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ta có:
\(4\frac{2}{7}=4+\frac{2}{7}\)
\(4\frac{2}{7}=\left(4+\frac{2}{7}\right)\cdot3=4\cdot3+\frac{2}{7}\cdot3=12+\frac{6}{7}=12\frac{6}{7}\)
`Answer:`
Bài 1:
a. \(\frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}-\frac{2}{3}x+\frac{1}{3}=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{6}-\frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow-\frac{2}{3}=\frac{2}{3}-\frac{5}{6}\)
\(\Leftrightarrow-\frac{2}{3}x=-\frac{1}{6}\)
\(\Leftrightarrow x=-\frac{1}{6}:-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b. \(\frac{3}{x+5}=15\%\left(ĐKXĐ:x\ne-5\right)\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{3}{20}\)
\(\Leftrightarrow\frac{60}{20\left(x+5\right)}=\frac{3\left(x+5\right)}{20\left(x+5\right)}\)
\(\Leftrightarrow60x=3x+15\)
\(\Leftrightarrow-3x=-45\)
\(\Leftrightarrow x=15\)
Bài 2:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
Câu a:
-4\(\frac38\) + (39\(\frac38\) - 5\(\frac{7}{32}\))
= -4 - \(\frac38\) + 39 + \(\frac38\) - 5 - \(\frac{7}{32}\)
= (-4 + 39 - 5) + (-3/8 + 3/8 - 7/32)
= (35 - 5) + (0 - 7/32)
= 30 - 7/32
= 953/32
Câu b:
2\(\frac12\) + 19\(\frac{3}{11}\).\(\frac{7}{26}\) -6\(\frac{3}{11}\).\(\frac{7}{26}\)
= 5/2 + 7/26.(19 + 3/11 - 6 - 3/11)
= 5/2 + 7/26.[(19 - 6) + (3/11 - 3/11)
= 5/2 + 7/26.[13 + 0]
= 5/2 + 7/26.13
= 5/2 + 7/2
= 6
a) 2A= 1+1/2^2+1/2^3+...+1/2^2015+1/2^2016
2A-A=(1+1/2+1/2^2+...+1/2^2015+1/2^2016)-(1/2+1/2^2+...+1/2^2016+1/2^2017)
A= 1-1/2^2017
b) B=5.(5/1.6+5/6.11+...+5/26.31)
B=5.(1/5-1/6+1/6-1/11+1/11...-1/26+1/26-1/31)
B= 5.(1/5-1/31)
B=5.26/155
B=26/31
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2016}{2017}\)
\(=\frac{1.2.3......2016}{2.3.4.......2017}\)
\(=\frac{1}{2017}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)\(\Leftrightarrow A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}\)
nhân A với 1/2
ta có: A= 1/2+1/2^2+1/2^3+...+1/2^10
1/2.A= 1/2^2+1/2^3+...+1/2^11
lấy A-1/2.A= 1/2-1/2^11
1/2.A=1/2-1/2^11
A= (...-...):1/2
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}_{ }+...+\frac{1}{2^{10}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}_{ }+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)