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Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=11\cdot\frac{13}{17}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{143}{17}\)
\(\Rightarrow\frac{a+b}{a+b}+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{a+c}{c+a}=\frac{143}{17}\)
\(\Rightarrow1+1+1+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=\frac{143}{17}\)
\(\Rightarrow A=\frac{143}{17}-3=\frac{92}{17}\)
a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
Số phần tử của tập B là: 7+6+5+4+3+2+1=7*8/2=28 phân số
\(A:B=11:13\)
=>\(\dfrac{A}{11}=\dfrac{B}{13}=k\)
=>A=11k; B=13k
\(\dfrac{1}{A}-\dfrac{1}{B}=\dfrac{1}{286}\)
=>\(\dfrac{1}{11k}-\dfrac{1}{13k}=\dfrac{1}{286}\)
=>\(\dfrac{13-11}{143k}=\dfrac{1}{286}\)
=>\(\dfrac{2}{143k}=\dfrac{1}{286}\)
=>\(\dfrac{2}{k}=\dfrac{1}{2}\)
=>k=4
=>\(A=11\cdot4=44;B=13\cdot4=52\)