8类RD Sharma解决方案–第3章平方根和平方根–练习3.2 |套装1

问题1.以下数字不是完美的平方。给出理由。

(i)1547

解决方案：

Number ending with 7 is not perfect square

(ii)45743

解决方案：

Number ending with 3 is not perfect square

(iii)8948

解决方案：

Number ending with 8 is not perfect square

(iv)333333

解决方案：

Number ending with 3 is not perfect square

问题2。证明以下数字不是完美的平方：

(i)9327

解决方案：

Number ending with 7 is not perfect square.

(ii)4058

解决方案：

Number ending with 8 is not perfect square

(iii)22453

解决方案：

Number ending with 3 is not perfect square

(iv)743522

解决方案：

Number ending with 2 is not perfect square

问题3.以下哪个数字的平方是旧数字?

(i)731

解决方案：

Square of an odd number is odd

731 is an odd number. Therefore, square of 731 is an odd number.

(ii)3456

解决方案：

Square of an even number is even

3456 is an even number. Therefore, square of 3456 is an even number.

(iii)5559

解决方案：

Square of an odd number is odd

5559 is an odd number. Therefore, square of 5559 is an odd number.

(iv)42008

解决方案：

Square of an even number is even

42008 is an even number. Therefore, square of 42008 is an even number.

问题4.以下数字的平方的单位位数是多少?

(i)52

解决方案：

Unit digit is 2

Therefore, unit digit of (52)² = (2²) = 4

(ii)977

解决方案：

Unit digit is 7

Therefore, unit digit of (977)² = (7²) = 49 = 9

(iii)4583

解决方案：

Unit digit is 3

Therefore, unit digit of (4583)² = (3²) = 9

(iv)78367

解决方案：

Unit digit is 7

Therefore, unit digit of (78367)² = (7²) = 49 = 9

(v)52698

解决方案：

Unit digit is 8

Therefore, unit digit of (52698)² = (8²) = 64 = 4

(vi)99880

解决方案：

Unit digit is 0

Therefore, unit digit of (99880)² = (0²) = 0

(vii)12796

解决方案：

Unit digit is 6

Therefore, unit digit of (12796)² =(6²) = 36 = 6

(viii)55555

解决方案：

Unit digit is 5

Therefore, unit digit of (55555)² =(5²) = 25 = 5

(ix)53924

解决方案：

Unit digit is 4

Therefore, unit digit of (53924)² =(4²) = 16 = 6

问题5.遵循以下模式
1 + 3 = 2 ²

1 + 3 + 5 = 3 ²

1 + 3 + 5 + 7 = 4 ²
并写出1 + 3 + 5 + 7 + 9 +………的值，最多n个项。

解决方案：

Number on the right-hand side is square of the number of terms present on the left-hand side.

1 + 3, These are two terms So, 1 + 3 = 2²

Therefore, The value of 1 + 3 + 5 + 7 + 9 +……… up to n terms = n² (as there are only n terms).

问题6：观察以下模式

2 ² – 1 ² = 2 +1

3 ² – 2 ² = 3 + 2

4 ² – 3 ² = 4 + 3

5 ² – 4 ² = 5 + 4

并找到价值

(i)100 ² – 99 ²

解决方案：

According to pattern right-hand side is the addition of two consecutive numbers on the left-hand side.

Therefore, 100² -99² =100 + 99 = 199

(ii)111 ² – 109 ²

解决方案：

According to pattern right-hand side is the addition og two numbers on the left-hand side.

But these two numbers are not consecutive

Therefore,

= (111² – 110²) + (110² – 109²)

= (111 + 110) + (100 + 109)

= 440

(iii)99 ² – 96 ²

解决方案：

According to pattern right-hand side is the addition og two numbers on the left-hand side.

But these two numbers are not consecutive

Therefore,

= 99² – 96²

= (99² – 98²) + (98² – 97²) + (97² – 96²)

= (99 + 98) + (98 + 97) + (97 + 96)

= 585

问题7.毕达哥拉斯有以下哪些三胞胎?

(i)(8、15、17)

解决方案：

(8, 15, 17)

As 17 is the largest number

LHS = 8² + 15²

= 289

RHS = 17²

       = 289

LHS = RHS

Therefore, the given triplet is a Pythagorean.
 

