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

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 -

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x.\left(x+1\right)}=1\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x.\left(x+1\right)}=1\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=1\)

MK chưa làm xong đợi mk ăn cơm xong làm nha

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 \(S=\left\{2005\right\}\)

\(\left(m:1-m.1\right):\left(m.2001.m+1\right)\)

\(=0:A=0\)

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Tìm m biết: ( m : 1 – m x 1) : ( m × 2001 + m + 1)

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

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

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

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

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

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