a) 2001 x 767 + 2002 x 233

= 2002 x 767 – 767 + 2002 x 233

= 2002 x (767+ 233) – 767

= 2002 x 1000 – 767

= 2002000 – 767

= 2001233

   b) (m : 1 – m x 1) : (m x 2001 + m + 1)

= (m – m) : (m x 2001 + m + 1)

= 0 : (m x 2001 + m + 1)                                 

= 0

- HT_ #lie -

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.

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

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

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

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

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

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

=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

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\\ \Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\\ \Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\\ \Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\\ \Leftrightarrow x+2004=0\\ \Leftrightarrow x=-2004\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy x = -2004

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

=> x+2004=0 => x=-2004