Tìm x, biết
\(\frac{\chi-1}{\chi+5}\)\(=\frac{6}{7}\)
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\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2013}{4030}\)
\(\frac{1}{x+1}=\frac{1}{2015}\)
=>x+1=2015
=>x=2014
\(25\%.x-\frac{1}{5}.x=\frac{-1}{20}\)
\(=>\frac{1}{4}.x-\frac{1}{5}.x=\frac{-1}{20}\)
\(=>x.\left(\frac{1}{4}-\frac{1}{5}\right)=\frac{-1}{20}\)
\(=>x.\frac{1}{20}=\frac{-1}{20}\)
\(=>x=\frac{-1}{20}:\frac{1}{20}\)
\(=>x=-1\)
5/x+y/3=1/6
<=>5/x=1/6-y/3
<=>5/x=1/6-2y/6=(1-2y)/6
=>x.(1-2y)=5.6=30
Từ đó lập bảng, tìm x, y tương ứng
5/x+y/3=1/6
<=>5/x=1/6-y/3
<=>5/x=1/6-2y/6=(1-2y)/6
=>x.(1-2y)=5.6=30
Từ đó lập bảng, tìm x, y tương ứng
a, B=[(x+3)/(x-3)+(2x^2-6)/(9-x^2)+x/(x+3)]:[(6x-12)/(2x^2-18)]
=[(x+3)/(x-3)+ -(2x^2-6)/(x^2-9)+x/(x+3)]:[(6x-12)/(2x^2-18)]
=[(x+3)/(x-3)+ -(2x^2-6)/(x-3)(x+3)+x/(x+3)]:[(6x-12)/2(x-3)(x+3)]
={[(x+3)^2-2x^2+6+x(x-3)]/(x-3)(x+3)}:[6(x-2)/2(x-3)(x+3)]
=(x^2+6x+9-2x^2+6+x^2-3x)/(x-3)(x+3): 6(x-2)/2(x-3)(x+3)
=3x+15/(x-3)(x+3): 6(x-2)/2(x-3)(x+3)
=3(x+5)/(x-3)(x+3): 6(x-2)/2(x-3)(x+3
=3(x+5)/(x-3)(x+3).2(x-3)(x+3)/6(x-2)
=3(x+5).6/(x-2)
=6(x+5)/6(x-2)
=x+5/x-2
b,Ta thay : x=1
=>x+5/x-2=1+5/1-2=-6
Ta thay : x=-3
=>x+5/x-2=-3+5/-3-2=-2/5
c, Ta co : x+5/x-2=0
x+5=(x-2).0
x+5=0
x=-5
Vậy : x=-5
Ta có: \(\left(\frac{3x}{7}+1\right):\left(-4\right)=\frac{-1}{7}\)
\(\Rightarrow\)\(\frac{3x}{7}+1=\frac{-1}{7}.\left(-4\right)\)
\(\Rightarrow\)\(\frac{3x}{7}+1=\frac{4}{7}\)\(\Rightarrow\frac{3x}{7}=\frac{-3}{7}\)
\(\Rightarrow3x=-3\)\(\Rightarrow\)\(x=-1\)
Vậy x=-1
đk: \(x\ne-5\)
\(\Leftrightarrow\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Leftrightarrow7x-7=6x+30\)
\(\Rightarrow x=37\)
Vậy x = 37
Bài làm
\(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow7x-7=6x+30\)
\(\Leftrightarrow x=37\)( tm )