Math 10 Chapter 3 Lesson 1: Introduction to Equations
1. Summary of theory
1.1. Equation one hidden
– Equation of an unknown \(x\) is a variable containing proposition of the form:
\(f(x) = g(x)\)(1)
Where \(f(x), g(x)\) are expressions with the same variable \(x\). We call \(f(x)\) the left side, \(g(x)\) the right side of the equation.
– The defining condition of the equation is the condition of the variable x so that the expressions on both sides have meaning.
– If there is a number \(x_0\) satisfying the CONCLUSIONS and \(f(x_0)= g(x_0)\) is a true statement, then we say the number \(x_0\) has the correct solution to equation (1) or \(x_0). \) is a solution of equation (1). An equation may or may not have a solution. For example, \(2\) is a solution of the equation: \(2 = 3x – x^2\)
1.2. Equivalence equation
– Two equations
\({f_1}\left( x \right) = {g_1}\left( x \right)\) (1)
\({f_2}\left( x \right) = {g_2}\left( x \right)\) (2)
is called equivalent, the notation \({f_1}\left( x \right) = {g_1}\left( x \right)⇔ {f_2}\left( x \right) = {g_2}\left( x \right)\) if the solution sets of (1) and (2) are equal.
Theorem:
a) If \(h(x)\) is an expression that satisfies the COC of the equation \(f(x) = g(x)\) then
\(f(x) + h(x) = g(x) + h(x) \)\(⇔ f(x) = g(x)\)
b) If \(h(x)\) satisfies the COC and is different from \(0\) for all \(x\) that satisfies the COC, then
\(f(x).h(x) = g(x).h(x) ⇔ f(x) = g(x)\)
\(\dfrac{f(x)}{h(x)}=\dfrac{g(x)}{h(x)} ⇔ f(x) = g(x)\).
1.3. Consequence Equation
The equation \({f_2}\left( x \right) = {g_2}\left( x \right)\) is a corollary of the equation \({f_1}\left( x \right) = {g_1 }\left( x \right)\), symbol
\({f_1}\left( x \right) = {g_1}\left( x \right)\) \(\Rightarrow \)\({f_2}\left( x \right) = {g_2}\left( x \right)\)
if the solution set of the first equation is a subset of the set of solutions of the second equation.
Eg: \(2x = 3 – x \Rightarrow (x – 1)(x + 2) = 0\)
2. Illustrated exercise
2.1. Form 1: Find the definite condition of the equation
Solution method
– The defining condition of the equation includes the conditions for the value of \(f\left( x \right),\,\,g\left( x \right)\) to be determined and the conditions other (if required in the topic)
– Condition for expression

\(\sqrt {f\left( x \right)} \) defines as \(f\left( x \right) \ge 0\)

\(\frac{1}{{f\left( x \right)}}\) defines as \(f\left( x \right) \ne 0\)

\(\frac{1}{{\sqrt {f\left( x \right)} }}\) defines as \(f\left( x \right) > 0\)
Question 1: Find the definite condition of the following equation: \(x + \frac{7}{{{x^2} – 25}} = 8\)
Solution guide
a) The defining condition of the equation is \({x^2} – 5 \ne 0 \Leftrightarrow {x^2} \ne 25 \Leftrightarrow x \ne \pm 5.\)
Verse 2: Find the definite condition of the following equation and then derive its solution set:
a) \(7x + \sqrt {7x – 6} = 2\sqrt {6 – 7x} + 6\)
b) \(\sqrt { – {x^2} + 4x – 4} + 2{x^3} = 16\)
Solution guide:
a) The defining condition of the equation is \(\left\{ {\begin{array}{*{20}{c}}{7{\rm{x}} – 6 \ge 0}\\{6 – 7{\rm{x}} \ge 0}\end{array}} \right \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x \ge \frac{ 6}{7}}\\{x \le \frac{7}{6}}\end{array} \Leftrightarrow x = \frac{6}{7}} \right.\)
Try to enter the equation and find that \(x = \frac{6}{7}\) is satisfied
So the karmic set of the equation is \({\rm{S}} = \left\{ {\frac{6}{7}} \right\}.\)
b) The defining condition of the equation is \( – {x^2} + 4x – 4 \ge 0 \Leftrightarrow – {\left( {x – 2} \right)^2} \ge 0 \Leftrightarrow x = 2\)
Substitute \({\rm{x}} = 2\) to find that the equation is satisfied
So the karmic set of the equation is \({\rm{S}} = \left\{ 2 \right\}.\)
2.2. Form 2: Solve the equation by equivalence transformation and corollary
Solution method
To solve the equation we perform the transformations to bring back the equation equivalent to the given equation, which is simpler to solve it. Some commonly used transformations

Adding (subtracting) both sides of the equation without changing the defining conditions of the equation we get an equation equivalent to the given equation.

