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\(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne-\frac{1}{2}\end{cases}}\)
a) \(A=\left(\frac{1}{x-1}+\frac{x}{x^3-1}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow A=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow A=\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow A=\frac{x+1}{x-1}\)
b) Thay \(x=\frac{1}{2}\)vào A, ta được :
\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3\)
a: \(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\cdot x+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{\left(x^2+1\right)\left(x+1\right)+x-1}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x^3+x^2+x+1+x-1}{\left(x-1\right)}\cdot\dfrac{x+1}{2x+1}\)
\(=\dfrac{x^3+x^2+2x}{x-1}\cdot\dfrac{x+1}{2x+1}=\dfrac{x\left(x^2+x+2\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)
b: Khi x=1/2 thì \(A=\dfrac{\dfrac{1}{2}\left(\dfrac{1}{4}+\dfrac{1}{2}+2\right)\left(\dfrac{1}{2}+1\right)}{\left(\dfrac{1}{2}-1\right)\left(2\cdot\dfrac{1}{2}+1\right)}=-\dfrac{33}{16}\)
\(a,\)\(đkxđ\)\(\hept{\begin{cases}3+2x\ne0\\3-2x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-\frac{3}{2}\\x\ne\frac{3}{2}\end{cases}}}\)
\(b,\)\(A=\left(\frac{1}{3+2x}+\frac{1}{3-2x}\right):\frac{1}{3+2x}\)
\(=\left(\frac{3-2x+3+2x}{\left(3-2x\right)\left(3+2x\right)}\right).\frac{3+2x}{1}\)
\(=\frac{6\left(3+2x\right)}{\left(3-2x\right)\left(3+2x\right)}=\frac{6}{3-2x}\)
\(c,\)Tại x = 3 \(\Rightarrow A=\frac{6}{3+2.3}=\frac{6}{9}=\frac{2}{3}\)
\(ĐKXĐ:x\ne\pm1\)
a) \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{4x^2}{1-x^2}\right):\frac{2x^2-2}{x^2-2x+1}\)
\(\Leftrightarrow A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x^2}{x^2-1}\right):\frac{2\left(x^2-1\right)}{\left(x-1\right)^2}\)
\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x+1-x^2+2x-1}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{4x-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{-4x\left(x-1\right)^3}{2\left(x-1\right)^2\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{-2x\left(x-1\right)}{\left(x+1\right)^2}\)
b) Thay x = -3 vào A, ta được :
\(A=\frac{\left(-2\right)\left(-3\right)\left(-3-1\right)}{\left(-3+1\right)^2}\)
\(\Leftrightarrow A=\frac{6.\left(-4\right)}{2^2}\)
\(\Leftrightarrow A=-6\)
c) Để A > -1
\(\Leftrightarrow-2x\left(x-1\right)>-\left(x+1\right)^2\)
\(\Leftrightarrow2x\left(x-1\right)< \left(x+1\right)^2\)
\(\Leftrightarrow2x^2-2x< x^2+2x+1\)
\(\Leftrightarrow x^2-4x-1< 0\)
\(\Leftrightarrow\left(x-2\right)^2-5< 0\)
\(\Leftrightarrow\left(x-2\right)^2< 5\)
Đoạn này bạn tự tìm giá trị x thỏa mãn là xong (Chú ý ĐKXĐ)
a: Thay x=5 vào B, ta được:
\(B=\dfrac{5-1}{5-3}=\dfrac{4}{2}=2\)
b: \(A=\dfrac{2x^2+6x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x+3\right)\left(x-3\right)}\)
\(A=\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-1\)
\(A=\frac{\left(x-1\right)^2}{\left(x-1\right)}+\frac{\left(x+1\right)^2}{\left(x+1\right)}-1\)
\(A=\left(x-1\right)+\left(x+1\right)-1\)
thay x = 1 vào A
\(\Rightarrow A=\left(1-1\right)+\left(1+1\right)-1\)
\(A=1\)