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Training creative thinking for high school students through solving some Algebra - Analysis exercises in many ways - 3


Method 3 : Rewrite the equation: math equivalent to:

( x m ) 2 m 2 m 2. Requirements of the problem

m 2m20

m2

Maybe you are interested!

m 1

m 2 m 2

Training creative thinking for high school students through solving some Algebra - Analysis exercises in many ways - 3

m

1

m 2 m 2

m1

m2

m 1

m103m2

m3


2

x x 2

x

1 x

Example 2 : Solve the equation: 1

3

Condition: 0 x 1

(1)

2

x x 2

x

1x

2 2

Method 1 : (1)

1

3


4( x x 2 ) 6 x x 2 0

x x 2

4 x x 2 6 0

x x 2

0

x 0

3

x x 2

x 1

2

Comparing with the conditions, we get the solution of the equation as

x 1, x 0


Method 2 : Place

t


x

x x 2

, 1 t

1 x

t 2 1

2

t 2 1


2


t 1

The equation becomes: 1

3

tt2


x

1 x

x 0

1x1

x

1 x

Method 3 : Set a ; b ; a 0, b 0


1 2 ab a b


3 2 ab 3( a b )

We have: 3

( a b ) 2 2 ab 1

a 2b 21


ab1

ab 0

ab2



ab 3



(a, b do not exist)

2

a0

b 1 x 0


x

Method 4 : Place


sin ,

a 1

b 0

0

1 sin 2

2

x 1

1 sin 2

The equation becomes: 1 2 sin

3

sin

(sin c os ) 2 3(sin c os ) 2 0

sin c os 1

0

x 0

sin c os 2

2

x 1

Through the above example, we have many ways to solve irrational equations.

Example 3:( Question II.2-A2010) Solve the inequality


x x

1 2( x 2 x 1)

1


(1)


Method 1 : Equivalent transformation method


Comment:


1

2 x 2x 1

x

2( x 2 x 1)

1

1x 22( due tox10,x(0;1)

x

2( x 1 ) 2 3

2

2

3

2

10x


Therefore, there are conditions.

0

or 0 x 1

(2)


x

x

2( x 2 x 1)

From that (1) x 1

2( x 2 x 1)

x 2x1


x1

x

x

x

1x 22( due tox10,x(0;1)


x

x 22 x

x2

10

x

x

x1 20


x 1

x0


x

1

2

5 x 3 5

2


Comment x


best of (1)

3 5 satisfies condition (2) so this is the only solution.

2

Method 2 : Method of placing auxiliary variables In the solution of method 1, we can place


x

t to solve now we consider 1 way

set another variable. Similar to method 1, we have


2( x 2 x 1)

x

(1)x1


Divide both sides of the inequality by 𝑥 to get

2( x 1 1

x

1

x

x

(1)) 1


Put

y

1 , has x 1 y 2 2

x

x


x

2( y 2 1)

We have the inequality y 1

y 1 0

2( y 2 1) y 2 2 y 1


( y 1) 2 0

y 1


y1


With y = 1, we can solve

x35

2

Method 3 : Method using inequality


We have conditions

x (0;1)

and (1)

x1


2( x 2 x 1)

x

Applying Bunyakovsky inequality, we have

x

2

1 1.(1 x ) 1 2 1 2x21 x 2


Good

2( x 2 x 1)

x

x1



The sign “=" occurs if and only if

1 x

x

1 1


good

x10 or

x

x35

2

Method 4 : Using conjugate quantities

Arguable


2( x 2 x 1)

(1) 1

x 0;1


x

x

2 x 2x 1

x

2 x x10

x

2( x 2 3 x 1)

2 x 2x 1 2 x

x10



x

Comment:

x 2 3 x 1 ( x 1) 2 x x 1  x 1 therefore

x

2 x 2x 1 2 x

(1)x 1 (1

2 x

x 1


3

) 0


x

x1

x

0


2 x 2x 1 (2 2 x )

2 x 2x 1 2 x

2( x 23 x1)

x1

2 x 2

x1 2

x

2 x 2

x 1 2 2 x 0

x

x1 2

2 x1

x

2 x 2x 1 2

x

2 x 2x 1 2 2 x

0



Notice the factors under the denominator and

2 x 1 is always positive for all

x

x

x

x 0;1so we have (1) ( x 1) 2 0 x 1 0



Exercise:

x ( x


1 x 2 )

x 3 5

2

1.

xx1

1

x 2 x 3


Answer: x

5 1

2


3 x 4 x 2

2. x 2 2 x 1


Answer: x = 1 5

2


x 1

2( x 2 5 x 8)

3. x 3


Answer: x = 5


Example 4 : Solve the following system of equations:

x 9

y9

4 (1)

7 y

7 x

4 (2)



Solution:Conditions:


Method 1:

9x7

9y7

From the given system of equations, we deduce:


x 9

7 y

y 9

7 x

x 9

7 y

y 9

7 x

Consider the function:


(*)


2 7 t

f ( t )

t9

7 t

(9t7 )

2t 9​

f ' ( t )

1 1

0 ,

t(9;7)

It follows that the function f(t) is increasing on9; 7

From (*) we have

fxfyx y


Substitute into (1) and we get the equation:


x 9

7 x

4

x97x2 x9 7x16

x 22 x630

x7y7

x9y9 So the system of equations has a solution:

x 9 ; x 7

y 7

9y


Method 2:

System of equations

x 9

y9

7 y 4

7 x

4

(

(

x 9

y9

7 y ) 2 16

7 x ) 2 16


x9 7y

y9 7x

2 2

x97y2

x97y2

x9 7y

x y2

y9 7x

x y2

x9 7y16

y9 7x16

0

0

7 xxy639 y7 yxy639 x

given is equivalent to the following equation:

16 x 16 y x y

Substitute into (1) and we get the equation:


x 9

7 x

x9 7x

4x97x2

x 22 x63

0

16

x 22 x630x7

x 9

y 7

y 9

So the system of equations has a solution:

x 9 ; x 7

y 7


Method 3:

9y

Solving the system of equations, we consider the following cases: Case 1 : If x>y then we have:


Hence:

and

x 9

y 9

7 y

7 x

x 9

7 y

y 9

7 x


y9

But

x9

4 so 4 4 (absurd)

7 y

7 x

4


x 9

y 9

7 y

7 x

Case 2 : If x y then we have: and


x 9

Hence:


But

7 y

7 y

x 9

y9

y 9

7 x

7 x

4 so 4 4 (absurd)

4

Case 3: If x y, substitute into (1) we get the equation:


x 9

7 x

4


x9 7x

x97x2

x 22 x63

0

16

x 22 x630x7

x 9

y 7

y 9

So the system of equations has a solution:

x 9 ; x 7

y 7


x 9

7 y

y 9

7 x

Method 4:

9y

Let u

0, v

0, z

0, t0


u v 4 (1)

u 2 v 2 16 (3)

z t 4 (2)

We get the system of equations:

z 2v 2 16 (4)

From (1) and (2) we deduce u v z t

From (3) and (4) we can deduce that

(5)

u 2 t 2 z 2 v 2 u 2 v 2 z 2 t 2

( u v )(u v) (z t)(z t)


(6)

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