Tháng Ba 29, 2024

Tìm \(a,b,c\) sao cho \(\begin{align}& a)\frac{4x-7}{{{x}^{2}}-3x+2}=\frac{a}{x-1}+\frac{b}{x-2} \\ & b)\frac{{{x}^{2}}+5}{{{x}^{3}}-3x-2}=\frac{a}{x-2}+\frac{b}{{{(x+1)}^{2}}} \\ \end{align}\)

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 \(\frac{{{x}^{2}}+5}{{{x}^{3}}-3x-2}=\frac{a}{x-2}+\frac{b}{{{(x+1)}^{2}}}\).