5³ : 5 + 4 : ( 49 - 5 × 3² )
5 5 5
_______ +_______+.....+_______
3×4 4×5 49×50
\(\dfrac{5}{3\times4}+\dfrac{5}{4\times5}+...+\dfrac{5}{49\times50}\)
\(=5\times\left(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{49\times50}\right)\)
\(=5\times\left(\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+...+\dfrac{50-49}{49\times50}\right)\)
\(=5\times\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=5\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)=5\times\left(\dfrac{50}{150}-\dfrac{3}{150}\right)\)
\(=\dfrac{5\times47}{150}=\dfrac{47}{30}\)
B= 4/1\(\times\)3+4/3\(\times\)5+4/5\(\times\)7+...|+4/47\(\times\)49+4/49\(\times\)51
\(B=\dfrac{4}{1\times3}+\dfrac{4}{3\times5}+\dfrac{4}{5\times7}+...+\dfrac{4}{47\times49}+\dfrac{4}{49\times51}\)
\(=2\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{47\times49}+\dfrac{2}{49\times51}\right)\)
\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=2\times\left(1-\dfrac{1}{51}\right)\)
\(=2\times\dfrac{50}{51}\)
\(=\dfrac{100}{51}\)
Câu 2:
a,-2/49-(5/3-2/49) b,C=5*4^15*9^9-4*3^20*8^9/5*2^9*6^19-7*2^29*27^6
a: =-2/49-5/3+2/49=-5/3
b: \(=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{27}\cdot2^2}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(2\cdot3-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\cdot\dfrac{6-9}{15-14}=2\cdot\left(-3\right)=-6\)
Câu 2:
a,-2/49-(5/3-2/49) b,C=5*4^15*9^9-4*3^20*8^9/5*2^9*6^19-7*2^29*27^6
1+5+5^2+5^3+5^4+.....+5^49+5^50
#)Giải :
Đặt \(A=1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\)
\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\)
\(\Rightarrow5A-A=4A=\left(5+5^2+5^3+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}\right)\)
\(\Rightarrow4A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
Đặt S = 1 + 5 + 52 + 53 + 54 + ........ + 549 + 550
5S = 5 + 52 +53+ 54 + 55 + ........ + 550 +551
5S - S = (5 + 52 +53+ 54 + 55 + ........ + 550 +551) - (1 + 5 + 52 + 53 + 54 + ........ + 549 + 550)
4S =551 - 1
S =(551- 1) : 4
~ Chúc bạn học giỏi ~
~TMT_Nhók~
Gọi \(A=1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\)
\(A=1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\)
\(5A=5+5^2+5^3+5^4+...+5^{49}+5^{50}+5^{51}\)
\(5A-A=\left(5+5^2+5^3+5^4+...+5^{49}+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+5^4+...+5^{49}+5^{50}\right)\)
\(4A=5^{51}-1\)
\(A=\frac{5^{51}-1}{4}\)
11/49×1/2+11/49÷5/4-11/49×3/4
\(\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}:\dfrac{5}{4}-\dfrac{11}{49}\times\dfrac{3}{4}\)
\(=\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}\times\dfrac{4}{5}-\dfrac{11}{49}\times\dfrac{3}{4}\)
\(=\dfrac{11}{49}\times\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{3}{4}\right)\)
\(=\dfrac{11}{49}\times\left(\dfrac{10}{20}+\dfrac{16}{20}-\dfrac{15}{20}\right)\)
\(=\dfrac{11}{49}\times\dfrac{11}{20}\)
\(=\dfrac{121}{980}\)
Chúc bạn học tốt
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}\div\dfrac{5}{4}-\dfrac{11}{49}\times\dfrac{3}{4}\)
`=`\(\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}\times\dfrac{4}{5}-\dfrac{11}{49}\times\dfrac{3}{4}\)
`=`\(\dfrac{11}{49}\times\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{3}{4}\right)\)
`=`\(\dfrac{11}{49}\times\dfrac{11}{20}\)
`=`\(\dfrac{121}{980}\)
\(\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}:\dfrac{5}{4}-\dfrac{11}{49}\times\dfrac{3}{4}\)
\(=\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}\times\dfrac{4}{5}-\dfrac{11}{49}\times\dfrac{3}{4}\)
\(=\dfrac{11}{49}\times\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{3}{4}\right)\)
\(=\dfrac{11}{49}\times\dfrac{11}{20}=\dfrac{121}{980}\)
Tính nhanh:
\(\frac{5-\frac{5}{7}-\frac{5}{49}}{4-\frac{4}{7}-\frac{4}{49}}+\frac{1,5+75\%-\frac{3}{8}}{0,625-\frac{5}{2}-125\%}\)
\(\frac{5-\frac{5}{7}-\frac{5}{49}}{4-\frac{4}{7}-\frac{4}{49}}+\frac{1,5+75\%-\frac{3}{8}}{0.625-\frac{5}{2}-125\%}\)
\(=\frac{5.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{49}\right)}{4\cdot\left(\frac{1}{4}-\frac{1}{7}-\frac{1}{49}\right)}+\frac{\frac{3}{2}+\frac{3}{4}-\frac{3}{8}}{\frac{5}{8}-\frac{5}{2}-\frac{5}{4}}\)
\(=\frac{5}{4}+\frac{3\cdot\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}\right)}{5\cdot\left(\frac{1}{8}-\frac{1}{2}-\frac{1}{4}\right)}\)
\(=\frac{5}{4}+\left(-\frac{3}{5}\right)\)
\(=\frac{13}{20}\)
2/35 : (3/5+3/7)
1,2 . 25/6 - (4/3+5/3)
5 5/7 - (3/7+2 3/5) : 4/5
1,4 . 15/49 - (4/5+2/3) : 2 1/5
CÂU 1:\(\frac{2}{35}:\left(\frac{3}{5}+\frac{3}{7}\right)\)
\(=\frac{2}{35}:\frac{36}{35}\)
\(=\frac{1}{18}\)
CÂU 2:\(\frac{1}{2}\cdot\frac{25}{6}-\left(\frac{4}{3}+\frac{5}{3}\right)\)
\(=\frac{25}{12}-3\)
\(=-\frac{11}{12}\)
CÂU 3:\(5\frac{5}{7}-\left(\frac{3}{7}+2\frac{3}{5}\right):\frac{4}{5}\)
\(=\frac{40}{7}-\left(\frac{3}{7}+\frac{13}{5}\right):\frac{4}{5}\)
\(=\frac{40}{7}-\frac{106}{35}:\frac{4}{5}\)
\(=\frac{40}{7}-\frac{53}{14}\)
\(=\frac{27}{14}\)
1.so sánh
1/2:2/3......2/3:1/2 4/7:2/5......4/7:3/5
4/15:4/7......2/5x10/3 5/6......15/18-11/18
2.viết số thích hợp vào ô trống
2/3=..../18 7/9=49/.... ..../5=18/15
..../3=10/15 5/9=..../45 49/56=7/....
6/8=42/.... 2/9=..../63 49/56=..../8
`1)1/2:2/3 .... 2/3 : 1/2`
`=>1/2xx3/2 .... 2/3xx2`
`=>3/4 .... 4/3`
Vì `3/4 < 1` và `4/3>1`
`=>3/4<4/3`
__
`4/7:2/5 ... 4/7 : 3/5`
`=>4/7xx5/2....4/7xx5/3`
`=>20/14...20/21`
`=>10/7...20/21`
Vì `10/7>1` và `20/21<1`
`=>10/7>20/21`
__
`4/15:4/7....2/5xx10/3`
`=>4/15xx7/4...20/15`
`=>7/15...20/15`
Vì `7<20` nên `7/15<20/15`
__
`5/6...15/18-11/18`
`=>5/6...4/18`
Ta có : MSC : `18`
`5/6 = 15/18`
Vì `15>4` nên `5/6 > 4/18`
`2)2/3=(2xx6)/(3xx6)=12/18`
`7/9=(7xx7)/(9xx7)=49/63`
`6/5=(6xx3)/(5xx3)=18/15`
`2/3=(2xx5)/(3xx5)=10/15`
`5/9=(5xx5)/(9xx5)=25/45`
`49/56=(49:7)/(56:7)=7/8`
`6/8=(6xx7)/(8xx7)=42/56`
`2/9=(2xx7)/(9xx7)=14/63`
`49/56=(49:7)/(56:7)=7/8`