Đặt \(A=\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(-A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

\(-A=1-\frac{1}{99}\)

\(-A=\frac{98}{99}\)

\(A=\frac{-98}{99}\)

Chúc bạn học tốt ~ 

Đặt A = \(\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

=> - A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

- A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

- A = \(1-\frac{1}{99}\)

- A = \(\frac{98}{99}\)

=> A = \(-\frac{98}{99}\)

Vậy A = \(-\frac{98}{99}\)

Hok tốt

\(A=\frac{1}{99.98}-...-\frac{1}{2.1}\)

\(-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(-A=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

\(-A=1-\frac{1}{99}\)

\(-A=\frac{98}{99}\Leftrightarrow A=\frac{-98}{99}\)

 -1/(1999.2000)= 1/2000-1/1999    ......    -1/2= 1/2-1

Vậy A= 1/2000-1/1999 +1/1999-1/1998+....+1/3-1/2+1/2-1 = -1+1/2000= -1999/2000

Đặt A = \(\frac{1}{99}-\frac{1}{99.98}-.....-\frac{1}{2.1}\)

\(A=\frac{1}{99}-\left[-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{98.99}\right)\right]\)

\(A=\frac{1}{99}+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{99}\right)\)

\(A=\frac{1}{99}+\left(1-\frac{1}{99}\right)\)

\(A=\frac{1}{99}+\frac{98}{99}=1\)

     \(\frac{1}{100.99}-\frac{1}{99.98}-......-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=-\left(-\frac{1}{100.99}+\frac{1}{99.98}+...........+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=-\left(-\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+......+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)

\(=-\left(-\frac{1}{100}-1\right)\)

\(=\frac{1}{100}+1\)

\(=\frac{101}{100}\)

Lần sau nhớ chọn đúng môn

Uk, Long cứ đợi ng ta trả lời xong long làm giống là đc ý mà!

Xin lỗi Long vì có hơi bực nên mới nói vậy, mong bn bỏ qua

\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=-\frac{98}{100}=-\frac{49}{50}\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

   \(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

    \(=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)

      \(=\frac{1}{100}-\left(\frac{1}{100}-1\right)\)

      \(=1\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{-49}{50}\)

Sửa đề: \(\left(m-1\right)x^2+3mx-4m+1=0\)

Ta có: \(\Delta=\left(3m\right)^2-4\cdot\left(-4m+1\right)\left(m-1\right)=9m^2-4\left(-4m^2+4m+m-1\right)\)

\(=9m^2+16m^2-20m+4\)

\(=25m^2-20m+4\)

\(=\left(5m-2\right)^2\ge0\forall m\)

hay phương trình luôn có nghiệm với mọi m

Áp dụng hệ thức Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-3m}{m-1}\\x_1\cdot x_2=\dfrac{-4m+1}{m-1}\end{matrix}\right.\)

Vì \(x_1+x_2=\dfrac{-3m}{m-1}\) và \(2x_1=3x_2\) nên ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-3m}{m-1}\\2x_1-3x_2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x_1+2x_2=\dfrac{-6m}{m-1}\\2x_1-3x_2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x_2=\dfrac{-6m}{m-1}\\x_1+x_2=\dfrac{-3m}{m-1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{-6m}{5m-5}\\x_1=\dfrac{-9m}{5m-5}\end{matrix}\right.\)

Ta có: \(x_1\cdot x_2=\dfrac{-4m+1}{m-1}\)

\(\Leftrightarrow\dfrac{-6m}{5m-5}\cdot\dfrac{-9m}{5m-5}=\dfrac{-4m+1}{m-1}\)

\(\Leftrightarrow\dfrac{54m^2}{5m-5}=\dfrac{-20m+5}{5m-5}\)

Suy ra: \(54m^2+20m-5=0\)

\(\Delta=20^2-4\cdot54\cdot\left(-5\right)=1480\)

Đến đây bạn tự làm tiếp nhé, chỉ cần tìm m và so sánh với ĐK m khác 1 thôi

Bài làm:

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

\(=\frac{1}{99.100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}\right)\)

\(=\frac{1}{99.100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{99.100}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{99.100}-\frac{98}{99}\)

\(=\frac{1-98.100}{99.100}=\frac{1-9800}{9900}=-\frac{9799}{9900}\)

Học tốt!!!!

\(\left(\frac{1}{100.99}\right)-\left(\frac{1}{99.98}\right)-\left(\frac{1}{98.97}\right)-...-\left(\frac{1}{3.2}\right)-\left(\frac{1}{2.1}\right)\)

\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)

\(=\frac{1}{99}-\frac{1}{100}-\left(\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1+\frac{1}{2}\right)\)

\(=\frac{1}{99}-\frac{1}{100}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{99}-\frac{1}{100}-1+\frac{1}{99}\)

\(=\frac{2}{99}-\frac{101}{100}\)