1.6/3.4+2.7/4.5+3/8+5.6+...+18.25/20.21
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a,ta thay hai so o duoi mau so la hai sop tu nhien lien tiep khoang cach chinh bang 1
ta co 1/3-1/4+1/4-1/5+.............................+1/20-1/21
ta co =1/3-1/21 vi co cac so doi to da the hien tren
=2/7
b vi khoang cach duoi mau kac tu mau la 2 con tu la 1 vay nhan 2 vao ca day so ta duoc
2/4.6+2/6.8+..............................+2/30.32
bay gio khoang cach duoi mau bang tu ta co
1/4-1/6+1/6-1/8+............................+1/30-1/32
nhu tren ta co =(1/4-1/32):2=7/64
\(=-3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)=-3\cdot\dfrac{5}{14}=-\dfrac{15}{14}\)
= 3.(3/3 - 3/4 + 3/4 - 3/5 + 3/5 - 3/6 +.....+ 3/277 - 3/278 + 3/278 - 3/279)
= 3.(3/3 - 3/279 )
= 3.92/93
= 187/93
Mình cũg không chắc chắn là 100% đâu bạn nên dò lại nhé
Đặt biểu thức là \(A\)
\(A=\dfrac{3}{3.4}+\dfrac{3}{4.5}+\dfrac{3}{5.6}+...+\dfrac{3}{278.279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}\left(\dfrac{3}{3.4}+\dfrac{3}{4.5}+\dfrac{3}{5.6}+...+\dfrac{3}{278.279}\right)\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{278.279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{278}-\dfrac{1}{279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}-\dfrac{1}{279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{93}{279}-\dfrac{1}{279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{92}{279}\)
\(\Leftrightarrow A=\dfrac{92}{279}:\dfrac{1}{3}\)
\(\Leftrightarrow A=\dfrac{92}{279}.3\)
\(\Leftrightarrow A=\dfrac{92}{93}\)
\(E=2.3+3.4+4.5+3.6+2.7+4.15=2\left(3+7\right)+3\left(4+6\right)+4\left(5+15\right)=2.10+3.10+4.20=20+30+80=130\)
\(F=3\left(12+13+14+15\right)+3\left(8+7+6+5\right)=3\left(12+8+13+7+14+6+15+5\right)=3\left(20+20+20+20\right)=3.80=240\)
f: Ta có: \(E=3\cdot\left(12+13+14+15\right)+3\left(8+7+6+5\right)\)
\(=3\left(12+13+14+15+8+7+6+5\right)\)
\(=3\cdot80=240\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)
Ta có: \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{3}{10}\)
\(\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{3}{10}\)
\(\dfrac{1}{x+1}=\dfrac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1\)
\(x=29\)
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