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!

\(A=\frac{65+891+135+909}{731-47+69-253}\)

\(A=\frac{65+135+891+909}{731+69-47-253}\)

\(A=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)

\(A=\frac{200+1800}{800-300}\)

\(A=\frac{2000}{500}=4\)

A = \(\frac{65+891+135+909}{731-47+69-253}\)

A = \(\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)

A = \(\frac{200+1800}{800-300}\)

A = \(\frac{2000}{500}\)

A = 4

\(A=\frac{65+891+135+909}{731-47+69-253}\)

\(A=\frac{5\times13+9\times9\times11+5\times27+9\times101}{17\times43-47+23\times3-11\times23}\)

\(A=\frac{5\times13+5\times27+9\times99+9\times101}{17\times43-47-11\times23+23\times3}\)

\(A=\frac{5\times\left(13+27\right)+9\times\left(99+101\right)}{17\times43-47-23\times\left(11-3\right)}\)

\(A=\frac{5\times40+9\times200}{17\times43-47-23\times8}\)

\(A=\frac{2000}{500}\)

\(A=4\)

no tính nhanh đc rùi! cho mk xin lỗi bn nhiều nha!!!!

\(A=\frac{595959}{454545}-\frac{14141414}{15151515}\)

\(A=\frac{595959:10101}{454545:10101}-\frac{14141414:1010101}{15151515:1010101}\)

\(A=\frac{59}{45}-\frac{14}{15}\)

\(A=\frac{59}{45}-\frac{42}{45}=\frac{17}{45}\)

\(B=\frac{65+891+135+909}{731-47+69-253}\)

\(B=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)

\(B=\frac{200+1800}{800-300}=\frac{2000}{500}=4\)

Giải theo cách lớp 5

\(A=\frac{595959}{454545}-\frac{14141414}{15151515}\\ A=\frac{595959:10101}{454545:10101}-\frac{14141414:1010101}{15151515:1010101}\\ A=\frac{59}{45}-\frac{14}{15}=\frac{17}{45}\)

\(B=\frac{65+891+135+909}{731-47+69-253}=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47-253\right)}=\frac{200+1800}{731+69-47+253}=\frac{200+1800}{500}=\frac{2000}{500}=4\)

65+891+135+909 phần 731-47+69-253 = 2000 phần 500 =4

1089.24350205

\(C=\frac{65+891+135+909}{731-47+69-253}\)

\(C=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)

\(C=\frac{200+1800}{800-300}\)

\(C=\frac{2000}{500}=4\)

Ta có:

\(C=\frac{65+891+135+909}{731-47+69-253}\)

\(\Leftrightarrow C=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)

\(\Leftrightarrow C=\frac{200+1800}{800-300}\)

\(\Leftrightarrow C=\frac{2000}{500}=4\)