a,\((\) 1\(-\) \(\dfrac{1}{3}\)\()\)x\((\)1\(-\)\(\dfrac{2}{5}\)\()\)x\((\)1\(-\)\(\dfrac{2}{7}\)\()\)x\((\)1\(-\)\(\dfrac{2}{9}\)\()\)
b,\(\dfrac{1}{1x3}\) + \(\dfrac{1}{3x5}\) + \(\dfrac{1}{5x7}\) + \(\dfrac{1}{7x9}\)
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\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}=\dfrac{49}{99}>\dfrac{49}{100}=A\)
\(I=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}\)
Lời giải:
\(2\times I=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{199\times 201}\)
\(=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+....+\frac{201-199}{199\times 201}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\)
\(=1-\frac{1}{201}=\frac{200}{201}\)
\(I=\frac{200}{201}:2=\frac{100}{201}\)
a: x+2/5=1/2
=>x=1/2-2/5=5/10-4/10=1/10
b; x-2/5=2/7
=>x=2/7+2/5=10/35+14/35=24/35
c: 3/5-x=1/10
=>x=3/5-1/10=6/10-1/10=5/10=1/2
d: x*3/4=9/20
=>x=9/20:3/4=9/20*4/3=36/60=3/5
e: x:1/7=14
=>x=14*1/7=2
f: =>x+1/4=2/5:1/2=4/5
=>x=4/5-1/4=16/20-5/20=11/20
g: =>x*2/3=9/12+2/3=3/4+2/3=9/12+8/12=17/12
=>x=17/12:2/3=17/12*3/2=51/24=17/8
a) \(\Leftrightarrow\dfrac{3}{2}:x=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{3}{2}:\dfrac{1}{2}\\ \Leftrightarrow x=3\)
b) \(\Leftrightarrow x=\dfrac{7}{9}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)
c) \(\Leftrightarrow x=\dfrac{8}{7}:\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{4}{3}\)
d) \(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{3}{7}\\ \Leftrightarrow x=\dfrac{48}{35}\)
a) x = 3
b) x = \(\dfrac{1}{9}\)
c) x = \(\dfrac{4}{3}\)
d)\(\dfrac{48}{35}\)
`a/`
` 7/5 + 3 2/5 - 1 1/2 `
`= (7/5 + 17/5) - 3/2`
`= 24/5 - 3/2 `
`= 48/10 - 15/10 `
`= 33/10 `
`b/`
` 3 xx 2 4/9 xx 3/2 `
` = 3 xx 22/9 xx 3/2 `
` = 22/3 xx 3/2`
`= 11.`
`c/`
` 5/9 xx ( 2 1/6 - 1 2/3 ) `
`= 5/9 xx ( 13/6 -5/3 )`
`= 5/9 xx ( 13/6 - 10/6 ) `
`= 5/9 xx 3/6 `
`= 5/9 xx 1/2 `
`= 5/18`
bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu
a: =>\(4x-5=2x-2+x=3x-2\)
=>x=3
b: \(\Leftrightarrow7x-35=3x+6\)
=>4x=41
=>x=41/4
c: =>(2x+5)(x+5)-2x^2=0
=>2x^2+10x+5x+25-2x^2=0
=>15x=-25
=>x=-5/3
e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)
=>11(x^2-3x-4)=x(11x-34)
=>11x^2-33x-44=11x^2-34x
=>x=44
a, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
\(\dfrac{-3}{x-1}+\dfrac{1}{x}+\dfrac{2}{x+2}=0\left(ĐKXĐ:x\notin\left\{1;0;-2\right\}\right)\\ \Leftrightarrow\dfrac{-3x\left(x+2\right)+\left(x-1\right)\left(x+2\right)+2x\left(x-1\right)}{\left(x-1\right).x.\left(x+2\right)}=0\\ \Leftrightarrow-3x^2-6x+x^2+2x-x-2+2x^2-2x=0\\ \Leftrightarrow-7x-2=0\\ \Leftrightarrow x=\dfrac{-2}{7}\)
Chọn B
a) \(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{2}{5}\right)\times\left(1-\dfrac{2}{7}\right)\times\left(1-\dfrac{2}{9}\right)\)
\(=\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{5}{5}-\dfrac{2}{5}\right)\times\left(\dfrac{7}{7}-\dfrac{2}{7}\right)\times\left(\dfrac{9}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{2}{3}\times\dfrac{3}{5}\times\dfrac{5}{7}\times\dfrac{7}{9}\)
\(=\dfrac{2\times3\times5\times7}{3\times5\times7\times9}\)
\(=\dfrac{2}{9}\)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(=1-\dfrac{1}{9}\)
\(=\dfrac{9}{9}-\dfrac{1}{9}\)
\(=\dfrac{8}{9}\)
Sửa câu b)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
Đặt \(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(2A=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(2A=1-\dfrac{1}{9}\)
\(2A=\dfrac{9}{9}-\dfrac{1}{9}\)
\(2A=\dfrac{8}{9}\)
\(A=\dfrac{8}{9}:2\)
\(A=\dfrac{8}{18}\)
\(A=\dfrac{4}{9}\)
Vậy : \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}=\dfrac{4}{9}\)