10类RD Sharma解决方案–第8章二次方程式–练习8.3 |套装2

问题21.(16 / x)– 1 = 15 /(x + 1)，x≠0，-1。

解决方案：

We have equation,

(16/x) – 1 = 15/(x + 1)

(16 – x)/x = 15/(x + 1)

15x = (x + 1) (16 – x)

15x = 16x – x² + 16 – x

15x – 16x + x² -16 + x = 0

x² – 16 = 0

(x – 4) (x + 4) = 0

Therefore, roots of the equation are 4 or -4.

问题22.(x + 3)/(x + 2)=(3x – 7)/(2x – 3)，x≠-2，3/2。

解决方案：

We have equation,

(x + 3) (2x – 3) = (3x – 7) (x + 2)

2x² – 3x + 6x – 9 = 3x² + 6x -7x -14

2x² + 3x -9 = 3x² -x -14

x² -4x -5 = 0

We can factorize this equation as:

x² – 5x + x -5 = 0

x (x – 5) + 1 (x – 5) = 0

(x + 1) (x – 5) = 0

Therefore, roots of the equation are 5 or -1.

问题23.(2x /(x – 4))+((2x – 5)/(x – 3))= 25/3，x≠3，4

解决方案：

We have equation,

(2x/(x – 4)) + ((2x – 5)/(x – 3)) = 25/3

((2x)(x – 3) + (2x – 5) (x – 4))/((x – 4) (x – 3)) = 25/3

25x² – 175x + 300 = 12x² – 57x + 60

13x² – 118x + 240 = 0

We can factorize this equation as:

13x² – 78x – 40x + 240 = 0

13x (x – 6) – 40 (x – 6) = 0

(13x – 40) (x – 6) = 0

Therefore, roots of the equation are 6 or 40/13.

问题24.(((x + 3)/(x-2))-((1-x)/(x))= 17/4，x≠0，2

解决方案：

We have equation,

((x + 3)/(x – 2)) + ((1 – x)/(x)) = 17/4

8x² + 8 = 17x² – 34x

9x² – 34x – 8 = 0

We can factorize this equation as:

9x² – 36x + 2x – 8 = 0

9x(x – 4) + 2(x – 4) = 0

(9x + 2) (x – 4) = 0

Therefore, roots of the equation are 4 or -2/9.

问题25.(((x – 3)/(x + 3))–((x + 3)/(x – 3))= 48/7，x≠3，-3

解决方案：

We have equation,

((x – 3)/(x + 3)) – ((x + 3)/(x – 3)) = 48/7

-84x = 48x² – 432

48x² + 84x – 432 = 0

4x² + 7x -36 = 0

We can factorize this equation as:

4x² + 16x – 9x – 36 = 0

4x (x + 4) – 9 (x + 4) = 0

(4x – 9) (x + 4) = 0

Therefore, roots of the equation are -4 or 9/4.

问题26.(1 /(x – 2))+(2 /(x – 1))= 6 / x，x≠0

解决方案：

We have equation,

(1/(x – 2)) + (2/(x – 1)) = 6/x

3x² – 5x = 6x² – 18x + 12

3x² – 13x + 12 = 0

We can factorize this equation as:

3x² – 9x -4x + 12 = 0

3x (x – 3) – 4 (x – 3) = 0

(3x – 4) (x – 3) = 0

Therefore, roots of the equation are 3 or 4/3.

问题27.((x + 1)/(x-1))-((x-1)/(x + 1))= 5/6，x≠1，-1

解决方案：

We have equation,

((x + 1)/(x – 1)) – ((x – 1)/(x + 1)) = 5/6

5x² – 5 = 24x

5x² – 24x – 5 = 0

We can factorize this equation as:

5x² – 25x + x – 5 = 0

5x (x – 5) + 1 (x – 5) = 0

(5x + 1) (x – 5) = 0

Therefore, roots of the equation are 5 or -1/5.

问题28.(((x – 1)/(2x +1))+((2x +1)/(x – 1))= 5/2，x≠1，-1/2

解决方案：

We have equation,

((x – 1)/(2x + 1)) + ((2x + 1)/(x – 1)) = 5/2

2 (5x² + 2x + 2) = 5 (2x² – x – 1)

9x + 9 = 0

9 (x + 1) = 0

Therefore, roots of the equation are -1.

问题29.(4 / x)– 3 = 5 /(2x + 3)，x≠0，-3/2

解决方案：

We have equation,

(4/x) – 3) = 5/(2x + 3)   

5x = (2x + 3) (4 – 3x)

5x = 8x – 6x² + 12 – 9x

6x² – 6x -12 = 0

x² – x -2 = 0

We can factorize this equation as:

x² + 2x – x – 2 = 0

x (x + 2) -1 (x + 2) = 0

(x – 1) (x + 2) = 0

Therefore, roots of the equation are 1 or -2.

问题30.((x-4)/(x-5))+((x-6)/(x-7))= 10/3，x≠5，7

解决方案：

We have equation,

((x – 4)/(x – 5)) + ((x – 6)/(x – 7)) = 10/3

4x² – 54x + 176 = 0

2x² – 27x + 88 = 0

We can factorize this equation as:

2x² – 16x -11x + 88 = 0

2x (x – 8) – 11 (x – 8) = 0

(2x – 11) (x – 8) = 0

Therefore, roots of the equation are 8 or 11/2.

