Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)

\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)

\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)

\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)

\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)

\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)

Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)

=> x - 2000 = 0 

=> x = 2000

A=1991x1999=(1995-4)1999=1995x1999-4x1999

B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A

vậy A<B

A=1991x1999=(1995-4)1999=1995x1999-4x1999

B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A

vậy A<B

bông bèooooooooooooooooooooooooooooooooooooooooo

A=1991x1999=(1995-4)1999=1995x1999-4x1999

B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A

vậy A<B

giúp mình lun nha. Không tính và so sánh kết quả . A = 1996 x1996 B= 1995 x1997

A=1991x1999=(1995-4)1999=1995x1999-4x1999

B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A

vậy A<B

A=1991x(1995+4)=1991x1995+1991x4

B=1995x(1991+4)=1991x1995+1995x4

Bớt cả A và B đi 1991x1995 ta chỉ cần so sánh 1991x4 và 1995x4

Kết quả A<B

A=1991x1999=1991x(1995+4)=1991x1995+1991x4

B1995x1995=(1991+4)x1995=1991x1995+4x1995

Vì vậy B>A

A=1991x[1995+4]

B=1995x[1991+4]

A=1991x1995+1991x4

B=1995x1991+1995x4

A=1991x4                                                   B=1995x4          [cùng bớt đi 1991x1995]

nên A<B

A=1991x1999=(1995-4)1999=1995x1999-4x1999

B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A

vậy A<B

A = 1991 x 1999                            B = 1995 x 1995

A = 1991 x ( 1995 + 4 )                  B = ( 1991 + 4 ) x 1995

A = 1991 x 1995 + 1991 x 4           B = 1991 x 1995 + 1995 x 4

Vì 1991 x 4 < 1995 x 4 => 1991 x 1999 < 1995 x 1995

Vậy A < B

A=1991x1999=(1995-4)1999=1995x1999-4x1999

B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A

vậy A<B

Ta có :A= 1995 x 1995

             =(1991 +4) x 1995

             =1991 x1995+4 x1995

             = 1991 x 1995+ 7980

Lại có:B= 1991 x1999

            =1991 x (1995 +4 )

            =1991 x 1995 + 1991 x 4

            =1991 x 1995 +7964

Nhận thấy: 7980>7964

Vậy A>B

1995.1995>1991.1999

\(A=1995.1995=1995\left(1999-4\right)=\)\(1995.1999-1995.4\)

\(B=1991.1999=\left(1995-4\right).1999\)\(=1995.1999-4.1999\)

\(\Leftrightarrow A>B\)

Ta có:

A = 1991 x 1999 và B = 1995 x 1995.

= 1991 x ( 1995 + 4 )

= 1991 x 1995 + 1991 x 4

B = 1995 x 1995

= 1995 x (1991 + 4 )

= 1995 x 1991 + 1995 x 4

Vì 1991 x 1995 = 1995 x 1991 và 1991 x 4 < 1995 x 4

Nên 1991 x 1999 < 1995 x 1995

A=1991x1999=(1995-4)(1995+4)=1995^2-16

B=1995x1995=1995^2

=>A<B

1991x1999=(1995-4)(1995+4)=1995^2-16

=> A<B

Ta có:        

A = 1991 x 1999 và B = 1995 x 1995.

                      = 1991 x ( 1995 + 4 )

                      = 1991 x 1995 + 1991 x 4

          B = 1995 x 1995

                     = 1995 x (1991 + 4 )

                     = 1995 x 1991 + 1995 x 4

Vì 1991 x 1995 = 1995 x 1991 và 1991 x 4 < 1995 x 4

Nên 1991 x 1999 < 1995 x 1995