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a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)
b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)
\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)
c: Thay x=2017 vào C, ta được:
\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)
\(=\left(\dfrac{x-1}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(\dfrac{x+1-x-3}{x+1}\right)\)
\(=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{-2}{x+1}\)
\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}\)
\(=\dfrac{-2x+2}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)
\(=\dfrac{-2\left(x-1\right)}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)
\(=\dfrac{1}{2}\)
Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)
a) ĐKXĐ: \(x\notin\left\{0;-\dfrac{1}{2};\dfrac{1}{2}\right\}\)
Ta có: \(A=\left(\dfrac{1}{2x-1}+\dfrac{3}{1-4x^2}-\dfrac{2}{2x+1}\right):\left(\dfrac{x^2}{2x^2+x}\right)\)
\(=\left(\dfrac{2x+1}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{3}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}\right):\left(\dfrac{x^2}{x\left(2x+1\right)}\right)\)
\(=\dfrac{2x+1-3-4x+2}{\left(2x-1\right)\left(2x+1\right)}:\dfrac{x}{2x+1}\)
\(=\dfrac{-2x}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x}\)
\(=\dfrac{-2}{2x-1}\)
b) Ta có: \(\left|2x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(nhận\right)\\x=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
Thay \(x=\dfrac{3}{2}\) vào biểu thức \(A=\dfrac{-2}{2x-1}\), ta được:
\(A=-2:\left(2\cdot\dfrac{3}{2}-1\right)=-2:\left(3-1\right)=-2:2=-1\)
Vậy: Khi \(\left|2x-1\right|=2\) thì A=-1
c) Để \(A=\dfrac{1}{3}\) thì \(\dfrac{-2}{2x-1}=\dfrac{1}{3}\)
\(\Leftrightarrow2x-1=-6\)
\(\Leftrightarrow2x=-5\)
hay \(x=-\dfrac{5}{2}\)(thỏa ĐK)
Vậy: Để \(A=\dfrac{1}{3}\) thì \(x=-\dfrac{5}{2}\)
\(a,ĐK\left(A\right):x\ne-\dfrac{3}{2};ĐK\left(B\right):x\ne-1;x\ne-3\\ b,A=\dfrac{-1+1}{2\left(-1\right)+3}=0\\ B=\dfrac{2\left(-\dfrac{2}{3}\right)+3}{1-\dfrac{2}{3}}+\dfrac{2-\dfrac{2}{3}}{3-\dfrac{2}{3}}=\dfrac{3-\dfrac{4}{3}}{\dfrac{1}{3}}+\dfrac{4}{3}:\dfrac{7}{3}=\dfrac{5}{3}:\dfrac{1}{3}+\dfrac{4}{7}=5+\dfrac{4}{7}=\dfrac{39}{7}\)
\(a,ĐK:x\ne\pm1\\ b,B=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\\ c,B=-\dfrac{1}{2}\Leftrightarrow2\left(x+1\right)=-2\Leftrightarrow x+1=-1\Leftrightarrow x=-2\left(tm\right)\)
a: Thay x=5 vào B, ta được:
\(B=\dfrac{5-1}{5-3}=\dfrac{4}{2}=2\)
b: \(A=\dfrac{2x^2+6x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x+3\right)\left(x-3\right)}\)
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
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
