8类RD Sharma解决方案-第六章代数表达式和恒等式-练习6.2

问题1.添加以下代数表达式：

(i)3a ² b，-4a ² b，9a ² b

解决方案：

3a²b, -4a²b, 9a²b
Now we have add the given expression
= 3a²b + (-4a²b) + 9a²b
= 3a²b – 4a²b + 9a²b
= 8a²b

(ii)2 / 3a，3 / 5a，-6 / 5a

解决方案：

We have to add the given expression
2/3a + 3/5a + (-6/5a)
2/3a + 3/5a – 6/5a
Now take LCM for 3 and 5 which will be 15
= (2×5)/(3×5)a + (3×3)/(5×3)a – (6×3)/(5×3)a
= 10/15a + 9/15a – 18/15a
= (10a+9a-18a)/15
= a/15

(iii)4xy ² – 7x ² y，12x ² y -6xy ² ，-3x ² y + 5xy ²

解决方案：

We have to add the given expression
4xy² – 7x²y + 12x²y – 6xy² – 3x²y + 5xy²
Now rearrange the expression:
12x²y – 3x²y – 7x²y – 6xy² + 5xy² + 4xy
3xy² + 2x²y

(iv)3 / 2a – 5 / 4b + 2 / 5c，2 / 3a – 7 / 2b + 7 / 2c，5 / 3a + 5 / 2b – 5 / 4c

解决方案：

3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
Now add the given expression
3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c
rearrange 
3/2a + 2/3a + 5/3a – 5/4b – 7/2b + 5/2b + 2/5c + 7/2c – 5/4c
Now take LCM of (2 and 3 is 6), (4 and 2 is 4), (5,2 and 4 is 20)
(9a+4a+10a)/6 + (-5b-14b+10b)/4 + (8c+70c-25c)/20
23a/6 – 9b/4 + 53c/20

(v)11 / 2xy + 12 / 5y + 13 / 7x，-11 / 2y – 12 / 5x – 137xy

解决方案：

11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 13/7xy
Now add the given expression
11/2xy + 12/5y + 13/7x + -11/2y – 12/5x – 13/7xy
Now rearrange

11/2xy – 13/7xy + 13/7x – 12/5x + 12/5y -11/2y
Now take LCM for (2 and 7 is 14), (7 and 5 is 35), (5 and 2 is 10)
(11xy-12xy)/14 + (65x-84x)/35 + (24y-55y)/10
51xy/14 – 19x/35 – 31y/10

(vi)7 / 2x ³ – 1 / 2x ³ + 5/3，3 / 2x ³ + 7 / 4x ² – x + 1/3，3 / 2x ² -5 / 2x -2

解决方案：

Now add the given expression
7/2x³ – 1/2x² + 5/3 + 3/2x³ + 7/4x² – x + 1/3 + 3/2x² -5/2x – 2
Now rearrange
=7/2x³ + 3/2x³ – 1/2x² + 7/4x² + 3/2x² – x – 5/2x + 5/3 + 1/3 – 2
=10/2x³ + 11/4x² – 7/2x + 0/6
=5x³ + 11/4x² -7/2x

问题2.减去：

(i)从12xy到-5xy

解决方案：

Subtract the given expression
= 12xy – (- 5xy)
= 5xy + 12xy
= 17xy

(ii)从-7a ² 2a ²上

解决方案：

Subtract the given expression

= (-7a²) – 2a²
= -7a² – 2a²
= -9a²

(iii)3a-5b中的2a-b

解决方案：

Subtract the given expression
=(3a – 5b) – (2a – b)
= 3a – 5b – 2a + b
= a – 4b

( iv)2x ³ – 4x ² + 3x + 5从4x ³ + x ² + x + 6

解决方案：

Subtract the given expression

(4x³ + x² + x + 6) – (2x³ – 4x² + 3x + 5)
4x³ + x² + x + 6 – 2x³ + 4x² – 3x – 5

2x³ + 5x² – 2x + 1

(v)2 / 3y ³ – 2 / 7y ² – 5 from 1 / 3y ³ + 5 / 7y ² + y – 2

解决方案：

Subtract the given expression
1/3y³ + 5/7y² + y – 2 – 2/3y³ + 2/7y² + 5
On rearranging,
1/3y³ – 2/3y³ + 5/7y² + 2/7y² + y – 2 + 5
We will group similar expression:
= -1/3y³ + 7/7y² + y + 3
= -1/3y³ + y² + y + 3

(vi)3 / 2x – 5 / 4y – 7 / 2z从2 / 3x + 3 / 2y – 4 / 3z

解决方案：

Subtract the given expression
2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z)
On rearranging,
2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z
We will group similar expression:
LCM of (3 and 2 is 6), (2 and 4 is 4), (3 and 2 is 6)
=(4x-9x)/6 + (6y+5y)/4 + (-8z+21z)/6
= -5x/6 + 11y/4 + 13z/6

