Hình như bạn chép sai đề, mình sửa nhé :

\(S=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{26.31}\\ =>\dfrac{S}{5}=\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{26.31}\\ =>\dfrac{S}{5}=\dfrac{6-1}{1.6}+\dfrac{11-6}{6.11}+\dfrac{16-11}{11.16}+...+\dfrac{31-26}{26.31}\\ =>\dfrac{S}{5}=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{26}-\dfrac{1}{31}=1-\dfrac{1}{31}=\dfrac{30}{31}\\ =>S=\dfrac{30}{31}.5=\dfrac{150}{31}\)

\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right).x=\dfrac{1}{5}\\ =>\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right).x=\dfrac{1}{5}\\ =>\left(\dfrac{1}{2}-\dfrac{1}{100}\right).x=\dfrac{1}{5}\\ =>\dfrac{49}{100}.x=\dfrac{1}{5}\\ =>x=\dfrac{1}{5}:\dfrac{49}{100}=\dfrac{1}{5}.\dfrac{100}{49}\\ =>x=\dfrac{20}{49}\)

ai cứu mik với

thì  a vuông góc với c nha

a) Do AC ⊥ AB

BD ⊥ AB

⇒ AC // BD

b) Do AC // BD

⇒ ∠ACD = ∠CDx = 60⁰ (so le trong)

Ta có:

∠ACK + ∠ACD = 180⁰ (kề bù)

⇒ ACK = 180⁰ - ACD

= 180⁰ - 60⁰

= 120⁰

Ta có:

  \(3^{100}=\left(3^2\right)^{50}=9^{50}\) 

Mà \(8^{50}< 9^{50}\)

Vậy \(8^{50}< 3^{100}\)

8⁵⁰=(8²)²⁵=64²⁵

3¹⁰⁰=(3⁴)²⁵=81²⁵

Mà 64²⁵<81²⁵

Vậy 8⁵⁰<3¹⁰⁰

Bạn cần ghi đầy đủ điều kiện về n cũng như yêu cầu đề bài để được hỗ trợ tốt hơn.

\(\left(x-1\right)^3-\left(\dfrac{2}{2023}-\dfrac{7}{247}+\dfrac{1}{8}\right)=\dfrac{7}{247}-\dfrac{2}{2023}\)

\(\Rightarrow\left(x-1\right)^3-\dfrac{2}{2023}+\dfrac{7}{247}-\dfrac{1}{8}=\dfrac{7}{247}-\dfrac{2}{2023}\)

\(\Rightarrow\left(x-1\right)^3=\dfrac{7}{247}-\dfrac{7}{247}-\dfrac{2}{2023}+\dfrac{2}{2023}+\dfrac{1}{8}\)

\(\Rightarrow\left(x-1\right)^3=\dfrac{1}{8}\)

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

\(\Rightarrow x-1=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}+1\)

\(\Rightarrow x=\dfrac{3}{2}\)

Lời gải:

$(x-1)^3=\frac{7}{247}-\frac{2}{2023}+\frac{2}{2023}-\frac{7}{247}+\frac{1}{8}=\frac{1}{8}$

$x-1=\frac{1}{2}$

$x=\frac{1}{2}+1=\frac{3}{2}$