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\(B=\left(\frac{2x}{x-3}-\frac{x-1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\left(ĐK:x\ne\pm3\right)\)
\(=\frac{2x\left(x+3\right)-\left(x-1\right)\left(x-3\right)-x^2-1}{x^2-9}:\frac{x+3-x+1}{x+3}\)
\(=\frac{2x^2+6x-x^2+3x+x-3-x^2-1}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}\)
\(=\frac{10x-4}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}=\frac{10x-4}{4\left(x-3\right)}\)
\(B=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)
\(=\left[\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+3-x+1}{x+3}\right)\)
\(=\left(\frac{2x^2+6x-x^2+3x-x+3-x^2-1}{\left(x+3\right)\left(x-3\right)}\right):\frac{4}{x+3}\)
\(=\frac{8x-1}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{4}\)\(=\frac{8x-1}{4\left(x-3\right)}\)
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)
\(=\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right):\frac{3x^3-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\left(\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3x+3}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\frac{\left(x+1\right)\left(x+2\right)}{3\left(x^2-x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2\left(x+1\right)}{3\left(x^2-x+1\right)^2}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(A=\frac{x^3+2x^2+x}{x^3+x}=\frac{x^3+3x}{x^2+x}=\frac{x^2+3}{x^2+1}\)
\(B=\frac{x^2-9}{3-x}=\frac{x^2-9}{-\left(3-x\right)}=\frac{\left(x-3\right)\left(x+3\right)}{x+3}=x-3\)
Em mới lớp 7 nên rút gọn bừa ạ !!!
\(A=\frac{x^3+2x^2+x}{x^3+x}=\frac{\left(x^3+x\right)+2x^2}{x^3+x}=\frac{x^3+x}{x^3+x}+\frac{2x^2}{x^3+x}=\frac{2x^2}{x^3+x}\)\(=\frac{x^2.x.x}{x.\left(x^2+1\right)}=\frac{x^2.x}{x^2+1}\)
Lm thử sức thôi ạ !!!
\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}-\frac{3}{x^2-x+1}\right)\cdot\frac{3x^2-3x+3}{\left(x+1\right).\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left(\frac{x+1}{x}-\frac{3}{\left(x+1\right).\left(x^2-x+1\right)}+\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\right)\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{\left(x+1\right)^2.\left(x^2-x+1\right)-3x+3x^2+3x}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{x^4+x^3+x+1+3x^2}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2x^3+2x^2-2x-2}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(=\frac{3x^4+x^3+7x^2+5x+5}{x.\left(x+1\right)^2.\left(x+2\right)}\)
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
\(ĐK:x\ne\pm3\)
\(P=\left[\frac{\left(2x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-10x}{\left(x-3\right)\left(x+3\right)}\right]\cdot\frac{x-3}{x+2}\)
\(=\frac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x+2}\)
\(=\frac{3x^2+6x}{x+3}\cdot\frac{1}{x+2}=\frac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}=\frac{3x}{x+3}\)