\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\) 

\(=\frac{1}{5}-\frac{1}{5}-\frac{3}{7}+\frac{3}{7}+\frac{5}{9}-\frac{5}{9}-\frac{2}{11}+\frac{2}{11}+\frac{7}{13}-\frac{7}{13}-\frac{9}{16}\) 

\(=0+0+0+0+0-\frac{9}{16}\) 

\(=\frac{-9}{16}\)

Cứu !!!!!!!!!!!..........

Ta có hình vẽ :

A O B M N

Bạn cứ dựa vào tính chất cơ bản mà suy ra, chứ mk làm sai thì lây cả bạn:<<

Các tính chất cơ bản đó là :

*Hai góc đối đỉnh là hai góc mà mỗi cạnh của góc này là tia đối của một cạnh của góc kia.

*Hai góc đối đỉnh thì bằng nhau.

\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)

\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{9^{97}}+\frac{1}{3^{98}}\)

\(\Rightarrow3C-C=1-\frac{1}{3^{98}}\)

\(\Rightarrow C=\frac{1-\frac{1}{3^{98}}}{2}\)

Nhầm một chút ==

\(3C-C=2C=1-\frac{1}{3^{99}}\)

\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}\)

\(\frac{x+10}{30}+\frac{x+14}{43}+\frac{x+5}{95}+\frac{x+130}{5}=0\)

\(\Leftrightarrow\left(\frac{x+10}{30}+3\right)+\left(\frac{x+14}{43}+2\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+130}{5}-6\right)=0\)

\(\Leftrightarrow\frac{x+10+90}{30}+\frac{x+14+86}{43}+\frac{x+5+95}{95}+\frac{x+130-30}{5}=0\)

\(\Leftrightarrow\frac{x+100}{30}+\frac{x+100}{43}+\frac{x+100}{95}+\frac{x+100}{5}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{5}\right)=0\)

Vì \(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{5}\ne0\)

=> x + 100 = 0

=> x = -100

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{1999}-1\right)\)

\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)\left(-\frac{3}{4}\right)...\left(-\frac{1998}{1999}\right)\)

\(=\frac{\left(-1\right)\left(-2\right)\left(-3\right)...\left(-1998\right)}{2\cdot3\cdot4\cdot...\cdot1999}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot1998}{2\cdot3\cdot4\cdot...\cdot1999}=\frac{1}{1999}\)

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

\(\Leftrightarrow\left(1+\frac{315-x}{101}\right)+\left(1+\frac{313-x}{103}\right)+\left(1+\frac{311-x}{105}\right)+\left(1+\frac{309-x}{107}\right)=0\)

\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)

\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\Leftrightarrow416-x=0\)

\(\Leftrightarrow x=416\)

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

\(\Leftrightarrow\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)

\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)

\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

Vì \(\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)\ne0\)

\(\Rightarrow416-x=0\Leftrightarrow x=416\)

-> He started meeting her when he was young

He started metting her since he was young