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a:\(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}=\dfrac{x+1}{x-1}\)
b: Khi x=1/2 thì \(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-1\right)=\dfrac{3}{2}:\dfrac{-1}{2}=-3\)
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
mk nghỉ bài này đề sai
a) điều kiện : \(x\ne0;x\ne-1;x\ne2\)
ta có : \(A=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}+\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow A=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{x+1+x+1+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{2x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2-x+1}{x\left(x-2\right)}\) \(\Leftrightarrow A=1+\dfrac{2x^2+4}{x\left(x+1\right)\left(x-2\right)}=\dfrac{2x^2+4+x\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)\(\Leftrightarrow A=\dfrac{x^3+x^2-2x+4}{x\left(x+1\right)\left(x-2\right)}\)
b) ta có : \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=\dfrac{-5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=\dfrac{-1}{2}\end{matrix}\right.\)
thế vào \(A\) ta có : \(A=\dfrac{41}{5}\)
vậy ...............................................................................................................
a)
2x-3=0 => x=3/2
b)
2x^2 +1 =0 => vô nghiệm
c) x^2 -25 =0 => x=5 loiaj
x=-5 nhân
d)
x^2 -25 =0 => x=5 loại
x=-5 loại
a) \(A=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right).\dfrac{x^2-1}{2}\left(ĐKXĐ:x\ne\pm1\right)\)
\(A=\left[\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\left[\dfrac{x^2+x-2x-2-\left(x^2-x+2x-2\right)}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\left[\dfrac{x^2-x-2-x^2-x+2}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\dfrac{-2x}{\left(x+1\right)^2\left(x-1\right)}.\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\dfrac{-2x}{2\left(x+1\right)}\)
P/s: câu b bn tự làm nha
a.
\(A=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right)\dfrac{x^2-1}{2}\)
\(A=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right).\dfrac{x^2-1}{2}\)
\(A=\left(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)^2}-\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}\right).\dfrac{x^2-1}{2}\)
\(A=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}.\dfrac{x^2-1}{2}\)
\(A=\dfrac{-2x}{\left(x-1\right)\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2}\)
\(A=\dfrac{-x}{x+1}\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
a)
\(\left\{{}\begin{matrix}x-1\ne0\\x+2\ne0\end{matrix}\right.\)
b)
x khác 1
c)
x khác 0; x khác 5
d) x khác 5 ; x khác -5
