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ĐK: x#0; x#-1
\(\frac{x^4}{1-x}\)+ x3 + x2 + 1
= \(\frac{x^4}{1-x}\)+ \(\frac{x^3\left(1-x\right)}{1-x}\)+ \(\frac{x^2\left(1-x\right)}{1-x}\)+ \(\frac{1-x}{1-x}\)
= \(\frac{x^4+x^3-x^4+x^2-x^3+1-x}{1-x}\)
= \(\frac{x+1}{1-x}\)
- Muốn cộng hai phân thức cùng mẫu, ta cộng các tử với nhau và giữ nguyên mẫu.
- Muốn cộng hai phân thức khác mẫu, ta quy đồng mẫu thức rồi cộng các phân thức cùng mẫu vừa tìm được.
\(\dfrac{3x}{x^3-1}+\dfrac{x-1}{x^2+x+1}\)
\(=\dfrac{3x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+x^2-2x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{1}{x-1}\)
a) \(\frac{1}{x^2-x+1}+1-\frac{x^2+2}{x^3+1}\)
+) Đkxđ: \(\hept{\begin{cases}x^2-x+1\ne0\\x^3+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\ne0\\x^3\ne-1\end{cases}\Leftrightarrow}\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\left(lđ\right)\\x\ne-1\end{cases}}}\)
+) \(A=\frac{1}{x^2-x+1}+1-\frac{x^2+2}{x^3+1}\)
\(=\frac{1}{x^2-x+1}+1-\frac{x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x+1+x^3+1-x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^3-x^2+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
P/s: ko chắc
Huhu luoi qua
a) \(\frac{1}{x^2-x+1}+1-\frac{x^2+2}{x^3+1}\)
\(=\frac{1}{x^2-x+1}+1-\left(\frac{x^2+2}{x^3+1}\right)\)
\(=\frac{x^5-2x^4+3x^3-2x^2+x}{x^5-x^4+x^3+x^2-x+1}\)
\(=\frac{x\left(x^4-2x^3+3x^2-2x+1\right)}{\left(x+1\right)\left(x^4-2x^3+3x^2-2x+1\right)}\)
\(=\frac{x}{x+1}\)
b) \(\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x^2+6x}\)
\(=\frac{-x^2+7x+78}{x^2+6x}\)
\(=\frac{\left(-x-6\right)\left(x-13\right)}{x\left(x+6\right)}\)
\(=\frac{-x+13}{x}\)
a,$\frac{5}{2x^2y}+\frac{3}{5xy^2}+\frac{x}{y^3}$52x2y +35xy2 +xy3
b,\(\frac{x+1}{2x+6}+\frac{2x+3}{x\left(x+3\right)}\)
Cộng vào sẽ ra kết quả nha !!!
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) \(N=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac+abc^2+abc}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac\left(1+bc+b\right)}\)
\(N=\frac{1+b}{b+1+bc}+\frac{bc}{1+bc+b}\)
\(N=\frac{1+b+bc}{b+1+bc}\)
\(N=1.\)