(ii)(18、80、82)

解决方案：

(18, 80, 82)

As 82 is the largest number

LHS = 18² + 80²

       = 6724

RHS = 82²

        = 6724

LHS = RHS

Therefore, the given triplet is a Pythagorean.

(iii)(14、48、51)

解决方案：

(14, 48, 51)

As 51 is the largest number

LHS = 14² + 48²

        = 2500

RHS = 51²

        = 2601

LHS ≠ RHS

Therefore, the given triplet is not a Pythagorean.

(iv)(10、24、26)

解决方案：

(10, 24, 26)

As 26 is the largest number

LHS = 10² + 24²

       = 676

RHS = 26²

       = 676

LHS = RHS

Therefore, the given triplet is a Pythagorean.

(v)(16、63、65)

解决方案：

(16, 63, 65)

As 65 is the largest number

LHS = 16² + 63²

       = 4225

RHS = 65²

       = 4225

LHS = RHS

Therefore, the given triplet is a Pythagorean.

(vi)(12、35、38)

解决方案：

(12, 35, 38)

As 38 is the largest number

LHS = 12² + 35²

       = 1369

RHS = 38²

        = 1444

LHS ≠RHS

Therefore, the given triplet is not a Pythagorean.

问题8.遵循以下模式

(1×2)+(2×3)=(2×3×4)/ 3

(1×2)+(2×3)+(3×4)=(3×4×5)/ 3

(1×2)+(2×3)+(3×4)+(4×5)=(4×5×6)/ 3

并找到价值

(1×2)+(2×3)+(3×4)+(4×5)+(5×6)

解决方案：

(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6) = (5 × 6 × 7)/3 = 70

问题9.遵循以下模式

1 = 1/2(1×(1 + 1))

1 + 2 = 1/2(2×(2 + 1))

1 + 2 + 3 = 1/2(3×(3 + 1))

1 + 2 + 3 + 4 = 1/2(4×(4 + 1))

并找到以下每个值：

(i)1 + 2 + 3 + 4 + 5 +…+ 50

解决方案：

R.H.S = 1/2 [No. of terms in L.H.S × (No. of terms + 1)] (if only when L.H.S starts with 1)

1 + 2 + 3 + 4 + 5 + … + 50 = 1/2 (5 × (5 + 1))

25 × 51 = 1275

(ii)31 + 32 +…。 + 50

解决方案：

R.H.S = 1/2 [No. of terms in L.H.S × (No. of terms + 1)] (if only when L.H.S starts with 1)

31 + 32 + …. + 50 = (1 + 2 + 3 + 4 + 5 + … + 50) – (1 + 2 + 3 + … + 30)

1275 – 1/2 (30 × (30 + 1))

1275 – 465

810

问题10：观察以下模式

1 ² = 1/6(1×(1 + 1)×(2×1 + 1))

1 ² +2 ² = 1/6(2×(2 + 1)×(2×2 + 1)))

1 ² +2 ² +3 ² = 1/6(3×(3 + 1)×(2×3 + 1)))

1 ² +2 ² +3 ² +4 ² = 1/6(4×(4 + 1)×(2×4 + 1)))

并找到以下每个值：

(i)1 ² + 2 ² + 3 ² + 4 ² +…+ 10 ²

解决方案：

RHS = 1/6 [(No. of terms in L.H.S) × (No. of terms + 1) × (2 × No. of terms + 1)]

 1² + 2² + 3² + 4² + … + 10² = 1/6 (10 × (10 + 1) × (2 × 10 + 1))

= 1/6 (2310)

= 385

(ii)5 ² + 6 ² + 7 ² + 8 ² + 9 ² + 10 ² + 11 ² + 12 ²

解决方案：

RHS = 1/6 [(No. of terms in L.H.S) × (No. of terms + 1) × (2 × No. of terms + 1)]

5² + 6² + 7² + 8² + 9² + 10² + 11² + 12² = 1² + 2² + 3² + … + 12² – (1²+2²+3²+4²)

1/6 (12×(12+1)×(2×12+1)) — 1/6 (4×(4+1)×(2×4+1))

= 650 – 30

= 620

第3章平方根和平方根–练习3.2 |套装2