a) 2001 x 757 + 2002 x 233

    = 2001 x 757 + (2001 + 1) x 233

    = 2001 x 757 + 2001 x 233 + 1 x 233

    = 2001 x (757+233) + 233

    = 2001 x 1000 + 233

    = 2001000 + 233

    = 2001233

a) 2001x757+2002x243

=2001 x 757+ (2001+1) x233

=2001 x757+2001x233+1 x233

=2001x(757+233)+233

=2001x1000 +233

=201000+233

=2001233

b)(m:1-m x1):(mx2001+m+1)

=(m-m):(mx2001+m+1)

=0:(mx2001+m+1)

=0

   Học tốt

=2001*(767+233)*2002

=2001*1000*2002

=4006002000

nha! sao bạn ko thay số 2001=2002 hoặc thay 2002=2001 để cho dễ tính hơn?

chúc bạn học giỏi

2001*( 767 + 233) + 2002

                                =2001 * 1000 + 2002

                               = 2001 * 3002

                               = 6007002

 chúc bạn học tốt

2001 x 767 + 2002 x 233 = 1534767 + 466466

                                     = 2001233

2) \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)

\(\Rightarrow x\left(1-2y\right)=40\)

Vì \(1-2y\) luôn là số lẻ nên \(1-2y\in\left\{\pm1;\pm5\right\}\)

\(\Rightarrow y=\left\{0;1;-2;3\right\}\)

\(\Rightarrow x\in\left\{40;-40;8;-8\right\}\)

Vậy các cặp số x,y thỏa mãn là \(\left(0;40\right);\left(1;-40\right);\left(-2;8\right);\left(3;-8\right)\)

Ta có :

\(B=\dfrac{2000+2001}{2001+2002}=\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}\)

Mặt khác :

\(\dfrac{2000}{2001}>\dfrac{2000}{2001+2002}\)

\(\dfrac{2001}{2002}>\dfrac{2001}{2001+2002}\)

\(\Leftrightarrow A=\dfrac{2000}{2001}+\dfrac{2001}{2002}>\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}=\dfrac{2000+2001}{2001+2002}=B\)

\(\Leftrightarrow A>B\)

Ta có: B =20002001+2002 +20012001+2002

Mặt khác: 20002001 >20002001+2002

20012002 >20012001+2002

Suy ra 20002001 +20012002 >20002001+2002 +20012001+2002

hay A> B

Vậy A > B.

1012 nhé bạn

diễn giải dùm mik đc k?

1012 nhé

B-)B-)B-)

- Ta có : \(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)

=> \(\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1\)

=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}=\frac{x-2005}{2002}+\frac{x-2005}{2001}\)

=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)

=> \(\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

=> \(x-2005=0\)

=> \(x=2005\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{2005\right\}\)

Sửa đề: \(\dfrac{x+1}{2000}+\dfrac{x+2}{1999}=\dfrac{x+3}{1998}+\dfrac{x+4}{1997}\)

\(\Rightarrow\left(\dfrac{x+1}{2000}+1\right)+\left(\dfrac{x+2}{1999}+1\right)=\left(\dfrac{x+3}{1998}+1\right)+\left(\dfrac{x+4}{1997}+1\right)\)

\(\Rightarrow\dfrac{x+2001}{2000}+\dfrac{x+2001}{1999}=\dfrac{x+2001}{1998}+\dfrac{x+2001}{1997}\)

\(\Rightarrow\dfrac{x+2001}{2000}+\dfrac{x+2001}{1999}-\dfrac{x+2001}{1998}-\dfrac{x+2001}{1997}=0\)

\(\Rightarrow\left(x+2001\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\)

\(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\Leftrightarrow x+2001=0\Leftrightarrow x=-2001\)

a ) \(\frac{2003\times14+1988+2001+2002}{2002+2002\times503+504\times2002}\)

= \(\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)

= \(\frac{2002\times14+14+1998+2001\times2002}{2002\times1008}\)

= \(\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)

= \(\frac{2002\times\left(14+1+2001\right)}{2002\times1008}\)

= \(\frac{2016}{1008}\)

= 2

b ) Đặt A = 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128

=> 2A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64

=> 2A - A = ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 ) - ( 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 )

=> A = 1/2 - 1/128

A = 63/128

\(b,\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+..+\frac{1}{128}\)

\(=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}=A\)

\(\Rightarrow2A=\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^6}\)

\(\Rightarrow2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^6}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^7}\)