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a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
=>1/x-1=3/4
=>x-1=4/3
=>x=7/3
ĐK: \(x\ne\left\{1;3;8;20\right\}\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\)\(\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}=\frac{3}{4}\)
\(\Rightarrow\)\(x-1=\frac{4}{3}\)
\(\Leftrightarrow\)\(x=\frac{7}{3}\)(t/m)
Vậy...
Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
\(\left(x-3\right):\dfrac{4}{5}=20:\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right).\dfrac{5}{4}-\left(x-3\right).\dfrac{1}{20}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{5}{4}-\dfrac{1}{20}\right)=0\)
\(\Leftrightarrow\left(x-3\right).\dfrac{6}{5}=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
ta có:
\(\dfrac{x}{5}=\dfrac{4x}{20}\)
Suy ra: \(\dfrac{4x}{20}=\dfrac{3}{20}\)
=> 4x = 3 => \(x=\dfrac{3}{4}\)
x/5=3/20
= x . 20 = 5 . 3
= x . 20 = 15
= x = 15 : 20
= x = 0,75