\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Rightarrow\frac{a}{2}=\frac{2b}{6}=\frac{c}{4}=\frac{a+2b+c}{12}\)

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a+b-c}{2+3-4}=a+b-c\)

\(\Rightarrow\frac{a+2b+c}{12}=a+b-c\Rightarrow H=\frac{a+2b+c}{a+b-c}=12\)

\(5x-2\left(x+3\right)=4\left(x-1\right)\)

\(\Rightarrow5x-2x-6=4x-4\)

\(\Rightarrow3x-6=4x-4\)

\(\Rightarrow4x-3x=-6+4\)

\(\Rightarrow x=-2\)

∫(5x+1)⋅dx(x+3)(x+2)(x−4)

=12⋅ln(x−4)+32⋅ln(x+2)−2ln(x+3)+C

Explanation:

I decomposed integrand into basic fractions,

5x+1(x+3)(x+2)(x−4)

=Ax+3+Bx+2+Cx−4

After expanding denominator,

A(x+2)(x−4)+B(x+3)(x−4)+C(x+3)(x+2)=5x+1

After setting x=−3, 7A=−14, so A=−2

After setting x=−2, −6B=−9, so B=32

After setting x=4, 42C=21, so C=12

Thus,

∫(5x+1)⋅dx(x+3)(x+2)(x−4)

=−2⋅∫dxx+3+32⋅∫dxx+2+12⋅∫dxx−4

=12⋅ln(x−4)+32⋅ln(x+2)−2ln(x+3)+C

nhấn vào tên bạn rồi làm j nữa

rồi đọc trong phần giới thiệu của mik nhé

a)1/1.2+1/2.3+1/3.4+....+1/1999.2000

=1-1/2+1/2-1/3+1/3-1/4+....+1/1999-1/2000

=1-1/2000

= Bn tự tính

b)=1/3.(1/1.4+1/4.7+1/7.10+...+1/100+103)

=1/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)

=1/3.(1-1/103)

= tự làm

c)8/9-1/72-1/56-1/42-...-1/6-1/2

=8/9-(1/2+1/6+...+1/42+1/56+1/72)

=làm tương tự phần trên. Gợi ý :72=8.9 . Nói đến thế r mà ko bt làm thì chịu. yên tâm, đảm bảo đ, t học đội tuyển mà

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2000-1999}{1999.2000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)

\(=1-\frac{1}{2000}=\frac{1999}{2000}\)

b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

\(=\frac{1}{3}\left(\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{103-100}{100.103}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{103}\right)\)

\(=\frac{34}{103}\)

c) \(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{8.9}+\frac{1}{7.8}+...+\frac{1}{2.3}+\frac{1}{1.2}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=0\)

Ta có: \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right).n.\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right).n.\left(n+1\right)}=\frac{1}{2}\left[\frac{1}{\left(n-1\right).n}-\frac{1}{n\left(n+1\right)}\right]\)

Áp dụng ta được: 

\(A< \frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{5.6}+\frac{1}{5.6}-\frac{1}{6.7}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{20.21}\right)\)

\(< \frac{1}{2}.\frac{1}{4.5}=\frac{1}{40}\)