问题31.((x-2)/(x-3))+((x-4)/(x-5))= 10/3，x≠3，5。

解决方案：

We have equation,

((x – 2)/(x – 3)) + ((x – 4)/(x – 5)) = 10/3

We can rewrite it as :

((x – 3 + 1)/(x – 3)) + ((x – 5 + 1)/(x – 5)) = 10/3

1 + 1 + (1/(x – 3)) + (1/( x – 5)) = 10 3

4 (x² – 8x + 15) = 6x – 24

4x² – 38x + 84 = 0

2x² -19x + 42 = 0

We can factorize this equation as:

2x² – 12x – 7x + 42 = 0

2x (x – 6) – 7 (x – 6) = 0

(2x – 7) (x – 6) = 0

Therefore, roots of the equation are 6 or 7/2.

问题32.(((5 + x)/(5-x))-((5-x)/(5 + x))= 15/4，x≠5，-5。

解决方案：

We have equation,

(( 5 + x)/(5 – x)) – ((5 – x)/(5 + x)) = 15/4

80x = 375 – 15x²

15x² +80x -375 = 0

3x² +16x – 75 = 0

We can factorize this equation as:

3x² + 25x – 9x – 75 = 0

x (3x + 25) – 3 (3x + 25) = 0

(3x + 25) (x – 3) = 0

Therefore, roots of the equation are 3 or -25/3.

问题33.(3 /(x + 1))–(1/2)= 2 /(3x – 1)，x≠-1，1/3

解决方案：

We have equation,

(3/(x + 1)) – (1/2) = 2/(3x – 1)

2 (2x + 2) = (5 – x) (3x – 1)

4x + 4 = 15x – 5 – 3x²  + x

3x² – 12x + 9 = 0

x² – 4x + 3 = 0

We can factorize this equation as:

x² – 3x – x + 3 = 0

x (x – 3) – 1 (x – 3) = 0

(x – 3) (x – 1) = 0

Therefore, roots of the equation are 1 or 3.

问题34.(3 /(x + 1))+(4 /(x – 1))= 29 /(4x – 1)，x≠-1，1，1/4

解决方案：

We have equation,

(3/(x + 1)) + (4/(x – 1)) = 29/(4x – 1)

(7x + 1) (4x – 1) = 29 (x² – 1)

28x² – 7x + 4x – 1 = 29x² – 29

x² + 3x -28 = 0

We can factorize this equation as:

x² + 7x – 4x – 28 = 0

x (x + 7) – 4 (x + 7) = 0

(x – 4) (x + 7) = 0

Therefore, roots of the equation are 4 or -7.

问题35.(2 /(x + 1))+(3 /(2(x – 2)))= 23 / 5x，x≠0，-1，2

解决方案：

We have equation,

(2/(x + 1)) + (3/(2(x – 2))) = 23/5x

35x² – 25x = 46 (x² – x – 2)

11x² – 21x – 92 = 0

We can factorize this equation as:

11x² – 44x + 23x – 92 = 0

11x (x – 4) + 23 (x – 4) = 0

(11x + 23) ( x – 4) = 0

Therefore, roots of the equation are 4 or -23/11.

问题36. x ² –(√3+1)x +√3= 0

解决方案：

We have equation,

x² – (√3 + 1) x + √3 = 0

x² – √3x – x + √3 = 0

We can factorize this equation as:

x (x – √3) – 1 (x – √3) = 0

(x -1) (x – √3) = 0

Therefore, roots of the equation are 1 or √3.

问题37.3√5x ² + 25x –10√5= 0

解决方案：

We have equation,

3√5 x² + 25x – 10√5 = 0

√5 (3x² + (25/√5) x – (10√5/√5)) = 0

√5 (3x² + 5√5x – 10) = 0

We can factorize this equation as:

3x² – √5x + 6√5x – 10 = 0

x (3x – √5) + 2√5 (3x – √5) = 0

(x + 2√5) (3x – √5) = 0

Therefore, roots of the equation are -2√5 or √5/3.

问题^38.√3×2 – 2√2X – 2√3= 0

解决方案：

We have equation,

√3x² – 2√2 x – 2√3 = 0 

Here a = √3, b = -2√2 and c = -2√3

Since, Discriminant D = b² – 4ac and x = (-b ± √D)/2a

Therefore,

D = 8 + 24 = 32, and

x = (-(-2√2) ± √32)/2√3

x = ( 2√2 ± 4√2)/2√3

Therefore, roots of the equations are (2√2 + 4√2)/2√3 or (2√2 – 4√2)/2√3.

问题^39.4√3×2 + 5× – 2√3= 0

解决方案：

We have equation,

4√3x² + 5x – 2√3 = 0 

Here a = 4√3, b = 5 and c = -2√3

Since, Discriminant D = b² – 4ac and x = (-b ± √D)/2a

Therefore,

D = 25 + 96 = 121, and

x = (-(5) ± √121)/8√3

x = (-5 ± 11)/8√3

Therefore, roots of the equations are -2 /√3 or √3/ 4.

问题^40.√2×2 – 3× – 2√2= 0

解决方案：

We have equation,

√2x² – 3x – 2√2 = 0

Here a = √2, b = -3 and c = -2√2

Since, Discriminant D = b² – 4ac and x = (-b ± √D)/2a

Therefore,

D = 9 + 16 = 25, and

x = (-(-3) ± √25)/2√2

x = (3 ± 5)/2√2

Therefore, roots of the equations are 2√2 or -1/ √2.