1)

\(\frac{1994.1993-1992.1993}{1992.1993+1944.7+1986}\)

\(=\frac{\left(1994-1992\right).1993}{2.1985028+2.6804+2.993}\)

\(=\frac{2.1993}{2.\left(1985028+6804+993\right)}\)

\(=\frac{2.1993}{2.1992825}\)

\(=\frac{193}{1992825}\)

2)

a) \(\frac{399.45+55.399}{1995.1996-1991.1995}\)

\(=\frac{399.\left(45+55\right)}{1995.\left(1996-1991\right)}\)

\(=\frac{399.100}{1995.5}\)\(=4\)

Câu b bài 2 mình không biết làm nữa, xin lỗi nhé!

\(\dfrac{1996.1995-996}{1000+1996.1994}=\dfrac{1996\left(1994+1\right)-996}{1000+1996.1994}=\dfrac{1996.1994+1996-996}{1000+1996.1994}\) 

\(=\dfrac{1000+1996.1994}{1000+1996.1994}=1\)

\(\dfrac{1993+1993.1994}{1992.1995+1995}\)

=\(\dfrac{1993.1+1993.1994}{1992.1995+1995.1}\)

=\(\dfrac{1993\left(1+1994\right)}{1995\left(1992+1\right)}\)

=\(\dfrac{1993.1995}{1995.1993}\)

=1

\(=\dfrac{1993.1+1993.1994}{1992.1995+1995.1}\)

\(=\dfrac{1993\left(1+1994\right)}{1995\left(1992+1\right)}\)

\(=\dfrac{1993.1995}{1995.1993}\)

=1

\(\dfrac{1993+1993\cdot1994}{1992\cdot1995+1995}\\ =\dfrac{1993\cdot1+1993\cdot1994}{1992\cdot1995+1995\cdot1}\\ =\dfrac{1993\left(1+1994\right)}{1995\left(1992+1\right)}\\ =\dfrac{1993\cdot1995}{1995\cdot1993}\\ =1\)

Tính nhanh :

A = \(2016.20152015-2015.20162016\)

= \(2016.2015.10001-2015.2016.1001\)

=0

\(A=2016.20152015-2015.20162016\)

\(=2016.2015.10001-2015.2016.10001\)

\(=0\)

Tính :

\(\dfrac{1993+1993.1994}{1992.1995+1995}\)= \(\dfrac{1993.1+1993.1994}{1992.1995+1995.1}\)=\(\dfrac{1993\left(1+1994\right)}{1995\left(1992+1\right)}\)=\(\dfrac{1993.1995}{1995.1993}=1\)

A = \(\dfrac{2}{35}\) + \(\dfrac{4}{77}\) + \(\dfrac{2}{143}\) + \(\dfrac{4}{221}\) + \(\dfrac{2}{323}\) + \(\dfrac{4}{437}\) + \(\dfrac{2}{575}\)

A =  \(\dfrac{2}{5\times7}\)+\(\dfrac{4}{7\times11}\)+\(\dfrac{2}{11\times13}\)+\(\dfrac{4}{13\times17}\)+\(\dfrac{2}{17\times19}\)+\(\dfrac{4}{19\times23}\)+\(\dfrac{2}{23\times25}\)

A = \(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{13}\)+\(\dfrac{1}{13}\)-\(\dfrac{1}{17}\)+\(\dfrac{1}{17}\)-\(\dfrac{1}{19}\)+\(\dfrac{1}{19}\)-\(\dfrac{1}{23}\)+\(\dfrac{1}{23}\)-\(\dfrac{1}{25}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{25}\)

A = \(\dfrac{4}{25}\)

Bài 2: 

b) Gọi \(d\inƯC\left(21n+4;14n+3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d\inƯ\left(1\right)\)

\(\Leftrightarrow d\in\left\{1;-1\right\}\)

\(\LeftrightarrowƯCLN\left(21n+4;14n+3\right)=1\)

hay \(\dfrac{21n+4}{14n+3}\) là phân số tối giản(đpcm)

Bài 1: 

a) Ta có: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)

\(=75\cdot\left(-4\right)+603\)

\(=603-300=303\)

Bài 1: 

c) Ta có: \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Leftrightarrow3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(\Leftrightarrow3B-B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow2B=1-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow B=\dfrac{3^{99}-1}{3^{99}\cdot2}\)

Bài 2: 

a) Vì tổng của hai số là 601 nên trong đó sẽ có 1 số chẵn, 1 số lẻ

mà số nguyên tố chẵn duy nhất là 2

nên số lẻ còn lại là 599(thỏa ĐK)

Vậy: Hai số nguyên tố cần tìm là 2 và 599

b,Gọi ƯCLN(21n+4,14n+3)=d

21n+4⋮d ⇒42n+8⋮d

14n+3⋮d ⇒42n+9⋮d

(42n+9)-(42n+8)⋮d

1⋮d ⇒ƯCLN(21n+4,14n+3)=1

Vậy phân số 21n+4/14n+3 là phân số tối giản

c,xy-2x+5y-12=0

xy-2x+5y-12+2=0+2

xy-2x+5y-10=2

xy-2x+5y-5.2=-2

x.(y-2)+5.(y-2)=2

(y-2).(x+5)=2

Sau đó bạn tự lập bảng