Multiplying (dividing) on both sides with a nonzero expression and without changing the defining conditions of the equation, we get an equation equivalent to the given equation.

Square both sides of the equation to get the corollary of the given equation.

Square both sides of the equation (both sides always have the same sign) we get an equation equivalent to the given equation.
Question 1: Find m so that the equation \(\dfrac{{2mx – 1}}{{x + 1}} = 3\) has a solution?
Solution guide
The specified condition \(x \ne – 1\) . Then
\(\dfrac{{2mx – 1}}{{x + 1}} = 3\)
\(\Leftrightarrow 2mx – 1 = 3x + 3 \)
\(\Leftrightarrow \left( {2m – 3} \right)x = 4{\rm{ }}(1)\)
Equation (1) has a solution if and only if \(2m – 3 \ne 0 \Leftrightarrow m \ne \dfrac{3}{2}\) .
The solution of (1) is \(x = \dfrac{4}{{2m – 3}}\) . This solution is the solution of the given equation when
\(\dfrac{4}{{2m – 3}} \ne – 1 \Leftrightarrow 4 \ne 3 – 2m\)
\(\Leftrightarrow m \ne – \dfrac{1}{2}\) .
So the given equation has a solution when \(m \ne \dfrac{3}{2}\) and \(m \ne – \dfrac{1}{2}\) .
3. Practice
3.1. Essay exercises
Question 1: Find the conditions for the following equation:
a) \(x + \frac{5}{{{4x^2} – 1}} = 9\)
b) \(2 – \sqrt {2 – x} = \sqrt {x – 1} \)
Verse 2: Find the definite condition of the following equation and then derive its solution set:
a) \(3x + \sqrt {3x – 2} = \sqrt {2 – 3x} + 2\)
b) \(\sqrt { – {x^2} + 4x – 4} + {x^3} = 8\)
Question 3: Find the solutions of the following equations:
a) \(2 + \frac{1}{{x – 2}} = \frac{3}{{{x^2} + x – 6}}\)
b) \(\frac{{{x^2}}}{{\sqrt {x – 4} }} = \frac{1}{{\sqrt {x – 4} }} – \sqrt {x – 4 } \)
Question 4: Find \(m\) so that the following pair of equations are equivalent
\(m{x^2} – 3\left( {m – 2} \right)x + m – 5 = 0\) (1) and \(\left( {m – 3} \right){x^ 2} – 6x + {m^2} +7 = 0\) (2)
3.2. Multiple choice exercises
Question 1: Find the number of solutions to the equations \(\frac{{{x^2}}}{{\sqrt {x – 2} }} = \frac{1}{{\sqrt {x – 2} }} – \ sqrt {x – 2} .\)
A. 1 unique solution
B. inexperienced.
C. 3 solutions
D. 5 solutions
Verse 2: Find the number of solutions to the equations \(\sqrt {\sqrt x – 1} ({x^2} – x – 2) = 0.\)
A. 1 unique solution
B. inexperienced.
C. 2 solutions
D. 5 solutions
Question 3: Find the definite condition of the equation \(\frac{5}{{{x^2} – x – 1}} = \sqrt[3]{x}.\)
A. \(x \ge 2\)
B. \(x \in \emptyset \)
C. \(\left\{ {\begin{array}{*{20}{c}}{x \ge 3}\\\begin{array}{l}x \ne 1\\x \ne 2\ end{array}\end{array}} \right.\)
D. \(x \ne \frac{{1 \pm \sqrt 5 }}{2}\)
Question 4: Find the definite condition of the equation \(1 + \sqrt {2x – 4} = \sqrt {2 – 4x} .\)
A. \(x \ge 2\)
B. \(x \in \emptyset \)
C. \(\left\{ {\begin{array}{*{20}{c}}{x \ge 3}\\\begin{array}{l}x \ne 1\\x \ne 2\ end{array}\end{array}} \right.\)
D. \(x \ne \frac{{1 \pm \sqrt 5 }}{2}\)
Question 5: Find the definite condition of the equation \(\sqrt { – {x^2} + x – 1} + x = 1.\)
A. \(x \ge \frac{3}{4}\)
B. \(x \in \emptyset \)
C. \(x = 2\)
D. \(\left[ {\begin{array}{*{20}{c}}{x = 1}\\{x = 2}\end{array}} \right.\)
4. Kết luận
Qua bài học này, các em cần nắm được những nội dung sau:
 Biết được ba loại phương trình: phương trình một ẩn, phương trình tương đương, phương trình hệ quả.
 Biết tìm điều kiện và giải phương trình.
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