\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{19.21}\right).x=\frac{9}{7}\)

\(\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{21-19}{19.21}\right).x=\frac{9}{7}\)

\(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)

 \(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)

 \(\left[\frac{1}{2}.\frac{2}{7}\right].x=\frac{9}{7}\)

\(\frac{1}{7}.x=\frac{9}{7}\)

\(\Rightarrow x=\frac{9}{7}\div\frac{1}{7}=9\)

\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)

\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)x=\frac{9}{7}\)

\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{21}\right)x=\frac{9}{7}\)

\(\Leftrightarrow\frac{2}{7}x=\frac{9}{7}\)

\(\Leftrightarrow x=\frac{9}{2}\)

2 ( 1/3.5 + 1/5.7 + 1/7.9+......+ 1/19.21) . x = 2.9/7

 (2/3.5 + 2/5.7 + 2/7.9+......+ 2/19.21). x = 18/7

(1/3-1/5 + 1/5-1/7+ 1/7-1/9 +.....+ 1/19 - 1/21).x = 18/7

(1/3-1/21) . x= 18/7

2/7 x = 18/7

x= 18/7 . 7/2

x= 9

\(c.x\left(\frac{2}{5}-\frac{3}{5}\right)=\frac{2}{35}\)

\(x=\frac{2}{35}:\frac{-1}{5}=-\frac{2}{7}\)

\(d.\left(2x+1\right)^2=49=7^2=\left(-7\right)^2\)

\(TH1:2x+1=7\Rightarrow x=3\)

\(TH2=2x+1=-7\Rightarrow x=-4\)

\(a.x=\frac{-3}{5}-\frac{4}{9}=\frac{-47}{45}\)

\(b.\frac{3}{5}:x=\frac{17}{10}-\frac{2}{5}\)

\(x=\frac{3}{5}:\frac{13}{10}=\frac{6}{13}\)

Mk lộn ở dòng a

\(\frac{4}{9}+x=\frac{-5}{3}\) mới đúng

Nhân 2 cả 2 vế lên:

\(\left(2x+\frac{2}{1x3}\right)+...+\left(2x+\frac{2}{23x25}\right)=22x+\frac{2}{3}+\frac{2}{9}+\frac{2}{81}+\frac{2}{243}\)2/243

\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)

\(24x+\frac{24}{25}=22x+\frac{224}{243}\)

\(2x=\frac{224}{243}-\frac{24}{25}\)

\(2x=-\frac{232}{6025}\)

\(x=\frac{-116}{6075}\)

\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}\right)\)

\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)

\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)

\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)

\(12x+\frac{12}{25}=11x+\frac{112}{243}\)

\(11x-12x=\frac{112}{243}-\frac{12}{25}\)

\(-1x=-\frac{116}{6075}\)

\(x=-\frac{116}{6075}\div\left(-1\right)\)

\(x=\frac{116}{6075}\)

\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}=\frac{x+3}{2008}\)

\(\Leftrightarrow\frac{x+4}{2007}=\frac{x+1}{2010}\)

\(\Leftrightarrow\left(x+4\right)2010=\left(x+1\right)2007\)

\(\Leftrightarrow2010x+8040=2007x+2007\)

\(\Leftrightarrow2010x-2007x=2007-8040\)

\(\Leftrightarrow3x=-6033\)

\(\Leftrightarrow x=-2011\)

\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}+\frac{x+3}{2008}\)

=>\(\left(\frac{x\text{+4}}{2007}+1\right)+\left(\frac{x+8}{2003}+1\right)=\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+3}{2008}+1\right)\)

=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}=\frac{x+2011}{2010}+\frac{x+2011}{2008}\)

=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}-\frac{x+2011}{2010}-\frac{x+2011}{2008}=0\)

=>\(x+2011\left(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\right)=0\)

Mà \(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\ne0\)

