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a: \(P=\dfrac{4x-6-x+1}{2x-3}:\left(\dfrac{6x+1}{2x^2-3x+2x-3}+\dfrac{x}{x+1}\right)\)
\(=\dfrac{3x-5}{2x-3}:\left(\dfrac{6x+1}{\left(x+1\right)\left(2x-3\right)}+\dfrac{x}{x+1}\right)\)
\(=\dfrac{3x-5}{2x-3}:\dfrac{6x+1+2x^2-3x}{\left(x+1\right)\left(2x-3\right)}\)
\(=\dfrac{3x-5}{\left(2x-3\right)}\cdot\dfrac{\left(2x-3\right)\left(x+1\right)}{2x^2+3x+1}\)
\(=\dfrac{3x-5}{2x+1}\)
b: \(P-\dfrac{3}{2}=\dfrac{3x-5}{2x+1}-\dfrac{3}{2}=\dfrac{6x-10-6x-3}{2\left(2x+1\right)}=\dfrac{-7}{2\left(2x+1\right)}\)
Câu 1:
\(A=\frac{x\left(1-x^2\right)}{1+x^2}:\left[\left(\frac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}+x\right)\left(\frac{\left(1+x\right)\left(x^2-x+1\right)}{1+x}+x\right)\right]\)
\(=\frac{x\left(1-x^2\right)}{x^2+1}:\left[\left(x^2+2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\frac{x\left(1-x^2\right)}{\left(1+x^2\right)\left(1+x\right)^2\left(x-1\right)^2}=\frac{x}{\left(1+x^2\right)\left(x^2-1\right)}=\frac{x}{x^4-1}\)
Câu 2: thay x vào A có :
\(A=\frac{-\frac{1}{2}}{\frac{1}{4}-1}=\frac{2}{3}\)
Câu c :
2A=1 => \(\frac{x}{x^4-1}=\frac{1}{2}\)ĐK \(\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)
\(\Leftrightarrow x^4-2x-1=0\Leftrightarrow\left(x+1\right)\left(x^3-x^2+x-1\right)=0\)
\(\left(x+1\right)\left(x^2+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)loại do điều kiện vậy ko có giá trị nào của x thỏa mãn
1/(x^2+6x+9)-1/(x^2-6x+9)=(x-3)/(x-3)(x+3)-(x+3)/(x-3)(x+3)= -6/(x-3)(x+3)
1/(x+3)+1/(x-3)=
a)\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}.\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\left(ĐKXĐ:x\ne0;-1\right)\)
\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{x+1}{x}\right)+\frac{1}{\left(x+1\right)^2}.\left(\frac{x^2+1}{x^2}\right)\right]:\frac{x-1}{x^3}\)
\(P=\left[\frac{2}{\left(x+1\right)^2x}+\frac{x^2+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{x^3}\)
\(P=\left[\frac{x^2+2x+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{3}\)
\(P=\frac{\left(x+1\right)^2}{x^2\left(x+1\right)^2}:\frac{x-1}{3}\)
\(P=\frac{3}{x^2\left(x-1\right)}\)
b)Bài này liên quan đến dấu lớn nên mk ko làm đc
a)\(Q=\left(\frac{1}{x+1}+\frac{6x+3}{x^3+1}-\frac{2}{x^2-x+1}\right):\left(x+2\right)\)\(\left(ĐKXĐ:x\ne-1\right)\)
\(Q=\left(\frac{x^2-x+1}{x^3+1}+\frac{6x+3}{x^3+1}-\frac{2\left(x+1\right)}{x^3+1}\right):\left(x+2\right)\)
\(Q=\left(x^2-x+1+6x+3-2x-2\right):\left(x+2\right)\)
\(Q=\left(x^2+3x+2\right):\left(x+2\right)\)
\(Q=\left(x+1\right)\left(x+2\right):\left(x+2\right)\)
\(Q=x+1\)
b)Tại \(Q=\frac{1}{3}\)ta được:\(\frac{1}{3}=x+1\)
\(\Rightarrow x=-\frac{2}{3}\)