A=3/4.5+3/5.6+.......+3/59.60
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\(A=1.2+2.3+3.4+4.5+...+59.60\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+59.60.\left(61-58\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+...+59.60.61-58.59.60\)
\(\Rightarrow3A=59.60.61\)
\(\Rightarrow A=\frac{59.60.61}{3}\)
\(3A=3.4.3+4.5.3+5.6.3+...+59.60.3\)
\(3A=3.4\left(5-2\right)+4.5\left(6-3\right)+5.6.\left(7-4\right)+...+59.60\left(61-58\right)\)
\(3A=3.4.5-2.3.4+4.5.6-3.4.5+...+59.60.61-58.59.60\)
\(3A=59.60.61-2.3.4\)
\(\Rightarrow A=59.20.61-2.4=...\)
\(=-3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)=-3\cdot\dfrac{5}{14}=-\dfrac{15}{14}\)
= 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 \(A\)
\(A=\dfrac{3}{3.4}+\dfrac{3}{4.5}+\dfrac{3}{5.6}+...+\dfrac{3}{278.279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}\left(\dfrac{3}{3.4}+\dfrac{3}{4.5}+\dfrac{3}{5.6}+...+\dfrac{3}{278.279}\right)\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{278.279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{278}-\dfrac{1}{279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}-\dfrac{1}{279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{93}{279}-\dfrac{1}{279}\)
\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{92}{279}\)
\(\Leftrightarrow A=\dfrac{92}{279}:\dfrac{1}{3}\)
\(\Leftrightarrow A=\dfrac{92}{279}.3\)
\(\Leftrightarrow A=\dfrac{92}{93}\)
\(P=\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.33....59.60\)
\(\text{Ta có:}\)
\(91=13.7\)
\(\rightarrow4.13+5.17=42.35⋮91\)
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.33....59.60\)
\(\rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.....60.42.35\)
\(\rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32....60.20.91⋮91\)
Bài này là cơ bản luôn đó:
= 3.(1/1.2 + 1/2.3+...)
= 3.(1/1-1/2+1/2-1/3...)
(tự viết nốt và tính)
\(A=\dfrac{3}{4\cdot5}+\dfrac{3}{5\cdot6}+...+\dfrac{3}{59\cdot60}\\ =3\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{59\cdot60}\right)\\ =3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{59}-\dfrac{1}{60}\right)\\ =3\left(\dfrac{1}{4}-\dfrac{1}{60}\right)=3\left(\dfrac{15}{60}-\dfrac{1}{60}\right)\\ =3\cdot\dfrac{7}{30}=\dfrac{7}{10}\)
sao bấm máy nhanh vậy