Leftrightarrow A=\frac{12x}{\left(2+x\right)\left(2-x\right)}\)
a) Để phân thức được xác định thì x2-4\(\ne\) 0
=>(x+2)(x-2)\(\ne\) 0
=>x\(\ne\)-2;x\(\ne\)2
Vậy tại x\(\ne\)-2;x\(\ne\)2 thì phân thức trên đc xác định.
b)y=\(\frac{3x+6}{x^2-4}\)=\(\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)=\(\frac{3}{x-2}\)
c)Thay x=5 vào phân thức ta có:
y=\(\frac{3}{5-2}\)=\(\frac{3}{3}\)=1
Vậy y=1 tại x=5