Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)

\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(A=\frac{11}{3}\)(1)

+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)

\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)

Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)

=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)

=> \(\frac{11}{3}-x=1\)

=> \(x=\frac{11}{3}-1=\frac{8}{3}\)

Vậy x = 8/3

A=\(\frac{1.2.3.4...8.9}{2.3.4.5...9.10}\)

A=\(\frac{1}{10}\)

mình làm đc 1 câu thôi. Bạn thông cảm nhé

ĐK:\(x\ne-1;-3;-5;-7;-9\)

\(pt\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}+\frac{2}{\left(x+7\right)\left(x+9\right)}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-...-\frac{1}{x+9}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+9}=\frac{2}{5}\)\(\Leftrightarrow\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

\(\Leftrightarrow2\left(x+1\right)\left(x+9\right)=40\)\(\Leftrightarrow x^2+10x-11=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+11=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}\) (thoả)

Vậy....

xin lỗi nha, bài đó bằng có một cái 1/5 thôi, tại viết sai

ĐK : \(X\ne-1;-3;-7;-9\)

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}+\frac{1}{x^2+16x+63}=\frac{1}{5}\)

\(\frac{1}{\left(x+2\right)^2-1}+\frac{1}{\left(x+4\right)^2-1}+\frac{1}{\left(x+6\right)^2-1}+\frac{1}{\left(x-8\right)^2-1}=\frac{1}{5}\)

\(\frac{1}{\left(x+2-1\right)\left(x+2+1\right)}+\frac{1}{\left(x+4-1 \right)\left(x+4+1\right)}+\frac{1}{\left(x+6-1\right)\left(x+6+1\right)}+\frac{1}{\left(x+8-1\right)\left(x+8+1\right)}=\frac{1}{5}\)

\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)

\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+....-\frac{1}{x+9}\right)=\frac{1}{5}\)

\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+9}\right)=\frac{1}{5}\)

\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{1}{5}:\frac{1}{2}=\frac{2}{5}\)

\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

\(2\left(x+1\right)\left(x+9\right)=40\)

\(2x^2+20x+18=40\Leftrightarrow x^2+10x+9=20\)

\(\Leftrightarrow x^2+10x-11=0\Leftrightarrow x^2+10x-10-1=0\)

\(\Leftrightarrow\left(x^2-1\right)+\left(10x-10\right)=0\Leftrightarrow\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+11\right)=0\)

\(\orbr{\begin{cases}x-1=0\\x++11=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}}\)( Thõa mãn ) 

Vậy ...............

x^2 + 4x + 3 = (x+1)(x+3)

x^2 + 8x + 15 = (x+3)(x+5)

x^2 + 12x + 35 = (x+5)(x+7)

x^2 + 16x + 63 = (x+7)(x+9)

Bạn phân tích ra quy luật rồi thì bạn giải tiếp sẽ có: 

             1/x+1 -1/x+9 = 2/5

             8/(x+1)(x+9) =2/5

             (x+1)(x+9) = 20

             x^2 +10x+9 = 20

             x^2 +10x -11 = 0

             (x-1)(x+11) = 0

Vậy x=1 hoặc x= -11(thỏa măn ĐKXĐ)          

câu 2:

= 6/13

Các bạn nêu rõ cách làm từng bài giúp mình nhé! Thanks ^-^!

Chị nói điêu thất mấy bài này mà chị bảo chị ko biết làm thế em làm linh tinh rồi chị phải k cho em 10 cái nha

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}\cdot\frac{14}{15}\)

\(=\frac{7}{15}\)

Sửa đề chút nhé:

\(\left(1+3+5+7+...+2009+2011\right).\left(125125.127-127127.125\right)\)

\(=\left(1+3+5+7+...+2009+2011\right).\left(125.1001.127-127.1001.125\right)\)

\(=\left(1+3+5+7+...+2009+2011\right).0\)

\(=0\)

Ý b tham khảo bài bạn nguyen thi thuy linh nhé

\(\text{Tính nhanh : }\)

\(a,\text{ }1+3+5+7+9+\text{...}+2007+2009+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=\left\{\left(2009-1\right)\text{ : }2+1\right\}\cdot\left(2009+1\right)\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1005\cdot2010\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=2020050\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1010025+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1010025+2011\cdot\left(15890875+15890875\right)\)

\(=1010025+2011\cdot15890875\cdot2\)

\(=1010025+31956549625\cdot2\)

\(=1010025+63913099250\)

\(=63914109275\)

\(b,\text{ }\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

1. Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath

1. Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath