## Math 8 Chapter 2 Lesson 9: Transformation of Rational Expressions and Values of Fractions

## 1. Theoretical Summary

### 1.1. Rational expression

Each expression is a fraction or represents a sequence of operations: addition, subtraction, multiplication, and division on fractions. We call such expressions rational expressions.

### 1.2. Convert a rational expression to a fraction

Thanks to the rules of addition, subtraction, multiplication, and division of fractions, we can transform a rational expression into a fraction.

### 1.3. The value of the fraction

When doing problems related to the value of fractions, we must first find the condition of the variable so that the corresponding value of the denominator is different from \(0\). That is the condition for the value of the fraction to be determined.

## 2. Illustrated exercise

### 2.1. Exercise 1

Convert the following expression into a fraction

\(B = \dfrac{{1 + \dfrac{2}{{x – 1}}}}{{1 + \dfrac{{2x}}{{{x^2} + 1}}}}\)

**Solution guide**

\(B = \dfrac{{1 + \dfrac{2}{{x – 1}}}}{{1 + \dfrac{{2x}}{{{x^2} + 1}}}}\)

\(\eqalign{& \Rightarrow B = \left( {1 + {2 \over {x – 1}}} \right):\left( {1 + {{2x} \over {{x^2} + 1}}} \right) \cr & = \left( {{{x – 1} \over {x – 1}} + {2 \over {x – 1}}} \right):\left( {{ {{x^2} + 1} \over {{x^2} + 1}} + {{2x} \over {{x^2} + 1}}} \right) \cr & = {{x – 1 + 2} \over {x – 1}}:{{{x^2} + 1 + 2x} \over {{x^2} + 1}} \cr& = {{x +1} \over {x – 1}}:{{(x+1)^2} \over {{x^2} + 1}} \cr & = {{x + 1} \over {x – 1}}. {{{x ^2} + 1} \over {{{(x + 1)}^2}}} \cr & = {{(x+1).({x^2} + 1)} \over {\left( {x – 1} \right).(x+1)^2}} \cr &= {{{x^2} + 1} \over {\left( {x – 1} \right)\left( { x + 1} \right)}}=\frac{{{x^2} + 1}}{{{x^2} – 1}} \cr} \)

### 2.2. Exercise 2

Given the fraction \(\dfrac{{x + 1}}{{{x^2} + x}}\)

a) Find the condition of \(x\) so that the value of the fraction is determined

b) Calculate the value of the fraction at \(x = 1 000 000\) and at \(x = – 1\).

**Solution guide**

a) We have: \({x^2} + x = x\left( {x + 1} \right)\)

This fraction value is determined with the condition \({x^2} + x ≠ 0\)

\( \Rightarrow x\left( {x + 1} \right) \ne 0\)

\(\Rightarrow x \ne 0 \) and \(x+1 \ne 0\)

\(\Rightarrow x \ne 0 \) and \(x \ne -1\)

b) With the condition \(x\ne 0, x\ne -1\). We have:

\(\dfrac{{x + 1}}{{{x^2} + x}} = \dfrac{{x + 1}}{{x\left( {x + 1} \right)}} = \ dfrac{1}{x}\)

At \(x = 1000000 \) (which satisfies the condition), we have:

\(\dfrac{{x + 1}}{{{x^2} + x}} = \dfrac{1}{x} = \dfrac{1}{{1000000}}\)

At \(x = – 1 \) does not satisfy the condition so the given fraction is undefined.

So there is no value of the fraction at \(x = -1.\)

### 2.3. Exercise 3

Calculate the value of expression A at x=-8

\(A = \frac{{3{x^2} – x}}{{9{x^2} – 6x + 1}}\)

**Solution guide**

We have:

\(\begin{array}{l} A = \frac{{3{x^2} – x}}{{9{x^2} – 6x + 1}}\\ {\rm{ }} = \ frac{{x\left( {3x – 1} \right)}}{{{\left( {3x – 1} \right)}^2}}} \end{array}\)

DKXD:

\(x \ne \frac{1}{3}\)

In \(x = – 8\) We have:

\(\begin{array}{l} \frac{x}{{3x – 1}}\\ = \frac{{ – 8}}{{3.\left( { – 8} \right) – 1} }\\ = \frac{8}{{25}} \end{array}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Convert the following expressions into fractions

a) \(\displaystyle {1 \over 2} + \displaystyle {x \over {1 – \displaystyle {x \over {x + 2}}}}\)

b) \(\displaystyle {{x – \displaystyle {1 \over {{x^2}}}} \over {x + \displaystyle {1 \over x} + {1 \over {{x^2}} }}}\)

**Verse 2: **Do the following calculations:

a) \(\displaystyle \left( {{{5x + y} \over {{x^2} – 5xy}} + {{5x – y} \over {{x^2} + 5xy}}} \right )\)\(.\displaystyle {{{x^2} – 25{y^2}} \over {{x^2} + {y^2}}}\)

b) \(\displaystyle {{4xy} \over {{y^2} – {x^2}}}\)\(:\displaystyle \left( {{1 \over {{x^2} + 2xy + {y^2}}} – {1 \over {{x^2} – {y^2}}}} \right)\)

c) \(\displaystyle \left[ {{1 \over {{{\left( {2x – y} \right)}^2}}} + {2 \over {4{x^2} – {y^2}}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right]\)\(. \displaystyle {{4{x^2} + 4xy + {y^2}} \over {16x}}\)

d) \(\displaystyle \left( {{2 \over {x + 2}} – {4 \over {{x^2} + 4x + 4}}} \right)\)\(:\displaystyle \left ( {{2 \over {{x^2} – 4}} + {1 \over {2 – x}}} \right)\)

**Question 3: **Find the condition of the variable so that the value of the fraction is determined:

a) \(\displaystyle {{5{x^2} – 4x + 2} \over {20}}\)

b) \(\displaystyle {8 \over {x + 2004}}\)

c) \(\displaystyle {{4x} \over {3x – 7}}\)

d) \(\displaystyle {{{x^2}} \over {x + z}}\)

**Question 4:** Factorize the denominators of the following fractions and then find the condition of \(x\) so that the value of the fraction is determined:

a) \(\displaystyle {5 \over {2x – 3{x^2}}}\)

b) \(\displaystyle {{2x} \over {8{x^3} + 12{x^2} + 6x + 1}}\)

### 3.2. Multiple choice exercises

**Question 1: **The result of performing the calculation \((2x+1-\frac{1}{1-2x}):(2x-\frac{4x^{2}}{2x-1})\) is:

A. 1-2x

B. 2x

C. -2x

D. 1+2x

**Verse 2: **Calculation result \((\frac{5x+y}{x^{2}-5xy}+\frac{5x-y}{x^{2}+5xy}).\frac{x^{ 2}-25y^{2}}{x^{2}+y^{2}}\) is:

A. \(\frac{x}{2}\)

B. \(\frac{5}{x}\)

C. \(\frac{10}{x}\)

D. Another answer

**Question 3: **The expression \(\frac{{x + \frac{1}{{{x^2}}}}}{{1 – \frac{1}{x} + \frac{1}{{{x^2) }}}}}\) is converted to the algebraic fraction of

A. \(\frac{1}{{x + 1}}\)

B. x + 1

C. x – 1

D. \(\frac{1}{{x – 1}}\)

**Question 4: **Given the expression \(B = \left( {\frac{1}{{x + 2}} – \frac{{2x}}{{4 – {x^2}}} + \frac{1}{{) 2 + x}}} \right).\left( {\frac{2}{x} – 1} \right)\). For what value of x does B determine

A. \(x \ne \left\{ { 0;2} \right\}\)

B. \(x \ne \left\{ { – 2;0;2} \right\}\)

C. \(x \ne \left\{ { – 2;2} \right\}\)

D. \(x \ne \left\{ { 0; -2} \right\}\)

**Question 5: **Given the expression \(B = \left( {\frac{1}{{x + 2}} – \frac{{2x}}{{4 – {x^2}}} + \frac{1}{{) 2 + x}}} \right).\left( {\frac{2}{x} – 1} \right)\). Find x so that \(B = \frac{1}{2}\)

A. x = 10

B. x = -10

C. x = -6

D. x = 6

## 4. Conclusion

Through this lesson, you will learn some of the main topics as follows:

- Understand the concepts of rational expressions, know that every fraction and every polynomial are rational expressions
- Perform math operations in the expression to turn it into an algebraic expression

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