3/2.4^x+5/3.4^x 2=3/2.4^8+5/3.4^10
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1, \(3\left(x+5\right)=3x+3.5=3x+15\)
2, \(5\left(2+x\right)=5.2+5x=10+5x\)
3, \(4\left(x-2\right)=4x-4.2=4x-8\)
4, \(2x+2.5=2\left(x+5\right)\)
5, \(3.4+4x=4\left(x+3\right)\)
6, \(2x-2.4=2\left(x-4\right)\)
Chúc học tốt
Sử dụng tính chất phân phối để điền vào chỗ trống
3(x +) =....( viết thiếu )
(2+x)5=2.5+ x.5
3.4+4x = (3+x).4
2x + 2.5 =2.(x+5)
4(x-2) =4.x-4.2
2x-2.4 =2.(x-4)
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
\(M=\dfrac{2^2.3^2.4^2.....20^2}{1.3.2.4.3.5.4.6.5.7.6.8.7.9....19.21}=\)
\(=\dfrac{2^2.3^2.4^2....20^2}{1.2.3^2.4^2....19^2.20.21}=\dfrac{2.20}{21}=\dfrac{40}{21}\)
\(N=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{10}{11}=\dfrac{1}{11}\)
\(\frac{3.7.8.13}{14\cdot15\cdot26}=\frac{3\cdot7\cdot8\cdot13}{2\cdot7\cdot3\cdot5\cdot2\cdot13}=\frac{8}{5\cdot2\cdot2}=\frac{8}{20}=\frac{2}{5}\)
\(\frac{3.7.8.13}{14.15.26}=\frac{3.7.2^3.13}{2.7.3.5.13.2}=\frac{1.7.2.1}{1.1.1.5.1.1}=\frac{14}{5}\)
\(\frac{11.8-11.3}{17-6}=\frac{11.\left(8-3\right)}{11}=\frac{11.5}{11}=\frac{1.5}{1}=\frac{5}{1}=5\)
\(\frac{-17.13+17.2}{11.2-11.9}=\frac{-17.13+\left(-17\right).2}{11.\left(2-9\right)}=\frac{-17.\left(-13+2\right)}{11.\left(-7\right)}=\frac{-17.\left(-11\right)}{11.\left(-7\right)}=\frac{-17.\left(-1\right)}{-1.\left(-7\right)}=\frac{17}{7}\)
\(\frac{7}{9.10^2-2.10^2}=\frac{7}{100\left(9-2\right)}=\frac{7}{100.7}=\frac{7}{700}=\frac{1}{100}\)
mình làm được tới đó thôi, hihi.
mình k viết lại đề nhé =)
câu A
A :5 =1/2.4+1/4.6+1/6.8+..+1/98.100
A:5 =1/2-1/4+1/4-1/6+1/6-1/8+...+1/98-1/100
A:5 =1/2-1/100 =49/100
A=49/100 x5 =49/20
câu B tươg tự nha =)
Ta có:
A =5/2(1/2-1/4 + 1/4-1/6+ 1/6..........1/98-1/100)
A =5/2 (1/2 -1/100)
A =5/2 x 49/100
A = 49/20
Sửa đề : `P=3/1.2+3/2.3+3/3.4+....+3/11.12`
`P=3/1.2+3/2.3+3/3.4+....+3/11.12`
`=3(1/1.2+1/2.3+1/3.4+...+1/11.12)`
`=3(1/1-1/2+1/2-1/3+1/3-1/4+...+1/11-1/12)`
`=3(1/1-1/12)`
`=3(12/12-1/12)`
`=3 . 11/12`
`=33/12`
`=11/4`
Vậy `P=11/4`
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
hình đề bị sai thì phải
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+\dfrac{3}{3\cdot4}+...+\dfrac{3}{11\cdot12}\) đề phải ntn chứ nhỉ?
\(=3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{11\cdot12}\right)\)
\(=3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)
\(=3\left(\dfrac{1}{1}-\dfrac{1}{12}\right)\)
\(=3\left(\dfrac{12}{12}-\dfrac{1}{12}\right)\\ =3\cdot\dfrac{11}{12}\\ =\dfrac{33}{12}\\ =\dfrac{11}{4}\)