Tìm \(a,b,c\) sao cho
\(\begin{align}& a)\frac{4x-7}{{{x}^{2}}-3x+2}=\frac{a}{x-1}+\frac{b}{x-2} \\ & b)\frac{{{x}^{2}}+5}{{{x}^{3}}-3x-2}=\frac{a}{x-2}+\frac{b}{{{(x+1)}^{2}}} \\ \end{align}\)
A. \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
a = -3\\
b = 1
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
a = 1\\
b = 2
\end{array} \right.
\end{array}\)
B. \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
a = -3\\
b = 1
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
a = 1\\
b = – 2
\end{array} \right.
\end{array}\)
C. \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
a = 3\\
b = 1
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
a = 1\\
b = 2
\end{array} \right.
\end{array}\)
D. \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
a = 3\\
b = 1
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
a = 1\\
b = – 2
\end{array} \right.
\end{array}\)
Hướng dẫn Chọn đáp án là: D
Phương pháp giải:
Sử dụng kiến thức cộng các phân thức khác mẫu, phân tích đa thức thành nhân tử và đồng nhất hệ số.
Lời giải chi tiết:
\(\begin{array}{l}a)\frac{{4x – 7}}{{{x^2} – 3x + 2}} = \frac{a}{{x – 1}} + \frac{b}{{x – 2}}\\\,\,\,\,\,\frac{a}{{x – 1}} + \frac{b}{{x – 2}} = \frac{{a(x – 2) + b(x – 1)}}{{(x – 1)(x – 2)}} = \frac{{ax – 2a + bx – b}}{{{x^2} – 3x + 2}} = \frac{{(a + b)x – 2a – b}}{{{x^2} – 3x + 2}} = \frac{{4x – 7}}{{{x^2} – 3x + 2}}\\ \Rightarrow \left\{ \begin{array}{l}a + b = 4\\ – 2a – b = – 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = 1\end{array} \right.\end{array}\)
Vậy \(a=3,b=1\) thì \(\frac{4x-7}{{{x}^{2}}-3x+2}=\frac{a}{x-1}+\frac{b}{x-2}\).
\(b)\frac{{{x}^{2}}+5}{{{x}^{3}}-3x-2}=\frac{a}{x-2}+\frac{b}{{{(x+1)}^{2}}}\)
\(\begin{array}{l}\frac{a}{{x – 2}} + \frac{b}{{{{(x + 1)}^2}}} = \frac{{a({x^2} + 2x + 1) + b(x – 2)}}{{(x – 2){{(x + 1)}^2}}} = \frac{{a{x^2} + 2ax + a + bx – 2b}}{{(x – 2){{(x + 1)}^2}}} = \frac{{a{x^2} + (2a + b)x + a – 2b}}{{(x – 2){{(x + 1)}^2}}} = \frac{{{x^2} + 5}}{{{x^3} – 3x – 2}}\\ \Rightarrow \left\{ \begin{array}{l}a = 1\\2a + b = 0\\a – 2b = 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = – 2.\end{array} \right.\end{array}\)
Vậy \(a=1;b=-2\) thì \(\frac{{{x}^{2}}+5}{{{x}^{3}}-3x-2}=\frac{a}{x-2}+\frac{b}{{{(x+1)}^{2}}}\).