\(\frac{37}{195}\)và\(\frac{73}{179}\)

Ta có 

\(\frac{37}{195}< \frac{37}{179}< \frac{73}{179}\)

\(\Rightarrow\)\(\frac{37}{195}< \frac{73}{179}\)

Vậy \(\frac{37}{195}< \frac{73}{179}\)

Vì 37/195<37/179<73/179

=>37/195<73/179

\(\frac{9}{x-5}=x+3\)

\(\left(x-5\right)\left(x+3\right)=9\)

\(x^2-5x+3x-15=9\)

\(x^2-2x=24\)

\(\left(x-2\right)x=24\)

\(x-2;x\inƯ\left(24\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm8;\pm12;\pm24\right\}\)

Mà x-2 và x cách nhau 2 đơn vị

\(\Rightarrow x\in\left\{-4;6\right\}\)

huỳnh phước mạnh ơi tao chả hiểu mày viết cái gì

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2007}-\frac{1}{2008}\)

\(A=\left(1+\frac{1}{3}+...+\frac{1}{2007}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2008}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2008}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1004}\)

\(A=\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}\)    (1)

\(B=\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}\)     (2)

\(\left(1\right)\left(2\right)\Rightarrow\frac{A}{B}=\frac{\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}}{\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}}=1\)

-7/60 nhé

\(\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\)

\(=\frac{20}{60}-\frac{15}{60}-\frac{12}{60}\)

\(=\frac{20-15-12}{60}\)

\(=\frac{-7}{60}\)

A=1/4+1/5+1/6+...+1/15

A=1.484

=>A<2

ai có cách nhanh hơn ko

giúp mk với

=12017+22016+32015+...+20162+2017112+13+...+12016+12017+12018A=12017+22016+32015+...+20162+2017112+13+...+12016+12017+12018

A=(12017+1)+(22016+1)+(32015+1)+...+(20162+1)+112+13+...+12016+12017+12018A=(12017+1)+(22016+1)+(32015+1)+...+(20162+1)+112+13+...+12016+12017+12018

A=20182017+20182016+20182015+...+20182+2018201812+13+...+12016+12017+12018A=20182017+20182016+20182015+...+20182+2018201812+13+...+12016+12017+12018

A=2018(12+13+...+12016+12017+12018)12+13+...+12016+12017+12018

thui chít ghi nhầm