Prove that 1/x + 1/y + 1/z = 0.

If 2x = 3y = 6-z, prove that 1/x + 1/y + 1/z = 0.

Solution:

Consider

2x = 3y = 6-z = k

Here

2x = k

We can write it as

2 = (k)1/x

3y = k

We can write it as

3 = (k)1/y

6-z = k

We can write it as

6 = (k)-1/z

So we get

2 × 3 = 6

(k)1/x × (k)1/y = (k)-1/z

By further calculation

(k)1/x + 1/y = (k)-1/z

We get

1/x + 1/y = – 1/z

1/x + 1/y + 1/z = 0

Therefore, it is proved.

More Solutions:

Prove that aq – r. br – p. cp – q = 1.

If a = xyp – 1, b = xyq – 1, and c = xyr – 1, prove that aq – r. br – p. cp – q = 1.

Solution:

It is given that

a = xyp – 1

Here

aq – r = (xyb – 1)q – r = xq – r. y(q – r) (p – 1)

b = xyq – 1

Here

br – p = (xyq – 1)r – p = xr – p. y(q – 1) (r – p)

c = xyr – 1

Here

cp – q = (xyr – 1)p – q = xp – q. y(r – 1) (p – q)

Consider

LHS = aq – r. br – p. cp – q

Substituting the values

= xq – r. y(q – r) (p – 1). xr – p. y(q – 1) (r – p). xp – q. y(r – 1) (p – q)

By further calculation

= xq – r + r – p – q. y(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)

So we get

= x0. ypq – pr – q + r + qr – pr – r + p + rp – qr – p + q

= x0. y0

= 1 × 1

= 1

= RHS

More Solutions:

If a = cz, b = ax and c = by, prove that xyz = 1.

If a = cz, b = ax and c = by, prove that xyz = 1.

Solution:

It is given that

a = cz, b = ax and c = by

We can write it as

a = (by)z where c = by

So we get

a = byz

Here

a = (ax)yz

a1 = axyz

By comparing both

xyz = 1

Therefore, it is proved.

More Solutions:

Prove the following: (i) (a + b)-1 (a-1 + b-1) = 1/ab

Prove the following:

(i) (a + b)-1 (a-1 + b-1) = 1/ab

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 60

Solution:

(i) (a + b)-1 (a-1 + b-1) = 1/ab

Here

LHS = (a + b)-1 (a-1 + b-1)

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 61

= RHS

Hence, proved.

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 62

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 63

= xyz

= RHS

Hence, proved.

More Solutions:

Solve the following : (i) (a-1 + b-1) ÷ (a-2 – b-2)

(i) (a-1 + b-1) ÷ (a-2 – b-2)

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 54

Solution:

(i) (a-1 + b-1) ÷ (a-2 – b-2)

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 55

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 56

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 57

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 58

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 59

= 1

More Solutions:

Solve Indices:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 49

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 50

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 51

= x0

= 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 52

= x0

= 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 53

More Solutions:

Simplify for the following indices:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 45

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 46

We can write it as

= (xm – n)l. (xn – 1)m. (x1-m)n

By further calculation

= (x)(m – n)l. (x)(n – 1)m. (x)(l – m)n

= xml – nl. xnm – lm. xln – mn

So we get

= xml – nl + nm – lm + ln – mn

= x0

= 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 47

We can write it as

= (xa + b – c)a – b. (xb + c – a)b – c. (xc + a – b)c – a

By further calculation

= x(a + b – c) (a – b). x(b + c – a) (b – c). x(c + a – b) (c – a)

So we getML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 48

= x0

= 1

More Solutions:

Evaluate for the following indices:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 41

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 42

= (3)1

= 3

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 43

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 44

More Solutions:

Solve indices: 7(2n+3) -49(n+2)/343(n+1)pow(2/3)

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 36

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 37

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 38

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 39

= 7 – 7 × 7

= 7 – 49

= – 42

(ii) (27)4/3 + (32)0.8 + (0.8)-1

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 40

= 98.25

More Solutions:

Evaluate following indices:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 31

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 32

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 33

= 4

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 34

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 35

= 56x – 2 – 6x

= 5-2

= 1/(5)2

= 1/25

More Solutions: