1/2 + 1/6 + 1/12 ..... 1/240
1/240-...-1/30-1/20-1/12-1/6-1/2
1/6+1/12+1/20+1/30+1/42...+1/240
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{240}\)
= \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{15.16}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{15}-\frac{1}{16}\)
= \(\frac{1}{2}-\frac{1}{16}\)
= \(\frac{7}{16}\)
tính nhanh
B = (x:1-x.1).(x.2015+x+1).(x là số tự nhiên)
F = 2/6+2/12+....+2/240
G = 1/3+1/9+1/27+1/81+1/243
H = 1/2+1/4+1/8+...+1/1024
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: \(2H=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Leftrightarrow H=1-\dfrac{1}{1024}\)
cú t ccau oiii ét ô ét giải giúp t bài này zới
4,7 : 0,25 + 5,3 x 4
3x ( a - 2) +150 = 240
1/9 + a + 7/12 = 17/18
( 1/2 x 1/3 + 1/3 x 1/4 + 1/4 x 1/5 + 1/5 x 1/6 + 1/6 x 1/7 + 1/7 x 1/8 ) x a = 9/16
\(4,7\div0,25+5,3\times4\)
\(=18,8+21,2\)
\(=40\)
\(3\times\left(a-2\right)+150=240\)
\(3\times\left(a-2\right)=90\)
\(a-2=30\)
\(a=32\)
\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)
\(\dfrac{1}{9}+a=\dfrac{13}{36}\)
\(a=\dfrac{1}{4}\)
\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)
\(a=\dfrac{3}{2}\)
Ti'nh nhanh :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.........+\frac{1}{210}+\frac{1}{240}\)
bạn vào câu hỏi tuỷong tự để tìm lời giải chi tiết
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{14.15}+\frac{1}{15.16}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{210}+\frac{1}{240}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{14\times15}+\frac{1}{15\times16}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{2}-\frac{1}{16}\)
\(=\frac{7}{16}\)
Tìm giá trị của 240÷1+240÷2+240÷3+240÷6
A,480 B,20 C,400 D,120
\(240:1+240:2+240:3+240:6\)
\(=240+120+80+40\)
\(=360+120\)
\(=480\)
⇒ 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
\(240:1+240:2+240:3+240:6\\=240+240\cdot \dfrac{1}{2}+240\cdot \dfrac{1}{3} +240\cdot \dfrac{1}{6}\\=240\cdot \Big(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} \Big)\\=240\cdot \Big(\dfrac{6}{6}+\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6} \Big)\\=240\cdot\dfrac{12}{6}\\=240\cdot 2\\=480\\\Rightarrow Chọn.A\)
Tính nhanh :
F=\(\dfrac{2}{6}\)+\(\dfrac{2}{12}+...+\dfrac{2}{240}\)
G= \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
H= \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
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)
1.Tính nhanh nếu có thể:
1) (+15)+(+17)
2) (-3)+(-7)
3) (-25)+(+4)
4) (-6)+(-54)
5) (-15)+20
6)(-5)+8+7+5
7) (-8)+(-11)+(-2)
8) 15+(-5)+(-14)+(-16)
9) (-20)+(-14)+3+(-86)
10)(-136)+123+(-264)+(-83)+240
11) (-596)+2001+1999+(-404+189)
12, ( 314 + ( -153) +64 +121+(-247)+218
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) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)
\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)
\(=\left(378+121\right)+\left(-400\right)+218\)
\(=499-400+218\)
\(=99+218\)
\(=317\)
\(\text{#}Toru\)
B =\(\frac{1}{240}\)- ...... - \(\frac{1}{30}\)- \(\frac{1}{20}\)- \(\frac{1}{12}\)- \(\frac{1}{6}\)- \(\frac{1}{2}\)
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 ~