第9章一个变量中的线性方程式–练习9.3 |套装1
问题13(7x-2)/(5x-1)=(7x + 3)/(5x + 4)
解决方案:
Given:
(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
(7x – 2) / (5x – 1) – (7x +3)/(5x + 4) = 0
By taking LCM as (5x – 1) (5x + 4)
((7x-2) (5x+4) – (7x+3)(5x-1)) / (5x – 1) (5x + 4) = 0
After cross-multiplying we will get,
(7x-2) (5x+4) – (7x+3)(5x-1) = 0
Now after simplification,
35x2 + 28x – 10x – 8 – 35x2 + 7x – 15x + 3 = 0
10x – 5 = 0
10x = 5
x = 5/10
x = 1/2
Now verify the given equation by substituting x = 1/2,
(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
x = 1/2
(7(1/2) – 2) / (5(1/2) – 1) = (7(1/2) + 3) /(5(1/2) + 4)
(7/2 – 2) / (5/2 – 1) = (7/2 + 3) / (5/2 + 4)
((7-4)/2) / ((5-2)/2) = ((7+6)/2) / ((5+8)/2)
(3/2) / (3/2) = (13/2) / (13/2)
1 = 1
Here, L.H.S. = R.H.S.,
Thus the given equation is verified.
问题14.((x + 1)/(x + 2)) 2 =(x + 2)/(x + 4)
解决方案:
Given:
((x+1)/(x+2))2 = (x+2) / (x + 4)
(x+1)2 / (x+2)2 – (x+2) / (x + 4) = 0
By taking LCM as (x+2)2 (x+4)
((x+1)2 (x+4) – (x+2) (x+2)2) / (x+2)2 (x+4) = 0
After cross-multiplying we will get,
(x+1)2 (x+4) – (x+2) (x+2)2 = 0
Now expand the equation as follows,
(x2 + 2x + 1) (x + 4) – (x + 2) (x2 + 4x + 4) = 0
x3 + 2x2 + x + 4x2 + 8x + 4 – (x3 + 4x2 + 4x + 2x2 + 8x + 8) = 0
x3 + 2x2 + x + 4x2 + 8x + 4 – x3 – 4x2 – 4x – 2x2 – 8x – 8 = 0
-3x – 4 = 0
x = -4/3
Now verify the given equation by substituting x = -4/3,
((x+1)/(x+2))2 = (x+2) / (x + 4)
x = -4/3
(x+1)2 / (x+2)2 = (x+2) / (x + 4)
(-4/3 + 1)2 / (-4/3 + 2)2 = (-4/3 + 2) / (-4/3 + 4)
((-4+3)/3)2 / ((-4+6)/3)2 = ((-4+6)/3) / ((-4+12)/3)
(-1/3)2 / (2/3)2 = (2/3) / (8/3)
1/9 / 4/9 = 2/3 / 8/3
1/4 = 2/8
1/4 = 1/4
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题15((x + 1)/(x-4)) 2 =(x + 8)/(x-2)
解决方案:
Given:
((x+1)/(x-4))2 = (x+8)/(x-2)
(x+1)2 / (x-4)2 – (x+8) / (x-2) = 0
By taking LCM as (x-4)2 (x-2)
((x+1)2 (x-2) – (x+8) (x-4)2) / (x-4)2 (x-2) = 0
After cross-multiplying we will get,
(x+1)2 (x-2) – (x+8) (x-4)2 = 0
After expansion we get,
(x2 + 2x + 1) (x-2) – ((x+8) (x2 – 8x + 16)) = 0
x3 + 2x2 + x – 2x2 – 4x – 2 – (x3 – 8x2 + 16x + 8x2 – 64x + 128) = 0
x3 + 2x2 + x – 2x2 – 4x – 2 – x3 + 8x2 – 16x – 8x2 + 64x – 128 = 0
45x – 130 = 0
x = 130/45
x = 26/9
Now verify the given equation by substituting x = 26/9
((x+1)/(x-4))2 = (x+8)/(x-2)
(x+1)2 / (x-4)2 = (x+8) / (x-2)
x = 26/9
(26/9 + 1)2 / (26/9 – 4)2 = (26/9 + 8) / (26/9 – 2)
((26+9)/9)2 / ((26-36)/9)2 = ((26+72)/9) / ((26-18)/9)
(35/9)2 / (-10/9)2 = (98/9) / (8/9)
(35/-10)2 = (98/8)
(7/2)2 = 49/4
49/4 = 49/4
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题16.(9x-7)/(3x + 5)=(3x-4)/(x + 6)
解决方案:
