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\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{48}{49}.\dfrac{49}{50}=\dfrac{1}{50}\)
`a)25/(x+1)-1 1/6=-1/3-0,5`
`=>25/(x+1)=-1/3-1/2+1+1/6`
`=>25/(x+1)=1/3`
`=>75=x+1`
`=>x=74`
Vậy `x=74`
`b)(2x+25 3/5)^2-9/25=0`
`=>(2x+128/5)=9/25`
`**2x+128/5=3/5`
`=>2x=-125/5=-25`
`=>x=-25/2`
`**2x+128/5=-3/5`
`=>2x=-131/5`
`=>x=-131/10`
Giải:
a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\)
\(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\)
\(\dfrac{25}{x+1}=\dfrac{1}{3}\)
\(\Rightarrow1.\left(x+1\right)=25.3\)
\(\Rightarrow x+1=75\)
\(\Rightarrow x=75-1\)
\(\Rightarrow x=74\)
b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\)
\(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\)
Chúc bạn học tốt!
a) x - 1/2 = 3/5
x = 3/5 + 1/2
x = 11/10
b) x - 1/2 = -2/3
x = -2/3 + 1/2
x = -1/6
c) 2/5 - x = 0,25
x = 2/5 - 0,25
x = 2/5 - 1/4
x = 3/20
\(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{100}\)
\(\Rightarrow A=\dfrac{99}{100}\)
=1/2+1/3+1/4+...+1/100
xét mẫu:có ssh là (100-2):1+1=99 số
tổng là (100+2)*99:2=5940
vậy ta có 1/5940
B = 1+1³+1⁴+...+1⁹⁸+1⁹⁹
B1 = 1³+1⁵+...+1⁹⁹+1¹⁰⁰
B1-B =(1³+1⁵+...+1⁹⁹+1¹⁰⁰) - ( 1+1³+1⁴+...+1⁹⁸+1⁹⁹ )
B0 = 1¹⁰⁰ - 1⁹⁹
B = 1
\(x^2+2x+4⋮x+1\)
\(\Leftrightarrow\left(x^2+x\right)+\left(x+1\right)+3⋮x+1\)
\(\Leftrightarrow x\left(x+1\right)+\left(x+1\right)+3⋮x+1\)
\(\Leftrightarrow3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;2;-4\right\}\)
Ta có: \(x^2+2x+4\)
\(=\left(x^2+x\right)+\left(x+1\right)+3\)
\(=x\left(x+1\right)+\left(x+1\right)+3\)
\(=\left(x+1\right)\left(x+1\right)+3\)
Để \(x^2+2x+4\) chia hết cho x + 1 thì 3 phải chia hết cho x + 1
\(\Rightarrow\left(x+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{-4;-2;0;2\right\}\)
\(\left(\frac{1}{2}-1\right)\times\left(\frac{1}{3}-1\right)\times\left(\frac{1}{4}-1\right)\times...\times\left(\frac{1}{1963}-1\right)\)
\(=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{1963}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1962}{1963}\)
\(=\frac{1}{1963}\)
Đây là bài tính nhanh ạ!!