Question

[Đi theo xe rác nhặt xác em yêu]

\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+......+\dfrac{3}{59.61}\)

\(\left(x^2-4\right)\left(6-2x\right)=0\)

\(\dfrac{3}{2}-2x^2=-\dfrac{1}{2}\)

So sánh

\(A=\dfrac{10^{100}+1}{10^{101}+1}\)và\(B=\dfrac{10^{101}+1}{10^{102}+1}\)

5³⁰và124¹⁰

27¹¹và81⁸

5³⁶và11²⁴

3³⁹ và 11²¹

Answer 1

[JakiNatsumi]

\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)

= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

=\(\dfrac{3}{2}.\dfrac{56}{305}\)

= \(\dfrac{78}{305}\)

Answer 2

[JakiNatsumi]

\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)

*Nếu \(x^2-4=0\)

⇒ x² = 4

⇒ x ∈ {2 ; -2}

*Nếu \(6-2x=0\)

⇒2x = 6

⇒ x = 6 : 2 = 3

Vậy x ∈ { -2 ; 2 ; 3 }