B = -1/2 - 2/3 +-3/4+2/3
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Bài 1
a) 3 2/5 - 1/2
= 17/5 - 1/2
= 34/10 - 5/10
= 29/10
b) 4/5 + 1/5 × 3/4
= 4/5 + 3/20
= 16/20 + 3/20
= 19/20
c) 3 1/2 × 1 1/7
= 7/2 × 8/7
= 4
d) 4 1/6 : 2 1/3
= 25/6 : 7/3
= 25/14
Bài 2
a) 3 × 1/2 + 1/4 × 1/3
= 3/2 + 1/12
= 18/12 + 1/12
= 19/12
b) 1 4/5 - 2/3 : 2 1/3
= 9/5 - 2/3 : 7/3
= 9/5 - 2/7
= 63/35 - 10/35
= 53/35
a)\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(A=1-\frac{1}{2^{50}}
Bài 1:
A = 1 + 3 + 32 + ... + 3100
=> 3A = 3 + 32 + ... + 3101
=> 2A = 3101 - 1
=> A = \(\frac{3^{101}-1}{2}\)
B = 1 + 42 + 44 + ... + 4100
=> 8B = 42 + 44 + ... + 4102
=> 7B = 4102 - 1
=> B = \(\frac{4^{102}-1}{7}\)
Bài 2:
a) S1 = 22 + 42 + ... + 202
=> S1 = 22(1+22+...+102)
=> S1 = 22.385
=> S1 = 1540
b) S2 = 1002 + 2002 + ... + 10002
=> S2 = 1002(1+22+...+102)
=> S2 = 1002.385
=> S2 = 3850000
`@` `\text {Ans}`
`\downarrow`
`a,`
\(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{6}{5}-x=\dfrac{2}{3}\)
\(x=\dfrac{6}{5}-\dfrac{2}{3}\)
\(x=\dfrac{8}{15}\)
Vậy, `x = \dfrac{8}{15}`
`b,`
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times3\dfrac{1}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div3\dfrac{1}{3}\)
\(x=\dfrac{4}{17}\)
Vậy, `x=`\(\dfrac{4}{17}\)
`c,`
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{34}{7}\)
Vậy, `x=`\(\dfrac{34}{7}\)
a,\(\dfrac{3}{2}.\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{4}{5}-x=\dfrac{2}{3}:\dfrac{3}{2}\)
\(\dfrac{4}{5}-x=\dfrac{4}{9}\)
\(x=\dfrac{4}{5}-\dfrac{4}{9}\)
\(x=\dfrac{16}{45}\)
a) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)
\(x-\dfrac{3}{4}=\dfrac{9}{4}\)
=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)
b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)
\(\dfrac{7}{8}:x=\dfrac{5}{2}\)
=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)
c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)
d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{6}{5}-x=\dfrac{2}{3}\)
=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)
e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)
f) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\)
\(\dfrac{17}{3}:x=\dfrac{5}{3}\)
=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)
a: =>x-3/4=18/8=9/4
=>x=9/4+3/4=12/4=3
b: =>7/8:x=5/2
=>x=7/8:5/2=7/8*2/5=14/40=7/20
c: x+1/2*1/3=3/4
=>x+1/6=3/4
=>x=3/4-1/6=9/12-2/12=7/12
d: =>12/10-x=2/3
=>6/5-x=2/3
=>x=6/5-2/3=18/15-10/15=8/15
e: =>x*10/3=10/3:17/4=10/3*4/17
=>x=4/17
f: =>17/3:x=13/3-5/2=26/6-15/6=11/6
=>x=17/3:11/6=17/3*6/11=34/11
`#040911`
a,
\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)
Vậy, \(x=-\dfrac{8}{21}\)
b,
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, \(x\in\left\{-2;3\right\}\)
c,
\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)
Bạn xem lại đề có sai kh nhỉ?
c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)
`B= -1/2 - 2/3 + (-3/4) +2/3`
`= (-1/2 + (-3)/4 ) + (-2/3 +2/3)`
`= ( -2/4 + (-3)/4 ) +0`
`= -5/4+0`
`=-5/4`