Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, ĐK: \(x\ne\pm1\)
b, Ta có:
\(A=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\) \(=\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2x^2-2}\) \(=\frac{x-1}{2\left(x-1\right)\left(x+1\right)}\) \(=\frac{1}{2\left(x+1\right)}\)
c, Với \(x\ne\pm1\), để \(A=-\frac{1}{2}\):
\(\Leftrightarrow\frac{1}{2\left(x+1\right)}=-\frac{1}{2}\) \(\Leftrightarrow x=-2\) (tm)
a) Để M có nghĩa thì \(x\ne1;x\ne-1\)
b) \(M=\dfrac{x}{2x-2}+\dfrac{x^2-1}{2-2x^2}\)
\(=\dfrac{x^2+x}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{2x-2}\)
c) Để \(M=\dfrac{1}{2}\) thì \(\dfrac{1}{2x-2}=\dfrac{1}{2}\)
\(\Leftrightarrow2x-2=2\)
\(\Leftrightarrow x=2\)
\(\text{a) }A=\dfrac{x}{2x+2}+\dfrac{x^2+1}{2-2x^2}\\ A=\dfrac{x}{2\left(x+1\right)}+\dfrac{x^2+1}{2\left(1-x^2\right)}\\ A=\dfrac{x}{2\left(x+1\right)}+\dfrac{x^2+1}{2\left(1-x\right)\left(1+x\right)}\\ A=\dfrac{x\left(1-x\right)}{2\left(x+1\right)\left(1-x\right)}+\dfrac{x^2+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{x-x^2+x^2+1}{2\left(x+1\right)\left(1-x\right)}\)
\(\Rightarrow\) Để \(A\) có nghĩa
\(\text{thì }\Rightarrow2\left(x+1\right)\left(1-x\right)\ne0\\ \Rightarrow\left\{{}\begin{matrix}x+1\ne0\\1-x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)
\(\text{b) }A=\dfrac{x-x^2+x^2+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{x+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{1}{2\left(1-x\right)}\\ A=\dfrac{1}{2-2x}\)
c) Để \(A=\dfrac{1}{2}\)
\(\text{thì }\Rightarrow\dfrac{1}{2-2x}=\dfrac{1}{2}\\ \Leftrightarrow2-2x=2\\ \Leftrightarrow2x=0\\ \Leftrightarrow x=0\)
Vậy......................
\(a,ĐKXĐ:x-1\ne0;1-x\ne0;1+x\ne0\)
\(\Rightarrow\left[{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
\(b,C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)
\(C=\dfrac{x^2+x}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-x^2-1}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{2x+2}\)
\(c,C=-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2x+2}=-\dfrac{1}{2}\Leftrightarrow-2x-2=2\Leftrightarrow x=0\)
a, Để C có nghĩa <=> \(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x\ne2\\2x^2\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)\(\Leftrightarrow x\ne\pm1\) thì C có nghĩa.
b, \(\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}=\dfrac{x}{2\left(x-1\right)}+\dfrac{-\left(x^2+1\right)}{2\left(x^2-1\right)}\)
\(=\dfrac{x}{2\left(x-1\right)}+\dfrac{-\left(x^2+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-\left(x^2+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\)
c, \(C=-0,5\Leftrightarrow\dfrac{1}{2\left(x+1\right)}=-0,5\)
\(\Leftrightarrow2\left(x+1\right)=\dfrac{1}{-0,5}=-2\Leftrightarrow x+1=-1\Leftrightarrow x=-2\)
Vậy....
a,
ĐKXĐ: 2x - 2 \(\ne\)0 <=> 2x \(\ne\)2 <=> x \(\ne\)1
2 - 2x2 \(\ne\)0 <=> 2( 1 - x2) \(\ne\)0 <=> (1 - x)(1 + x) \(\ne\)0
<=> x \(\ne\)-1
b,
C = \(\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)
= \(\dfrac{x}{2x-2}-\dfrac{x^2+1}{2\left(x^2-1\right)}\)
= \(\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x+1\right)\left(x-1\right)}\)
= \(\dfrac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{1}{2\left(x+1\right)}\)
c,
Để C = \(-\dfrac{1}{2}\)
<=> \(\dfrac{1}{2\left(x+1\right)}=\dfrac{-1}{2}\)
<=> \(\dfrac{1}{x+1}=-1\)
<=> x + 1 = -1
<=> x = -2
d,
Để C > 0
<=> \(\dfrac{1}{2\left(x+1\right)}\)> 0
<=> 2(x + 1) > 0
<=> x + 1 > 0
<=> x > -1
Lm 1 bài góp vui thôi ! Thấy @Nguyễn Trần Thành Đạt lm hoài à?
a) Đê C có nghĩa => \(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-1\ne0\\\left(x-1\right)\left(x+1\right)\ne0\end{matrix}\right.\)
=> \(x\ne\pm1\)
b) Ta có: C = \(\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\) => C = \(\dfrac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
=> C = \(\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\) = \(\dfrac{1}{2\left(x+1\right)}\)
c) Để C = 0,5 => \(\dfrac{1}{2\left(x+1\right)}=0,5\)
=> \(2\left(x+1\right)=2\)
<=> \(x+1=1\)
<=> \(x=0\) (TM ĐKXĐ)
Thay x = 0 vào C ta được C = 0,5 (TM)
Vậy nghiệm của PT: \(S=\left\{0\right\}\)
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 \(x-1< 0\Leftrightarrow x< 1\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a, ĐKXĐ:\(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1\ne0\\1-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x^2\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\Leftrightarrow x\ne\pm1\)
b, \(C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)
\(\Rightarrow C=\dfrac{x}{2\left(x-1\right)}+\dfrac{x^2+1}{2\left(1-x^2\right)}\)
\(\Rightarrow C=\dfrac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow C=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow C=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow C=\dfrac{1}{2\left(x+1\right)}\)
c, \(C=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2\left(x+1\right)}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x+1}=1\\ \Rightarrow x+1=1\\ \Rightarrow x=0\)
a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b: \(C=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2x+2}\)
c: Để C=1/2 thì 2x+2=2
hay x=0