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a) C có nghĩa ⇔\(\left\{{}\begin{matrix}2x-2\ne0\\2x^2-2\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
b)C= \(\dfrac{x}{2x-2}-\dfrac{x^2+1}{2x^2-2}\)
= \(\dfrac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)-\(\dfrac{x^2+1}{2\left(x+1\right)\left(x-1\right)}\)
= \(\dfrac{x^2+x}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{1}{2\left(x+1\right)}\)
c) Ta có x2-x=0 ⇒ \(\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Thay x=0 vào C= \(\dfrac{1}{2\left(x+1\right)}\) ⇒ C= \(\dfrac{1}{2}\)
Thay x= 1 vào C = \(\dfrac{1}{2\left(x+1\right)}\) ⇒ C= \(\dfrac{1}{4}\)
d) C= \(\dfrac{1}{2\left(x+1\right)}\)= \(\dfrac{-1}{2}\)
⇔-2(x+1)=2 ⇔ x=-2
a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))
\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)
\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)
b) Thay x = 2003 ta có:
\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)
c) \(C>0\) khi:
\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\) (đpcm)
a: \(C=\dfrac{5x+1+\left(2x-1\right)\left(x-1\right)+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2x^2+7x+3+2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{4}{x-1}\)
b: x=4 thì C=4/(4-1)=4/3
Khi x=-4 thì C=4/(-4-1)=-4/5
c: C>0
=>x-1>0
=>x>1
a) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x^2+10x}\)
\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)
\(=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}\)
\(=\dfrac{x-1}{2}\)
b) Để B=0 thì \(\dfrac{x-1}{2}=0\)
\(\Leftrightarrow x-1=0\)
hay x=1(nhận)
Vậy: Để B=0 thì x=1
Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)(nhận)
Vậy: Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)
c) Thay x=3 vào biểu thức \(B=\dfrac{x-1}{2}\), ta được:
\(B=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)
Vậy: Khi x=3 thì B=1
d) Để B<0 thì \(\dfrac{x-1}{2}< 0\)
\(\Leftrightarrow x-1< 0\)
\(\Leftrightarrow x< 1\)
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)
Vậy: Để B<0 thì \(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)
Để B>0 thì \(\dfrac{x-1}{2}>0\)
\(\Leftrightarrow x-1>0\)
hay x>1
Kết hợp ĐKXĐ, ta được: x>1
Vậy: Để B>0 thì x>1
a: \(A=\dfrac{-\left(x+2\right)^2-2x\left(x-2\right)-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(=\dfrac{-x^2-4x-4-2x^2+4x-4x^2}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}\)
\(=\dfrac{-7x^2-4}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}=\dfrac{7x^2+4}{\left(x+2\right)\left(x-3\right)}\)
b: Khi x=1/3 thì \(A=\dfrac{7\cdot\dfrac{1}{9}+4}{\left(\dfrac{1}{3}-2\right)\left(\dfrac{1}{3}-3\right)}=\dfrac{43}{40}\)
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
a) Ta có: \(A=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\cdot\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)
\(=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{\left(1-x\right)}+x\right)\cdot\left(\dfrac{\left(1+x\right)\left(1-x+x^2\right)}{1+x}-x\right)\right]\)
\(=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(x^2+x+x+1\right)\left(x^2-x-x+1\right)\right]\)
\(=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(x^2+2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\dfrac{x\left(x-1\right)^2\cdot\left(x+1\right)^2}{1+x^2}\cdot\dfrac{1}{\left(x+1\right)^2\cdot\left(x-1\right)^2}\)
\(=\dfrac{x}{1+x^2}\)
b) Thay \(x=-\dfrac{1}{2}\) vào biểu thức \(A=\dfrac{x}{x^2+1}\), ta được:
\(A=\dfrac{-1}{2}:\left[\left(-\dfrac{1}{2}\right)^2+1\right]\)
\(\Leftrightarrow A=-\dfrac{1}{2}:\left(\dfrac{1}{4}+1\right)\)
\(\Leftrightarrow A=-\dfrac{1}{2}:\dfrac{5}{4}\)
\(\Leftrightarrow A=-\dfrac{1}{2}\cdot\dfrac{4}{5}\)
\(\Leftrightarrow A=\dfrac{-4}{10}\)
hay \(A=\dfrac{-2}{5}\)
Vậy: Khi \(x=-\dfrac{1}{2}\) thì \(A=\dfrac{-2}{5}\)
c) Để 2A=1 thì \(A=\dfrac{1}{2}\)
hay \(\dfrac{x}{x^2+1}=\dfrac{1}{2}\)
\(\Leftrightarrow2x=x^2+1\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
hay x=1(không nhận)
Vậy: Không có giá trị nào của x để 2A=1
a: Khi x=1 thì\(P=\dfrac{1-2}{1+2}=\dfrac{-1}{2}\)
b: \(=\dfrac{3x+6+5x-6+2x^2-4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x-2}\)
c: \(P=A\cdot B=\dfrac{2x}{x-2}\cdot\dfrac{x-2}{x+1}=\dfrac{2x}{x+1}\)
\(P-2=\dfrac{2x-2x-2}{x+1}=\dfrac{-2}{x+1}\)
P<=2
=>x+1>0
=>x>-1
\(C>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)
Vậy .......
b) Ta có: C<1
nên C-1<0
\(\Leftrightarrow\dfrac{x-2}{x+1}-1< 0\)
\(\Leftrightarrow\dfrac{x-2-x-1}{x+1}< 0\)
\(\Leftrightarrow\dfrac{-3}{x+1}< 0\)
\(\Leftrightarrow x+1>0\)
hay x>-1
c) Để \(C=\dfrac{1}{4}\) thì \(\dfrac{x-2}{x+1}=\dfrac{1}{4}\)
\(\Leftrightarrow4x-8-x-1=0\)
\(\Leftrightarrow3x=9\)
hay x=3