10^x : 5^y =20^y

=>\(2^{x^{ }}.5^x:5^y=4^9.5^9\)

=>\(2^x.5^{x.y}=2^{2y}.5^y\)

=>\(x=2y;x-y=y\)

=>\(x=2y\)

Vậy x=2y

1. \(\dfrac{5}{7}+\dfrac{2}{3}.x=\dfrac{3}{10}\)

<=>\(\dfrac{2}{3}.x=\dfrac{3}{10}-\dfrac{5}{7}=-\dfrac{29}{70}\)

<=>\(x=-\dfrac{29}{70}:\dfrac{2}{3}=-\dfrac{87}{140}\)

Vậy x=\(-\dfrac{87}{140}\)

2.

\(\dfrac{b+c}{bc}=\dfrac{2}{a}< =>\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{2}{a}\)

\(< =>\dfrac{1}{b}-\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{a}=0\)

\(< =>\dfrac{a-b}{ab}+\dfrac{a-c}{ac}=0\)

\(< =>\dfrac{a-b}{ab}=\dfrac{a-c}{ac}\)

\(< =>\dfrac{ab}{ac}=\dfrac{a-b}{c-a}< =>\dfrac{b}{c}=\dfrac{a-b}{c-a}\)

Ta thấy:

1.4 = 1.(1 + 3) = 1.(1 + 1 + 2) = 1.(1 + 1)+ 2.1

2.5 = 2.(2 + 3) = 2.(2 + 1 + 2) = 2.(2 + 1)+ 2.2

3.6 = 3.(3 + 3) = 3.(3 + 1 + 2) = 3.(3 + 1)+ 2.3

4.7 = 4.(4 + 3) = 4.(4 + 1 + 2) = 4.(4 + 1)+ 2.4

. . . . . . . . . . .

n(n + 3) = n(n + 1) + 2n

Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + . . . + n(n + 1) + 2n

= 1.2 + 2 +2.3 + 4 + 3.4 + 6 + . . . + n(n + 1) + 2n

= [1.2 +2.3 +3.4 + . . . + n(n + 1)] + (2 + 4 + 6 + . . . + 2n)

Mà 1.2 + 2.3 + 3.4 + … + n.(n + 1) = \(\dfrac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Và 2 + 4 + 6 + . . . + 2n = \(\dfrac{\left(2n+2\right).n}{2}\)

⇒C = \(\dfrac{n.\left(n+1\right).\left(n+2\right)}{3}+\dfrac{\left(2n+2\right).n}{2}-\dfrac{n.\left(n+1\right).\left(n+5\right)}{3}\)

\(\dfrac{\left(-2\right)^n}{-128}\)=4

Ta có:

(-2)ⁿ = -128 . 4 = -512

Mà -512 = (-2)⁹ => n=9

Vậy n=9

(\(\dfrac{1}{32}\))ⁿ . 16ⁿ = 1024^-1

<=> (\(\dfrac{1}{32}\) . 16)ⁿ = \(\dfrac{1}{1024}\)

<=> (\(\dfrac{1}{2}\))ⁿ = \(\dfrac{1}{1024}\)

<=> n=10

Vậy n=10