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a) \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)
\(=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{4x^2-3x+17+2x^2-x-2x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=-\frac{12}{x^2+x+1}\)
b) \(\frac{1}{x^2-x+1}-\frac{x^2+2}{x^3+1}+1=\frac{x+1-x^2-2+x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x-x^2+x^3}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x}{x+1}\)
c) \(N=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac+abc^2+abc}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac\left(1+bc+b\right)}\)
\(N=\frac{1+b}{b+1+bc}+\frac{bc}{1+bc+b}\)
\(N=\frac{1+b+bc}{b+1+bc}\)
\(N=1.\)
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 ??
\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}-\frac{a-2}{a^3+1}=\frac{\left(a+1\right)^2}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{a^3+1}\\ \)
\(=\frac{a^2+2a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{\left(a+1\right).\left(a^2-a+1\right)}\)
\(=\frac{a^2+2a+1-\left(a^2-a+1\right)-\left(a-2\right)}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2a+2}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2}{a^2-a+1}\)
a) Ta có:
B = \(\frac{1}{x+3}-\frac{x}{x-1}-\frac{4x}{x^2+2x-3}\)
=> B = \(\frac{x-1}{\left(x+3\right)\left(x-1\right)}-\frac{x\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4x}{\left(x+3\right)\left(x-1\right)}\)
=> B = \(\frac{\left(x-1\right)-x\left(x+3\right)-4x}{\left(x+3\right)\left(x-1\right)}\)
=> B = \(\frac{x-1-x^2-3x-4x}{\left(x+3\right)\left(x-1\right)}\)
=> B = \(\frac{-6x-1-x^2}{\left(x+3\right)\left(x-1\right)}\)
b) xem lại đề
\(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)
\(=\left[\left(\frac{1}{2}a+b\right)+\left(\frac{1}{2}a-b\right)\right].\left[\left(\frac{1}{2}a+b\right)^2-\left(\frac{1}{2}a+b\right).\left(\frac{1}{2}a-b\right)+\left(\frac{1}{2}a-b\right)^2\right]\)
Tới đấy phân tích ra là được nhé . Mình đang bận .
\(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)
\(=\left(\frac{1}{2}a+b+\frac{1}{2}a-b\right)\left[\left(\frac{1}{2}a+b\right)^2-\left(\frac{1}{2}a+b\right)\left(\frac{1}{2}a-b\right)+\left(\frac{1}{2}a-b\right)^2\right]\)
\(=a\left(\frac{1}{4}a^2+ab+b^2-\frac{1}{4}a^2+b^2+\frac{1}{4a^2}-ab+b^2\right)\)
\(=a\left(\frac{1}{4}a^2+3b^2\right)\)
\(=\frac{1}{4}a^3+3ab^2\)