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ND
7
3 tháng 7 2019
1. \(\frac{x+1}{x+5}=\frac{x+3}{x+2}\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=\left(x+3\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)x+\left(x+1\right).2=\left(x+3\right)x+\left(x+3\right).5\)
\(\Leftrightarrow x^2+x+2x+2=x^2+3x+5x+15\)
\(\Leftrightarrow x^2+3x+2=x^2+8x+15\)
\(\Leftrightarrow x^2+3x-x^2-8x=15-2\)
\(\Leftrightarrow-5x=13\)
\(\Leftrightarrow x=\frac{-13}{5}\)
Vậy ...
BM
0
14 tháng 8 2020
7/4.x+3/2=-4/5
7/4.x=-4/5-3/2
7/4.x=-23/10
x=-23/10:7/4
x=-46/35
vậy x=-46/35
1/4+3/4.x=3/4
1.x=3/4
x=3/4:1
x=3/4
vậy x=3/4
x.(1/4+1/5)-(1/7+1/8)=0
x.9/20-15/56=0
x.51/280=0
x=0:51/280
x=0
vậy x=0
3/35-(3/5+x)=2/7
(3/5+x)=3/35-2/7
(3/35+x)=-1/5
x=-1/5-3/5
x=-4/5
vậy x=-4/5
14 tháng 8 2020
\(a,1\frac{3}{4}.x+1\frac{1}{2}=\frac{4}{5}\)
\(\frac{7}{4}.x=\frac{4}{5}-\frac{3}{2}\)
\(\frac{7}{4}.x=\frac{-7}{10}\)
\(x=\frac{-7}{10}:\frac{7}{4}\)
\(x=\frac{-2}{5}\)
\(b,\frac{1}{4}+\frac{3}{4}.x=\frac{3}{4}\)
\(\frac{3}{4}.x=\frac{3}{4}-\frac{1}{4}\)
\(\frac{3}{4}.x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{3}{4}\)
\(x=\frac{2}{3}\)
\(c,x.\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
\(x.\frac{9}{20}-\frac{15}{56}=0\)
\(x.\frac{9}{20}=\frac{15}{56}\)
\(x=\frac{15}{56}:\frac{9}{20}\)
\(x=\frac{25}{42}\)
\(d,\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{-1}{5}\)
\(x=\frac{-1}{5}-\frac{3}{5}\)
\(x=\frac{-4}{5}\)
Học tốt
A
18 tháng 8 2021
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.\)
\(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)(cộng 2 vế cho 3)
\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}+\frac{x+3}{2007}+\frac{2007}{2007}=\frac{x+10}{2000}+\frac{2000}{2000}+\frac{x+11}{1999}+\frac{1999}{1999}+\frac{x+12}{1998}+\frac{1998}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
x+2010=0
x=-2010
18 tháng 8 2021
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)
\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)
\(=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
26 tháng 8 2018
a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0
=> x + 100 = 0 => x = -100
b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)
\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)
\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)
\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)
Vì 1/47 + 1/48 - 1/49 khác 0
Nên x -50 = 0 => x = 50