A = 19,20

A Thuoc { 19,20 }

2/3x = -7/12 - 1/4 

2/3x = -5/6

x      = -5/6 : 2/3

x      = -5/4

mk mới hok lớp 5 thôi

khó quá đấy

bố trí một cái

 3/4 + 1/4 : x = -3

1/4 : x = -3 - 3/4

1/4 : x = -15/4

        x = -1/15

\(|x+2|=5\)

\(\Rightarrow x+2=\pm5\)

\(\Rightarrow x+2=5\)hoặc    \(x+2=-5\)

    x=5-2                                x=-5-2

   x=3                                   x=-7

     Vậy x = 3 ; x=-7

Chúc bn hk tốt !

bạn làm 2 trương hợp 

tường hợp 1:

x+3=5 =>x=2

trường hợp 2:

x+3=-5=>x=-7

chắc bạn phai biết làm chứ

\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)

\(3x.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\right)=\frac{1}{21}\)

\(3x.\frac{1}{7}=\frac{1}{21}\)

\(\frac{3}{7}x=\frac{1}{21}\)

\(x=\frac{1}{21}:\frac{3}{7}\)

\(x=\frac{7}{81}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\)

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2015.2016.2017}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...-\frac{1}{2016.2017}\)

\(=\frac{1}{2}-\frac{1}{2016.2017}< \frac{1}{2}\)

\(\Rightarrow2A< \frac{1}{2}\Rightarrow A< \frac{1}{4}\)

4/5.31/2-4/5.46/3

=4/5.(31/2-46/3)

=4/5.(93/6-92/6)

=4/5.1/6

=4/30=2/15

k cho tui na

\(\frac{4}{5}\cdot\frac{31}{2}-\frac{4}{5}\cdot\frac{46}{3}\)

\(=\frac{4}{5}\cdot\left(\frac{31}{2}-\frac{46}{3}\right)\)

\(=\frac{4}{5}\cdot\left(\frac{93}{6}-\frac{92}{6}\right)\)

\(=\frac{4}{5}\cdot\frac{1}{6}\)

\(=\frac{4.1}{5\cdot6}=\frac{4}{30}=\frac{4:2}{30:2}=\frac{2}{15}\)

hình bn tự vẽ nha 

       a, Có \(\widehat{BAC}+\widehat{CAM}=\widehat{BAM}\)

mà \(\widehat{BAC}=40^0;\widehat{BAM}=80^0\)

\(\Rightarrow40^0+\widehat{CAM}=80^0\)\(\Rightarrow\widehat{CAM}=40^0\)

\(\Rightarrow\widehat{CAM}=\widehat{BAC}\)=> AC là tia phân giác góc BAM

b, Có AX là tia phân giác góc BAC 

=> CAX =XAB=1/2 ^BAC 

Mà góc BAC=\(40^0\)=> XAB =\(40^0.\frac{1}{2}=20^0\)

có ^XAC+^CAM=^MAX

mà ^XAC=20*;^CAM=40*

=> ^MAX=20*+40*

=> ^MAX=60*

Vậy ^MAX=60*

Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)

           \(B=75\%=\frac{3}{4}\)

Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)

                \(=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\right)< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)

                                                                                            \(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2018}=\frac{3}{4}-\frac{1}{2018}< \frac{3}{4}\)

\(\Rightarrow A< B\)

Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)

           \(B=75\%=\frac{3}{4}\)

Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)

                \(=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\right)< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)

                                                                                            \(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2018}=\frac{3}{4}-\frac{1}{2018}< \frac{3}{4}\)

\(\Rightarrow A< B\)