\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2019}{3^{2019}}\)

\(\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{2019}{3^{2020}}\)

\(A-\frac{1}{3}A=\frac{2}{3}A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}-\frac{2019}{3^{2020}}\)

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\)

\(\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^{2020}}\)

\(B-\frac{1}{3}B=\frac{2}{3}B=\frac{1}{3}-\frac{1}{3^{2020}}\)\(\Rightarrow B=\frac{1}{2}-\frac{1}{3^{2019}.2}\)

\(\frac{2}{3}A=\frac{1}{2}-\frac{1}{3^{2019}.2}-\frac{2019}{3^{2020}}\)

\(A=\frac{3}{4}-\frac{1}{3^{2019}.4}-\frac{2019.3}{3^{2020}.2}\)

=>A max < 3/4=0,75

\(1.\)

\(\left|-0,75\right|+\frac{1}{4}-2\frac{1}{2}\)

\(=0,75+\frac{1}{4}-\frac{5}{2}\)

\(=\frac{3}{4}+\frac{1}{4}-\frac{10}{4}\)

\(=\frac{4}{4}-\frac{10}{4}\)

\(=\frac{-6}{4}=\frac{-3}{2}\)

\(2.\)

\(a,3\frac{1}{2}-\frac{1}{2}x=\frac{2}{3}\)

\(\frac{7}{2}-\frac{1}{2}x=\frac{2}{3}\)

\(\frac{1}{2}x=\frac{7}{2}-\frac{2}{3}\)

\(\frac{1}{2}x=\frac{17}{6}\)

\(x=\frac{17}{6}:\frac{1}{2}\)

\(x=\frac{17}{3}\)

Vậy x = \(\frac{17}{3}\)

\(b,3,2x+\left(-1,2\right)x+2,7\)\(=-4,9\)

\(x\cdot\left[3,2++\left(-1,2\right)\right]+2,7=-4,9\)

\(x\cdot2+2,7=-4,9\)

\(x\cdot2=-4,9-2,7\)

\(x\cdot2=-7,6\)

\(x=-7,6:2\)

\(x=-3,8\)

Vậy x=-3,8

\(3.\)

\(Có:y=f\left(x\right)\)\(=2x+\frac{1}{2}\)

\(\Rightarrow f\left(0\right)=2\cdot0+\frac{1}{2}\)\(=0+\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow f\left(1\right)=2\cdot1+\frac{1}{2}=2+\frac{1}{2}=\frac{4}{2}+\frac{1}{2}=\frac{5}{2}\)

\(\Rightarrow f\left(\frac{1}{2}\right)=2\cdot\frac{1}{2}+\frac{1}{2}\)\(=\frac{2}{2}+\frac{1}{2}=\frac{3}{2}\)

\(\Rightarrow f\left(-2\right)=2\cdot\left(-2\right)+\frac{1}{2}=-4+\frac{1}{2}=\frac{-8}{2}+\frac{1}{2}=\frac{-7}{2}\)

CA=CB

CA=CB

\(\frac{1}{3}.\frac{3}{5}+\frac{4}{5}.\frac{1}{3}+\frac{1}{3}.\frac{8}{5}\)

\(=\frac{1}{3}.\left(\frac{3}{5}+\frac{4}{5}+\frac{8}{5}\right)\)

\(=\frac{1}{3}.3\)

\(=1\)

Linz

Answer:

a. Xét tam giác ABC và tam giác DMC

CA = CD

CB = CM

Góc ACB = góc DCM

=> Tam giác ABC = tam giác DMC (c.g.c)

b. Từ chứng minh ở phần a) => Góc ABC = góc CDM hay góc BAD = góc ADM

Mà  hai góc ở vị trí so le trong

=> AB//MB

c. bạn thông cảm, ý này mình không biết làm ^^.

ta có 3/5*x-6/7=1/7

3/5*x=1/7+6/7

3/5*x=1

x=1/(3/5)

x=5/3