ta thấy vế thứ hai có kết quả bằng 0
=>(1999x1998+1998x1997)x0
chằng cần tìm kết quả mà =>B=0

bạn học nhà cô Hà dạy Toán cấp 2 đt đúng ko 

bn ở tỉnh nào vậy

Bỏ 1/3 ở cuối nhé

Ta có:(1+1999/2)+(1+1998/3)+...(2/1999)(có 1998 tổng<=>1998 số 1)+(2000 - 1998)+400

        = 2001/2+2001/3+...+2001/1999+402

        =2001.(1/2+1/3+...+1/1999)+402(1)

      Thay (1) vào biểu thức trên và tính(tự tính nha!,tk cho mk!!!)

\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)

\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

\(=>x+2001=0\)

\(x=-2001\)

\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)

\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

\(=>x+1998=0\)

\(x=-1998\)

dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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

\(\Leftrightarrow\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\) \(\left(\frac{x+4}{1997}+1\right)=0\)

\(\Leftrightarrow\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(\Leftrightarrow\left(x+2001\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

Mà : \(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\ne0\)

\(\Rightarrow x+2001=0\)

\(\Leftrightarrow x=-2001\)

b) \(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+...+\frac{87}{1000}+\frac{99}{1000}\)

\(=\frac{1+13+25+...+85+97}{1000}=\frac{\left(97+1\right).\left[\left(97-1\right):12+1\right]:2}{1000}\)

\(=\frac{49.9}{1000}=\frac{441}{1000}.\) ( Đề bài sai nhé bạn tử số : 1; 13; 25; 37; 49 ; 61; 73; 85 ; 97. )

Ta có

\(D=\frac{2^{2x+1}}{2^{2x}-2}+\frac{2^{2\left(1-x\right)+1}}{2^{2\left(1-x\right)}-2}=\frac{2^{2x}}{2^{2x-1}-1}+\frac{2^{2\left(1-x\right)}}{2^{1-2x}-1}\)

Mà \(2^{1-2x}=\frac{1}{2^{2x-1}}\)(do 1-2x+2x-1=0)

=>\(D=\frac{2^{2x}}{2^{2x-1}-1}+\frac{2^{2\left(1-x\right)}}{\frac{1}{2^{2x-1}}-1}=\frac{2^{2x}-2^{2\left(1-x\right)}.2^{2x-1}}{2^{2x-1}-1}=\frac{2^{2x}-2^1}{2^{2x-1}-1}=\frac{2\left(2^{2x-1}-1\right)}{2^{2x-1}-1}=2\)

Áp dụng D ta được

\(P\left(\frac{1}{1998}\right)+P\left(\frac{1997}{1998}\right)=2\)

\(P\left(\frac{2}{1998}\right)+P\left(\frac{1996}{1998}\right)=2\)

..............................................................

Do \(x\ne\frac{1}{2}\)nên không có \(P\left(\frac{999}{1998}\right)\)

\(P\left(\frac{998}{1998}\right)+P\left(\frac{1000}{1998}\right)=2\)

=> \(A=1997+2+2+....+2\)(998 số 2)

=> \(A=1997+2.998=3993\)

Vậy A=3993

#)Trả lời :

\(a,\frac{2}{3}:\frac{5}{7}.\frac{5}{7}:\frac{2}{3}+1934\)

\(=\left(\frac{2}{3}:\frac{2}{3}\right).\left(\frac{5}{7}:\frac{5}{7}\right)+1934\)

\(=1.1+1934\)

\(=1935\)

              #~Will~be~Pens~#

Trả lời : 

\(a,\text{ }\frac{2}{3}\text{ : }\frac{5}{7}\text{ x }\frac{5}{7}\text{ : }\frac{2}{3}+1934\)

\(=\left(\frac{2}{3}\text{ : }\frac{2}{3}\right)\text{ x }\left(\frac{5}{7}\text{ : }\frac{5}{7}\right)+1934\)

\(=1\text{ x }1+1934\)

\(=1935\)

#)Next :

\(\left(30:7\frac{1}{2}+0,5.3-1,5\right)\left(4\frac{1}{2}-\frac{9}{2}\right):\left(14,5.1000\right)\)

\(=\left(30:7\frac{1}{2}+0,5.3-1,5\right)\left(\frac{9}{2}-\frac{9}{2}\right):\left(14,5.1000\right)\)

\(=\left(30:7\frac{1}{2}+0,5.3-1,5\right).0:\left(14,5.1000\right)\)

\(=0\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(\dfrac{4}{3}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times0\)

\(=0\)