Given:
(9x-7)/(3x+5) = (3x-4)/(x+6)
(9x-7)/(3x+5) – (3x-4)/(x+6) = 0
By taking LCM as (3x+5) (x+6)
((9x-7) (x+6) – (3x-4) (3x+5)) / (3x+5) (x+6) = 0
After cross-multiplying we will get,
(9x-7) (x+6) – (3x-4) (3x+5) = 0
Upon expansion we will get,
9x2 + 54x – 7x – 42 – (9x2 + 15x – 12x – 20) = 0
44x – 22 = 0
44x = 22
x = 22/44
= 2/4
x = 1/2
Now verify the given equation by substituting x =1/2,
(9x-7)/(3x+5) = (3x-4)/(x+6)
x = 1/2
(9(1/2) – 7) / (3(1/2) + 5) = (3(1/2) – 4) / ((1/2) + 6)
(9/2 – 7) / (3/2 + 5) = (3/2 – 4) / (1/2 + 6)
((9-14)/2) / ((3+10)/2) = ((3-8)/2) / ((1+12)/2)
-5/2 / 13/2 = -5/2 / 13/2
-5/13 = -5/13
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题17.(x + 2)/(x + 5)= x /(x + 6)
解决方案:
Given:
(x+2)/(x+5) = x/(x+6)
(x+2)/(x+5) – x/(x+6) = 0
By taking LCM as (x+5) (x+6)
((x+2) (x+6) – x(x+5)) / (x+5) (x+6) = 0
After cross-multiplying we will get,
(x+2) (x+6) – x(x+5) = 0
Upon expansion we will get
x2 + 8x + 12 – x2 – 5x = 0
3x + 12 = 0
3x = -12
x = -12/3
x = -4
Now verify the given equation by substituting x = -4,
(x+2)/(x+5) = x/(x+6)
x = -4
(-4 + 2) / (-4 + 5) = -4 / (-4 + 6)
-2/1 = -4 / (2)
-2 = -2
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题18:2x –(7-5x)/ 9x –(3 + 4x)= 7/6
解决方案:
Given:
2x – (7-5x) / 9x – (3+4x) = 7/6
(2x – 7 + 5x) / (9x – 3 – 4x) = 7/6
(7x – 7) / (5x – 3) = 7/6
After cross-multiplying we will get,
6(7x – 7) = 7(5x – 3)
42x – 42 = 35x – 21
42x – 35x = -21 + 42
7x = 21
x = 21/7
x = 3
Now verify the given equation by substituting
2x – (7-5x) / 9x – (3+4x) = 7/6
(7x – 7) / (5x – 3) = 7/6
x = 3
(7(3) -7) / (5(3) – 3) = 7/6
(21-7) / (15-3) = 7/6
14/12 = 7/6
7/6 = 7/6
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题19.(15(2-x)– 5(x + 6))/(1-3x)= 10
解决方案:
Given:
15(2-x) – 5(x+6) / (1-3x) = 10
(30-15x) – (5x + 30) / (1-3x) = 10
After cross-multiplying we will get,
(30-15x) – (5x + 30) = 10(1- 3x)
30- 15x – 5x – 30 = 10 – 30x
30- 15x – 5x – 30 + 30x = 10
10x = 10
x = 10/10
x = 1
Now verify the given equation by substituting x =1,
(15(2-x) – 5(x+6)) / (1-3x) = 10
x = 1
(15(2-1) – 5(1+6)) / (1- 3) = 10
(15 – 5(7))/-2 = 10
(15-35)/-2 = 10
-20/-2 = 10
10 = 10
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题20.(x + 3)/(x-3)+(x + 2)/(x-2)= 2
解决方案:
Given:
(x+3)/(x-3) + (x+2)/(x-2) = 2
By taking LCM as (x-3) (x-2)
((x+3)(x-2) + (x+2) (x-3)) / (x-3) (x-2) = 2
After cross-multiplying we will get,
(x+3)(x-2) + (x+2) (x-3) = 2 ((x-3) (x-2))
Upon expansion we will get,
x2 + 3x – 2x – 6 + x2 – 3x + 2x – 6 = 2(x2 – 3x – 2x + 6)
2x2 – 12 = 2x2 – 10x + 12
2x2 – 2x2 + 10x = 12 + 12
10x = 24
x = 24/10
x = 12/5
