1012 nhé bạn

diễn giải dùm mik đc k?

1012 nhé

B-)B-)B-)

\(\frac{2003\times4+1998+2001\times2002}{2002+2002\times1002+2002\times1003}\)

\(=\frac{2003\times4+2\times999+2001\times2\times1001}{2002.\left(1+1002+1003\right)}\)

\(=\frac{2\times\left(2003\times2+999+2001\times1001\right)}{1001\times2\times\left(1+1002+1003\right)}\)

\(=\frac{2003\times2+999+2001\times1001}{1001\times\left(1+1002+1003\right)}\)

\(=1\)

mk ko bít

Lời giải:

Dãy $x,x+1, x+2,..., 2002$ có số số hạng là:

$\frac{2002-x}{1}+1=2003-x$
Tổng $x+(x+1)+....+2001+2002=\frac{(2002+x)(2003-x)}{2}$

Do đó:

$\frac{(2002+x)(2003-x)}{2}=2002$

$\Rightarrow (2002+x)(2003-x)=4004$

$2002.2003+x-x^2=4004$

$x^2-x-4006002=0$

$(x-2002)(x+2001)=0$

$\Rightarrow x=2002$ hoặc $x=-2001$

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

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

\(=\frac{2002\times\left(14+1+2001\right)}{2002\times1008}=\frac{2016}{1008}=2\)

\(\dfrac{1}{\left(x+2000\right)\left(x+2001\right)}+\dfrac{1}{\left(x+2001\right)\left(x+2002\right)}+...+\dfrac{1}{\left(x+2009\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2001}+\dfrac{1}{x+2001}-\dfrac{1}{x+2002}+...+\dfrac{1}{x+2009}-\dfrac{1}{x+2010}=\dfrac{10}{11}\)

\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{x+2010-x-2000}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\)

\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{10}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\left(x+2000\right)\left(x+2010\right)=11\\ \Leftrightarrow...\)

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.

\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

<=> \(\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)

<=> \(\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)

<=> (x - 2004)(1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4) = 0

<=> x - 2004 = 0 (vì 1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4 khác 0)

<=> x = 2004

Vậy S = {2004}

đề bài \(=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

 \(\Leftrightarrow\frac{x}{2000}-\frac{4}{2000}+\frac{x}{2001}-\frac{3}{2001}+\frac{x}{2002}-\frac{2}{2002}=\frac{x}{2}-\frac{2002}{2}+\frac{x}{3}-\frac{2001\\}{3}+\frac{x}{4}-\frac{2000}{4}\)

\(\Leftrightarrow\frac{x}{2000}-\frac{1}{500}+\frac{x}{2001}-\frac{1}{667}+\frac{x}{2002}-\frac{1}{1001}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}+1001+667+500=0\)

\(\Leftrightarrow\left(\frac{x}{2000}+\frac{x}{2001}+\frac{x}{2002}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}\right)+\left(1001+667+500-\frac{1}{500}-\frac{1}{667}-\frac{1}{1001}\right)=0\)

=> x=1