(vii)x ² y – 4 / 5xy ² + 4 / 3xy从2 / 3x2y + 3 / 2xy ² – 1 / 3xy

解决方案：

Subtract the given expression
2/3x²y + 3/2xy² – 1/3xy – (x²y – 4/5xy² + 4/3xy)
on rearrange
2/3x²y – x²y + 3/2xy² + 4/5xy² – 1/3xy – 4/3xy
We will group similar expression:
LCM of (3 and 1 is 3), (2 and 5 is 10), (3 and 3 is 3)
-1/3x²y + 23/10xy² – 5/3xy

(viii)ab / 7 – 35 / 3bc + 6 / 5ac从3 / 5bc – 4 / 5ac

解决方案：

Subtract the given expression
3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac)
On rearrange
3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7
We will group similar expression:
LCM of (5 and 3 is 15), (5 and 5 is 5)
(9bc+175bc)/15 + (-4ac-6ac)/5 – ab/7
184bc/15 + -10ac/5 – ab/7
– ab/7 + 184bc/15 – 2ac

问题3：带走：

^(ⅰ)6/5×2 – ^4/5×3 + 5/6 + 3/2×从^×3/3 – ^5/2×2 + 3/5×+ 1/4

解决方案：

Subtract the given expression
1/3x³ – 5/2x² + 3/5x + 1/4 – (6/5x² – 4/5x³ + 5/6 + 3/2x)
On rearrange
1/3x³ + 4/5x³ – 5/2x² – 6/5x² + 3/5x – 3/2x + 1/4 – 5/6
By grouping similar expressions we get,
LCM of (3 and 5 is 15), (2 and 5 is 10), (5 and 2 is 10), (4 and 6 is 24)
17/15x³ – 37/10x² – 9/10x – 14/24
17/15x³ – 37/10x² – 9/10x – 7/12

(ⅱ)5A ^2/2 + 3A ^3/2 + A / 3 – 6/5 1 / 3A ³ – 3 / 4A ² – 5/2

解决方案：

Subtract the given expression
1/3a³ – 3/4a² – 5/2 – (5/2a² + 3/2a³ + a/3 – 6/5)
On rearrange
1/3a⁵ – 3/2a³ – 3/4a² – 5/2a² – a/3 – 5/2 + 6/5
By grouping similar expressions we get,
LCM of (3 and 2 is 6), (4 and 2 is 4), (2 and 5 is 10)
= (2a³ – 9a³)/6 – (3a² + 10a²)/4 – a/3 + (-25+12)/10
= -7/6a³ – 13/4a² – a/3 – 13/10

^(III)7/4×3 + ^3/5×2 + 1 / 2X + 9/2 7/2从- X / 3 – X ^4/10

解决方案：

Subtract the given expression

7/2 – x/3 – 1/5x² – (7/4x³ + 3/5x² + 1/2x + 9/2)

On rearranging,

-7/4x³ – 1/5x² – 3/5x² – x/3 – x/2 + 7/2 – 9/2

By grouping similar expressions we get,

LCM of (3 and 2 is 6)

-7/4x³ – 4/5x²– (2x-3x)/6 + (7-9)/2

-7/4x³ – 4/5x² – 5/6x – 1

(iv)y ^3/3 + 7 / 3y ² + 1 / 2y + 1/2从1/3 – 5 / 3y ²

解决方案：

Subtract the given expression
1/3 – 5/3y² – (1/3y² + 7/3y² + 1/2y + 1/2)
On rearrange
-1/3y³ – 5/3y² – 7/3y² – 1/2y + 1/3 – 1/2
By grouping similar expressions we get,
LCM of (3 and 3 is 3), (3 and 2 is 6)
-1/3y³ + (-5y² – 7y²)/3 – 1/2y + (2-3)/6 
-1/3y³ – 12/3y² – 1/2y – 1/6

(v)2 / 3ac – 5 / 7ab + 2 / 3bc从3 / 2ab -7 / 4ac – 5 / 6bc

解决方案：

Subtract the given expression
3/2ab – 7/4ac – 5/6bc – (2/3ac – 5/7ab + 2/3bc)
On rearrange
3/2ab + 5/7ab – 7/4ac – 2/3ac – 5/6bc – 2/3bc
By grouping similar expressions we get,
LCM of (2 and 7 is 14), (4 and 3 is 12), (6 and 3 is 6)
(21ab+10ab)/14 – (21ac-8ac)/12 – (5bc-4bc)/6
31/14ab – 29/12ac – 3/2bc

