\(\frac{x}{3}=\frac{2}{3}+-\frac{1}{7}\) Tìm x again
Help me please >~<
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DP
24 tháng 6 2017
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{19.21}\right).x=\frac{9}{7}\)
\(\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{21-19}{19.21}\right).x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\frac{2}{7}\right].x=\frac{9}{7}\)
\(\frac{1}{7}.x=\frac{9}{7}\)
\(\Rightarrow x=\frac{9}{7}\div\frac{1}{7}=9\)
24 tháng 6 2017
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\frac{2}{7}x=\frac{9}{7}\)
\(\Leftrightarrow x=\frac{9}{2}\)
1 tháng 3 2020
\(c.x\left(\frac{2}{5}-\frac{3}{5}\right)=\frac{2}{35}\)
\(x=\frac{2}{35}:\frac{-1}{5}=-\frac{2}{7}\)
\(d.\left(2x+1\right)^2=49=7^2=\left(-7\right)^2\)
\(TH1:2x+1=7\Rightarrow x=3\)
\(TH2=2x+1=-7\Rightarrow x=-4\)
1 tháng 3 2020
\(a.x=\frac{-3}{5}-\frac{4}{9}=\frac{-47}{45}\)
\(b.\frac{3}{5}:x=\frac{17}{10}-\frac{2}{5}\)
\(x=\frac{3}{5}:\frac{13}{10}=\frac{6}{13}\)
23 tháng 6 2017
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
\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)
\(24x+\frac{24}{25}=22x+\frac{224}{243}\)
\(2x=\frac{224}{243}-\frac{24}{25}\)
\(2x=-\frac{232}{6025}\)
\(x=\frac{-116}{6075}\)
DP
23 tháng 6 2017
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)
\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)
\(12x+\frac{12}{25}=11x+\frac{112}{243}\)
\(11x-12x=\frac{112}{243}-\frac{12}{25}\)
\(-1x=-\frac{116}{6075}\)
\(x=-\frac{116}{6075}\div\left(-1\right)\)
\(x=\frac{116}{6075}\)
IM
8 tháng 9 2016
\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}=\frac{x+3}{2008}\)
\(\Leftrightarrow\frac{x+4}{2007}=\frac{x+1}{2010}\)
\(\Leftrightarrow\left(x+4\right)2010=\left(x+1\right)2007\)
\(\Leftrightarrow2010x+8040=2007x+2007\)
\(\Leftrightarrow2010x-2007x=2007-8040\)
\(\Leftrightarrow3x=-6033\)
\(\Leftrightarrow x=-2011\)
8 tháng 9 2016
\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}+\frac{x+3}{2008}\)
=>\(\left(\frac{x\text{+4}}{2007}+1\right)+\left(\frac{x+8}{2003}+1\right)=\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+3}{2008}+1\right)\)
=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}=\frac{x+2011}{2010}+\frac{x+2011}{2008}\)
=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}-\frac{x+2011}{2010}-\frac{x+2011}{2008}=0\)
=>\(x+2011\left(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\ne0\)
=> x+2011=0
=>x=-2011
Vậy x = -2011
TM
1 tháng 8 2016
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
18 tháng 8 2016
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
HA
26 tháng 2 2020
\(\frac{-5}{x}=\frac{-y}{8}=\frac{18}{72}\)
\(\Leftrightarrow\frac{-5}{x}=\frac{-y}{8}=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{x}=\frac{1}{4}\\-\frac{y}{8}=\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5.4:1\\-y=8.1:4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-20\\-y=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-20\\y=-2\end{cases}}}\)
vậy x=-20 và y=-2
10 tháng 10 2016
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 \(x=\frac{1}{2};y=\frac{5}{6};z=\frac{-5}{6}\)
\(\frac{x}{3}=\frac{2}{3}+\frac{-1}{7}\)
\(\frac{x}{3}=\frac{11}{21}\)
\(x=\frac{11.3}{21}\)(Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\)khi \(a.d=b.c\))
\(\Rightarrow x=\frac{11}{7}\)
~Học tốt~
\(\frac{x}{3}=\frac{2}{3}+\frac{-1}{7}\)
\(\frac{x}{3}=\frac{2.7+\left(-1\right).3}{21}\)
\(\frac{x}{3}=\frac{11}{21}\)
\(\Leftrightarrow21x=33\)
\(\Leftrightarrow x=\frac{33}{21}=\frac{11}{7}\)