Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) Phân thức M xác định khi :
+) \(x\ne0\)
+) \(x-2\ne0\Leftrightarrow x\ne2\)
b) \(M=\left(\frac{2}{x}-\frac{2}{x-2}\right):\frac{3x}{x-2}\)
\(M=\left(\frac{2\left(x-2\right)}{x\left(x-2\right)}-\frac{2x}{x\left(x-2\right)}\right)\cdot\frac{x-2}{3x}\)
\(M=\left(\frac{2x-4-2x}{x\left(x-2\right)}\right)\cdot\frac{x-2}{3x}\)
\(M=\frac{-4\cdot\left(x-2\right)}{x\left(x-2\right)\cdot3x}\)
\(M=\frac{-4}{3x^2}\)
c) Thay x = -2 ta có :
\(M=\frac{-4}{3\cdot\left(-2\right)^2}=\frac{-1\cdot4}{3\cdot4}=\frac{-1}{3}\)
Vậy........
a, \(M=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\left(x-1\right)\left(x+1\right)=2x+1\)
b, Thay x = 1/2 vào biểu thức trên ta được : \(2.\frac{1}{2}+1=1\)
c, Để M luôn dương hay \(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)
Vậy với x \(\ge-\frac{1}{2}\)thì \(M\ge0\)
a)
2x-4=2(x-2)
2x+4=2(x+2)
x
Để P xác định thì
[2(x-2) => [2(x+2)
[2(x+2) =>[ 2(x-2)
[ (x-2)(x+2) => [(x+2)(x-2)
Vay 2(x+2) , 2(x-2), (x+2)(x-2) thi P xác định
M xác định
\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x^2-x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\left(x-1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne0;x\ne1\end{cases}}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)
Vậy ĐKXĐ của M là \(\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)
\(M=\frac{3}{x-1}+\frac{1}{x^2-x}=\frac{3}{x-1}+\frac{1}{x\left(x-1\right)}=\frac{3x}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}=\frac{3x+1}{x\left(x-1\right)}\)
Thay x=5 ta có:
\(M=\frac{3.5+1}{5\left(5-1\right)}=\frac{15+1}{5.4}=\frac{16}{20}=\frac{4}{5}\)
Vậy \(M=5\)tại x=5
\(M=0\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)( thỏa mãn đkxđ)
Vậy với \(x=-\frac{1}{3}\)thì \(M=0\)
\(M=-1\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=-1\Leftrightarrow3x+1=-x^2+x\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
Vậy với \(x=-1\)thì \(M=-1\)
\(a,ĐKXĐ:x\ne\pm2\)
\(b,P=\left(\frac{x+2}{2x-4}+\frac{x-2}{2x+4}+\frac{-8}{x^2-4}\right):\frac{4}{x-2}\)
\(=\left(\frac{x+2}{2\left(x-2\right)}+\frac{x-2}{2\left(x+2\right)}+\frac{-8}{\left(x-2\right)\left(x+2\right)}\right).\frac{x-2}{4}\)
\(=\left(\frac{\left(x+2\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{\left(-8\right).2}{2\left(x-2\right)\left(x+2\right)}\right)\)\(.\frac{x-2}{4}\)
\(=\left(\frac{x^2+4x+4+x^2-4x+4-16}{2\left(x-2\right)\left(x+2\right)}\right).\frac{x-2}{4}\)
\(=\frac{2x^2-8}{2\left(x-2\right)\left(x+2\right)}.\frac{x-2}{4}\)
\(=\frac{2\left(x-2\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}.\frac{x-2}{4}=1.\frac{x-2}{4}=\frac{x-2}{4}\)