\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{240}\)

= \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{15.16}\)

= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{15}-\frac{1}{16}\)

= \(\frac{1}{2}-\frac{1}{16}\)

= \(\frac{7}{16}\)

b: \(F=2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{240}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(=2\cdot\dfrac{7}{16}=\dfrac{7}{8}\)

c: \(3G=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\)

\(\Leftrightarrow2G=1-\dfrac{1}{243}=\dfrac{242}{243}\)

hay G=121/243

d: \(2H=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Leftrightarrow H=1-\dfrac{1}{1024}\)

\(4,7\div0,25+5,3\times4\)

\(=18,8+21,2\)

\(=40\)

\(3\times\left(a-2\right)+150=240\)

\(3\times\left(a-2\right)=90\)

\(a-2=30\)

\(a=32\)

\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)

\(\dfrac{1}{9}+a=\dfrac{13}{36}\)

\(a=\dfrac{1}{4}\)

\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)

\(a=\dfrac{3}{2}\)

bạn vào câu hỏi tuỷong tự để tìm lời giải chi tiết

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{14.15}+\frac{1}{15.16}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)

\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{210}+\frac{1}{240}\)

\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{14\times15}+\frac{1}{15\times16}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)

\(=\frac{1}{2}-\frac{1}{16}\)

\(=\frac{7}{16}\)

\(240:1+240:2+240:3+240:6\)

\(=240+120+80+40\)

\(=360+120\)

\(=480\)

⇒ Chọn A 

240:1+240:2+240:3+240:6

=240.1+240.1/2+240.1/3+240.1/6

=240(1+1/2+1/3+1/6)

=240.(1+1)

=240.2=480

Chọn A

\(240:1+240:2+240:3+240:6\\=240+240\cdot \dfrac{1}{2}+240\cdot \dfrac{1}{3} +240\cdot \dfrac{1}{6}\\=240\cdot \Big(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} \Big)\\=240\cdot \Big(\dfrac{6}{6}+\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6} \Big)\\=240\cdot\dfrac{12}{6}\\=240\cdot 2\\=480\\\Rightarrow Chọn.A\)

Giải:

a) \(F=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{240}\)

\(\Leftrightarrow F=\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{15.16}\)

\(\Leftrightarrow F=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{15}-\dfrac{1}{16}\)

\(\Leftrightarrow F=\dfrac{1}{2}-\dfrac{1}{16}\)

\(\Leftrightarrow F=\dfrac{7}{16}\)

Vậy ...

b) \(G=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

\(\Leftrightarrow G=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\)

\(\Leftrightarrow\dfrac{1}{3}G=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}+\dfrac{1}{3^6}\)

\(\Leftrightarrow\dfrac{2}{3}G=\dfrac{1}{3}-\dfrac{1}{3^6}\)

\(\Leftrightarrow G=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^6}}{\dfrac{2}{3}}\)

\(\Leftrightarrow G=\dfrac{\left(\dfrac{1}{3}-\dfrac{1}{3^6}\right)3}{2}\)

\(\Leftrightarrow G=\dfrac{1-\dfrac{1}{3^5}}{2}\)

\(\Leftrightarrow G=\dfrac{\dfrac{3^5-1}{3^5}}{2}\)

\(\Leftrightarrow G=\dfrac{3^5-1}{3^5.2}\)

Vậy ...

c) Tương tự b)

0 

0

1) \(\left(+15\right)+\left(+17\right)=15+17=32\)

2) \(\left(-3\right)+\left(-7\right)=-3-7=-\left(3+7\right)=-10\)

3) \(\left(-25\right)+\left(+4\right)=-25+4=-\left(25-4\right)=-21\)

4) \(\left(-6\right)+\left(-54\right)=-6-54=-\left(6+54\right)=-60\)

5) \(\left(-15\right)+20=20-15=5\)

6) \(\left(-5\right)+8+7+5\)

\(=\left(-5+5\right)+\left(8+7\right)\)

\(=15\)

7) \(\left(-8\right)+\left(-11\right)+\left(-2\right)\)

\(=\left[\left(-8\right)+\left(-2\right)\right]+\left(-11\right)\)

\(=\left(-10\right)+\left(-11\right)\)

\(=-21\)

8) \(15+\left(-5\right)+\left(-14\right)+\left(-16\right)\)

\(=\left[15+\left(-5\right)\right]+\left[\left(-14\right)+\left(-16\right)\right]\)

\(=10+\left(-30\right)\)

\(=-20\)

9) \(\left(-20\right)+\left(-14\right)+3+\left(-86\right)\)

\(=\left[\left(-20\right)+3\right]+\left[\left(-14\right)+\left(-86\right)\right]\)

\(=\left(-17\right)+\left(-100\right)\)

\(=-117\)

10) \(\left(-136\right)+123+\left(-264\right)+\left(-83\right)+240\)

\(=\left[\left(-136\right)+\left(-264\right)\right]+\left[123+\left(-83\right)\right]+240\)

\(=\left(-400\right)+40+240\)

\(=\left(-360\right)+240\)

\(=-120\)

11) \(\left(-596\right)+2001+1999+\left(-404+189\right)\)

\(=\left(-596\right)+2001+1999-404+189\)

\(=\left[\left(-596\right)-404\right]+\left(2001+189\right)+1999\)

\(=\left(-1000\right)+2190+1999\)

\(=1190+1999\)

\(=3189\)

12) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)

\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)

\(=\left(378+121\right)+\left(-400\right)+218\)

\(=499-400+218\)

\(=99+218\)

\(=317\)

\(\text{#}Toru\)

Ta có : 

\(B=\frac{1}{240}-...-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(B=\frac{1}{240}-\left(\frac{1}{210}+...+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)

\(B=\frac{1}{240}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{210}\right)\)

\(B=\frac{1}{240}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}\right)\)

\(B=\frac{1}{240}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}\right)\)

\(B=\frac{1}{240}-\left(1-\frac{1}{15}\right)\)

\(B=\frac{1}{240}-1+\frac{1}{15}\)

\(B=\frac{1-240+16}{240}\)

\(B=\frac{-223}{240}\)

Vậy \(B=\frac{-223}{240}\)

Chúc bạn học tốt ~