Math, asked by mahimachoudhary1978, 1 month ago

if two positive integers a and b are written as a=x⁵y². and b=x²y³( x and y are prime numbers) then HCF(a,b)is (a)xy. (b)x⁵y³. (c)x³y³. (d)x²y²​

Answers

Answered by bkbbrainlyuser
1

Step-by-step explanation:

The first equation given,

x-\frac{1}{x}=5x−

x

1

=5 …….. (i)

If we do the cube of equation (i) we get,

\left(x-\frac{1}{x}\right)^{3}=5^{3}(x−

x

1

)

3

=5

3

x^{3}-\left(\frac{1}{x}\right)^{3}-3 \times x \times\left(\frac{1}{x}\right) \times\left(x-\frac{1}{x}\right)=125x

3

−(

x

1

)

3

−3×x×(

x

1

)×(x−

x

1

)=125

x^{3}-\left(\frac{1}{x}\right)^{3}-3\times(x-\frac{1}{x})=125x

3

−(

x

1

)

3

−3×(x−

x

1

)=125

x^{3}-\left(\frac{1}{x}\right)^{3}-3 \times 5=125x

3

−(

x

1

)

3

−3×5=125

Grouping the terms,

x^{3}-\left(\frac{1}{x}\right)^{3}=125+15x

3

−(

x

1

)

3

=125+15

x^{3}-\left(\frac{1}{x}\right)^{3}=140x

3

−(

x

1

)

3

=140

Answered by piyush9695224208
1

Answer:

The actual length of path covered by moving body is called distance.

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