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NB
3
8 tháng 9 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{51}{153}-\frac{3}{153}\right)\)
\(=\frac{1}{2}.\frac{48}{153}\)
\(=\frac{24}{153}\)
8 tháng 9 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{17}{51}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
13 tháng 9 2020
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.\left(5^2\right)^4}=7.\frac{5^8}{5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3.3^8.5^2.5^3}{3.5.5^4.3^8}=\frac{5^5}{5^5}=1\)
c) Đề hơi sai roi bạn oi
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2=\frac{1}{100}+\frac{121}{100}=\frac{61}{50}\)
LN
3
AH
Akai Haruma
Giáo viên
28 tháng 6 2023
Lời giải:
a.
$\frac{-2}{3}+2x=\frac{4}{3}$
$2x=\frac{4}{3}-\frac{-2}{3}=2$
$x=2:2=1$
b.
$\frac{5}{8}-5:x=\frac{-3}{8}$
$5:x=\frac{5}{8}-\frac{-3}{8}=1$
$x=5:1=5$
c.
$\frac{2}{3}-x=\frac{-1}{2}$
$x=\frac{2}{3}-\frac{-1}{2}=\frac{7}{6}$
d.
$\frac{5}{7}-4x=\frac{-51}{7}$
$4x=\frac{5}{7}-\frac{-51}{7}=8$
$x=8:4=2$
28 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`-2/3 + 2x = 4/3`
`=> 2x = 4/3 - (-2/3)`
`=> 2x = 2`
`=> x=2 \div 2`
`=> x=1`
Vậy, `x=1`
`b,`
`5/8 - 5 : x = -3/8`
`=> 5 \div x = 5/8 - (-3/8)`
`=> 5 \div x = 1`
`=> x= 5 \div 1`
`=> x=5`
Vậy, `x=5`
`c,`
`2/3 - x = -1/2`
`=> x=2/3 - (-1/2)`
`=> x=7/6`
Vậy, `x=7/6`
`d,`
`5/7 - 4x = -51/7`
`=> 4x = 5/7 - (-51/7)`
`=> 4x=8`
`=> x=8 \div 4`
`=> x=2`
Vậy, `x=2.`
`@` `\text {Kaizuu lv u}`
NC
16 tháng 8 2020
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
NN
0
\(1.3+3.5+5.7+...+49.51\)
\(=1.\left(1+2\right)+3.\left(3+2\right)+5\left(5+2\right)+...+49\left(49+2\right)\)
\(=1^2+1.2+3^2+3.2+5^2+5.2+...+49^2+49.2\)
\(=\left(1^2+3^2+5^2+...+49^2\right)+2\left(1+3+5+...+49\right)\)
Có: \(1^2+3^2+5^2+...+49^2\)
\(=\left(1^2+2^2+3^2+...+49^2\right)-\left(2^2+4^2+...+48^2\right)\)
\(=\left(1^2+2^2+3^2+...+49^2\right)-2^2\left(1^2+2^2+3^2+...+24^2\right)\)
\(=\frac{49\left(49+1\right)\left(2.49+1\right)}{6}-4.\frac{24\left(24+1\right)\left(2.24+1\right)}{6}\)
= 40425 - 19600 =20825
\(1+3+5+...+49=\frac{\left(49+1\right)\left[\left(49-1\right):2+1\right]}{2}=625\)
=> \(1.3+3.5+5.7+...+49.51\)
\(=\left(1^2+3^2+5^2+...+49^2\right)+2\left(1+3+5+...+49\right)\)
\(=20825+625.2=22075\)