=> x+2011=0

=>x=-2011

Vậy x = -2011

 x-1/65-1-x-3/63-1+x-5/61-1+x-7/59-1                                                                                                                                                    x-66/65-x-66/63+x-66/61+x-66/59 =0                                                                                                                                                   suy ra (x-66).(1/65-1/63+1/61+1/59)=0                                                                                                                                                  vi 1/65-1/63+1/61+1/59khong thuoc 0                                                                                                                                                            nen x-66+66=0                                                                                                                                                                                              suy ra x =132

a-b=2.(a+b) tương đương a-b =2a + 2b tương đương -3b=a

a-b=a.b suy ra -3b-b=-3b.b tương đương -4b=-3b.b tương đương b=4/3 suy ra a=-4

với a=-4 ; b=4/3 thì a-b = 2.(a+b)= a.b

\(\frac{-5}{x}=\frac{-y}{8}=\frac{18}{72}\)

\(\Leftrightarrow\frac{-5}{x}=\frac{-y}{8}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{x}=\frac{1}{4}\\-\frac{y}{8}=\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5.4:1\\-y=8.1:4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-20\\-y=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-20\\y=-2\end{cases}}}\)

vậy x=-20 và y=-2

\(-\frac{1}{3}-x=\frac{1}{2}-\frac{1}{-4}\)

\(-\frac{1}{3}-x=\frac{1}{2}-\frac{-1}{4}\)

\(-\frac{1}{3}-x=\frac{2}{4}-\frac{-1}{4}\)

\(-\frac{1}{3}-x=\frac{3}{4}\)

\(x=-\frac{1}{3}-\frac{3}{4}\)

\(x=-\frac{4}{12}-\frac{9}{12}\)

\(x=-\frac{13}{12}\)

\(x=-\frac{18}{24}+\frac{15}{21}\)

\(x=-\frac{3}{4}+\frac{5}{7}\)

\(x=-\frac{21}{28}+\frac{20}{28}\)

\(x=-\frac{1}{28}\)

1) Ta có:

\(\frac{1+2y}{18}=\frac{1+4y}{24}\)\(\Rightarrow\left(1+2y\right).24=\left(1+4y\right).18\)

=> 24 + 48y = 18 + 72y

=> 72y - 48y = 24 - 18

=> 24y = 6

\(\Rightarrow y=\frac{6}{24}=\frac{1}{4}\)

Thay \(y=\frac{1}{4}\) vào đề bài ta có:

\(\frac{1+2.\frac{1}{4}}{18}=\frac{1+6.\frac{1}{4}}{6x}\)

\(\Rightarrow\frac{1+\frac{1}{2}}{18}=\frac{1+\frac{3}{2}}{6x}\)

\(\Rightarrow\frac{3}{2}.\frac{1}{18}=\frac{5}{2}:6x\)

\(\Rightarrow\frac{1}{12}=\frac{5}{2}:6x\)

\(\Rightarrow6x=\frac{5}{2}:\frac{1}{12}=\frac{5}{2}.12=30\)

=> x = 30 : 6 = 5

Vậy \(x=5;y=\frac{1}{4}\)

2) Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(x+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\frac{2.\left(x+y+z\right)}{x+y+z}=2\)

                                                                                  \(=\frac{1}{x+y+z}\) (theo đề bài)

\(\Rightarrow x+y+z=\frac{1}{2}\)

Ta có: \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=2\)

\(\Rightarrow\frac{y+z+1}{x}+1=\frac{x+z+2}{y}+1=\frac{x+y-3}{z}+1=2+1\)

\(\Rightarrow\frac{x+y+z+1}{x}=\frac{x+y+z+2}{y}=\frac{x+y+z-3}{z}=3\)

\(\Rightarrow\frac{\frac{1}{2}+1}{x}=\frac{\frac{1}{2}+2}{y}=\frac{\frac{1}{2}-3}{z}=3\)

\(\Rightarrow\frac{3}{2}:x=\frac{5}{2}:y=\frac{-5}{2}:z=3\)

\(\Rightarrow\begin{cases}x=\frac{3}{2}:3=\frac{1}{2}\\y=\frac{5}{2}:3=\frac{5}{6}\\z=\frac{-5}{2}:3=\frac{-5}{6}\end{cases}\)

Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=\frac{-5}{6}\) 

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