问题4.从x – 3y + 2z和-4x + 9y – 11z的总和中减去3x – 4y – 7z。

解决方案：

First we will find the sum:
The sum of x – 3y + 2z and -4x + 9y – 11z is
(x – 3y + 2z) + (-4x + 9y – 11z)
On rearrange
x – 4x – 3y + 9y + 2z – 11z
= -3x + 6y – 9z
Now Let’s subtract it from -3x + 6y – 9z
(-3x + 6y – 9z) – (3x – 4y – 7z)
On rearranging again 
= -3x – 3x + 6y + 4y – 9z + 7z
= -6x + 10y – 2z

问题5.从9l + 2m – 3n ²和-3l + m + 4n ²的总和中减去3l – 4m – 7n ²和2l + 3m – 4n ²的总和。

解决方案：

Sum of 3l – 4m – 7n² and 2l + 5m – 4n²
3l – 4m – 7n² + 2l + 3m – 4n²
On rearrange
3l + 2l – 4m + 3m – 7n² – 4n²
5l – m – 11n² ……………………..eq. (1)
Sum of 9l + 2m – 3n² and -3l + m + 4n²
9l + 2m – 3n² + (-3l + m + 4n²)
On rearrange
9l – 3l + 2m + m – 3n² + 4n²
6l + 3m + n² ……………………….eq. (2)
Let us subtract equ (i) from (ii), we get
6l + 3m + n² – (5l – m – 11n²)
On rearrange
6l – 5l + 3m + m + n2 + 11n2
l + 4m + 12n²

问题6.从5中减去2x – x ² + 5和-4x – 3 + 7x ^{2的总和。}

解决方案：

Sum of 2x – x² + 5 and -4x – 3 + 7x² is
2x – x² + 5 + (-4x – 3 + 7x²)
2x – x² + 5 – 4x – 3 + 7x²
On rearrange
– x² + 7x² + 2x – 4x + 5 – 3
6x² -2x + 2 …………eq (i)
Let subtract eq (i) from 5 we will get,
5 – (6x² -2x + 2)
5 – 6x² + 2x – 2
3 + 2x – 6x²

问题7.简化以下各项：

(i)x ² – 3x + 5 – 1/2(3x ² – 5x + 7)

解决方案：

x² – 3x + 5 – 1/2(3x² – 5x + 7)
On rearrange
x² – 3/2x² – 3x + 5/2x + 5 – 7/2
We will group similar expression:
LCM of (1 and 2 is 2)
= (2x² – 3x²)/2 – (6x + 5x)/2 + (10-7)/2
= -1/2x² – 1/2x + 3/2

(ii)[5 – 3x + 2y –(2x – y)] –(3x – 7y + 9)

解决方案：

5 – 3x + 2y – 2x + y – 3x + 7y – 9
On rearrange
= – 3x – 2x – 3x + 2y + y + 7y + 5 – 9
We will group similar expression:
= -8x + 10y – 4

(iii)11 / 2x ² y – 9 / 4xy ² + 1 / 4xy – 1 / 14y ² x + 1 / 15yx ² + 1 / 2xy

解决方案：

On rearrange
11/2x²y + 1/15x²y – 9/4xy² – 1/14xy² + 1/4xy + 1/2xy
We will group similar expression:
LCM of (2 and 15 is 30), (4 and 14 is 56), (4 and 2 is 4)
= (165x²y + 2x²y)/30 + (-126xy² – 4xy²)/56 + (xy + 2xy)/4 
= 167/30x²y – 130/56xy² + 3/4xy
= 167/30x²y – 65/28xy² + 3/4xy

(iv)(1 / 3y ² – 4 / 7y + 11)–(1 / 7y – 3 + 2y ² )–(2 / 7y – 2 / 3y ² + 2)

解决方案：

On rearrange
1/3y² – 2y² – 2/3y² – 4/7y – 1/7y – 2/7y + 11 + 3 – 2
We will group similar expression:
LCM of (3, 1 and 3 is 3), (7, 7 and 7 is 7)
= (y² – 6y² + 2y²)/3 – (4y – y – 2y)/7 + 12
= -3/3y² – 7/7y + 12
= -y² – y + 12

(v)-1 / 2a ² b ² c + 1 / 3ab ² c – 1 / 4abc ² – 1 / 5cb ² a ² + 1 / 6cb ² a – 1 / 7c ² ab + 1 / 8ca ² b

解决方案：

On rearrange
-1/2a²b²c – 1/5a²b²c + 1/3ab²c + 1/6ab²c – 1/4abc² – 1/7abc² + 1/8a²bc
We will group similar expression:
LCM of (2 and 5 is 10), (3 and 6 is 6), (4 and 7 is 28)
-7/10a²b²c + 1/2ab²c – 11/28abc² + 1/8a²bc