Now verify the given equation by substituting x = 12/5,
(x+3)/(x-3) + (x+2)/(x-2) = 2
x = 12/5
(12/5 + 3)/(12/5 – 3) + (12/5 + 2)/(12/5 – 2) = 2
((12+15)/5)/((12-15)/5) + ((12+10)/5)/((12-10)/5) = 2
(27/5)/(-3/5) + (22/5)/(2/5) = 2
-27/3 + 22/2 = 2
((-27×2) + (22×3))/6 = 2
(-54 + 66)/6 = 2
12/6 = 2
2 = 2
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题21.((x + 2)(2x-3)-2x 2 + 6)/(x-5)= 2
解决方案:
We have,
((x+2) (2x-3) – 2x2 + 6)/(x-5) = 2
After cross-multiplying we will get,
(x+2) (2x-3) – 2x2 + 6) = 2(x-5)
2x2 – 3x + 4x – 6 – 2x2 + 6 = 2x – 10
x = 2x – 10
x – 2x = -10
-x = -10
x = 10
Now verify the given equation by substituting x = 10
((x+2) (2x-3) – 2x2 + 6)/(x-5) = 2
x = 10
((10+2) (2(10) – 3) – 2(10)2 + 6)/ (10-5) = 2
(12(17) – 200 + 6)/5 = 2
(204 – 194)/5 = 2
10/5 = 2
2 = 2
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题22.(x 2 –(x + 1)(x + 2))/(5x + 1)= 6
解决方案:
Given:
(x2 – (x+1) (x+2))/(5x+1) = 6
After cross-multiplying we will get,
(x2 – (x+1) (x+2)) = 6(5x+1)
x2 – x2 – 2x – x – 2 = 30x + 6
-3x – 2 = 30x + 6
30x + 3x = -2 – 6
33x = -8
x = -8/33
Now verify the given equation by substituting x = -8/33
(x2 – (x+1) (x+2))/(5x+1) = 6
x = -8/33
((-8/33)2 – ((-8/33)+1) (-8/33 + 2))/(5(-8/33)+1) = 6
(64/1089 – ((-8+33)/33) ((-8+66)/33)) / (-40+33)/33) = 6
(64/1089 – (25/33) (58/33)) / (-7/33) = 6
(64/1089 – 1450/1089) / (-7/33) = 6
((64-1450)/1089 / (-7/33)) = 6
-1386/1089 × 33/-7 = 6
1386 × 33 / 1089 × -7 = 6
6 = 6
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题23(((2x + 3)–(5x-7))/(6x + 11)= -8/3
解决方案:
Given:
((2x+3) – (5x-7))/(6x+11) = -8/3
After cross-multiplying we will get,
3((2x+3) – (5x-7)) = -8(6x+11)
3(2x + 3 – 5x + 7) = -48x – 88
3(-3x + 10) = -48x – 88
-9x + 30 = -48x – 88
-9x + 48x = -88 – 30
39x = -118
x = -118/39
Now verify the given equation by substituting x = -118/39
((2x+3) – (5x-7))/(6x+11) = -8/3
x = -118/39
((2(-118/39) + 3) – (5(-118/39) – 7)) / (6(-118/39) + 11) = -8/3
((-336/39 + 3) – (-590/39 – 7)) / (-708/39 + 11) = -8/3
(((-336+117)/39) – ((-590-273)/39)) / ((-708+429)/39) = -8/3
(-219+863)/39 / (-279)/39 = -8/3
644/-279 = -8/3
-8/3 = -8/3
Here, L.H.S. = R.H.S.,
Thus, the given equation is verified.
问题24.找到满足给定方程的x的正值:
(i)(x 2 – 9)/(5 + x 2 )= -5/9
解决方案:
Given:
(x2 – 9)/(5+x2) = -5/9
After cross-multiplying we will get,
9(x2 – 9) = -5(5+x2)
9x2 – 81 = -25 – 5x2
9x2 + 5x2 = -25 + 81
14x2 = 56
x2 = 56/14
x2 = 4
x = √4
x = 2
(ii)(y 2 + 4)/(3y 2 + 7)= 1/2
解决方案:
Given:
(y2 + 4)/(3y2 + 7) = 1/2
After cross-multiplying we will get,
2(y2 + 4) = 1(3y2 + 7)
2y2 + 8 = 3y2 + 7
3y2 – 2y2 = 7 – 8
y2 = -1
y = √-1
= 1