a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)

\(\Rightarrow x=-100\)

b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)

\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)

\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)

b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)

=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0

\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0

\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))

\(\Leftrightarrow\)x=-1999

Vậy x=-1999

câu b sai đề bạn ơi

a)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{n+1}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n}{n+1}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot n}{2\cdot3\cdot4\cdot...\cdot\left(n+1\right)}\)

\(=\frac{1}{n+1}\)

a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\)

\(=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right)....\left(\frac{n+1}{n+1}-\frac{1}{n+1}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{n}{n+1}=\frac{1}{n+1}\)

=(3994002+1998+1997)x(\(\frac{2}{3}\)-\(1\frac{1}{3}\))

=3997997x\(\frac{-2}{3}\)

=-2665331,333

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!!!)

\(\)Sửa lại đề câu a:

\(a.\frac{x-13}{2006}+\frac{x-22}{1997}+\frac{x-21}{1998}=3\\ \Leftrightarrow\frac{x-13}{2006}-1+\frac{x-22}{1997}-1+\frac{x-21}{1998}-1=0\\\Leftrightarrow \frac{x-2019}{2006}+\frac{x-2019}{1997}+\frac{x-2019}{1998}=0\\ \Leftrightarrow\left(x-2019\right)\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1998}\right)=0\\\Leftrightarrow x-2019=0\left(Vi\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1998}\ne0\right)\\\Leftrightarrow x=2019\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{2019\right\}\)

Đặt \(y=x^2+x\) ta có:

\(y^2+4y=12\\\Leftrightarrow y^2+4y-12=0\\\Leftrightarrow y^2+4y+4-16=0\\ \Leftrightarrow\left(y+2\right)^2-4^2=0\\\Leftrightarrow \left(y+2-4\right)\left(y+2+4\right)=0\\ \Leftrightarrow\left(y-2\right)\left(y+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y-2=0\\y+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=2\\y=-6\end{matrix}\right.\)

Thay \(y=x^2+x\) vào ta có:

\(x^2+x=2\\ \Leftrightarrow x^2+x-2=0\\ \Leftrightarrow x^2-x+2x-2=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

\(x^2+x=-6\\ \Rightarrow x^2+x+6\ge0\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{1;-2\right\}\)

ta có \(A=\frac{2}{0,\left(1998\right)}+\frac{2}{0,0\left(1998\right)}+\frac{2}{0,00\left(1998\right)}=\frac{2}{0,\left(1998\right)}+\frac{2}{0,\left(1998\right)}.\frac{1}{10}+\frac{2}{0,\left(1998\right)}.\frac{1}{100}\)

                                                                          \(=\frac{2}{0,\left(1998\right)}.\left(1+\frac{1}{10}+\frac{1}{100}\right)=\frac{2}{0,\left(1998\right)}.1\frac{11}{100}=\frac{222}{0,00\left(1998\right)}\)

a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)

\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)

\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)

\(=\frac{x^2+4x+4}{x^2}\)

\(\left(\frac{x+2}{x}\right)^2\)

=>phép chia = 1 với mọi x # 0 và x#-1

b)Cm tương tự

khó quá

Ta thấy A gồm có 99 số hạng nên ta nhóm mỗi nhóm 3 số hạng.

Ta có: A = 1 + 5 + 5² + 5³ + 5⁴ + 5⁵ +...+ 5⁹⁷ + 5⁹⁸ + 5⁹⁹

             = (1 + 5 + 5² )+ (5³ + 5⁴ + 5⁵ )+...+( 5⁹⁷ + 5⁹⁸ + 5⁹⁹ )

             =(1 + 5 + 5² )+ 5³(1 + 5 + 5² ) +...+ 5⁹⁷(1 + 5 + 5² )

             = ( 1 + 5 + 5²)(1 + 5³+....+5⁹⁷)

             = 31(1 + 5³+....+5⁹⁷)

Vì có một thừa số là 31 nên A chia hết cho 31.

 P/s Đừng để ý câu trả lời của mình

Tiếp

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