Tìm x :
a) (x-2)3=(x-2)2
b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
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TK
1
VT
4 tháng 4 2022
\(x\) \((\)\(\dfrac{3}{2.5}\) \(+
\) \(\dfrac{3}{5.8}\) \(+\) \(\dfrac{3}{8.11}\) \(+\) \(\dfrac{3}{11.14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{5}\) \(+\) \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{8}\) \(+\) \(\dfrac{1}{8}\) \(-\) \(\dfrac{1}{11}\) \(+\) \(\dfrac{1}{11}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) x \(\dfrac{3}{7}\) \(=\) \(\dfrac{1}{21}\)
\(x\) \(=\) \(\dfrac{1}{21}\) \(:\) \(\dfrac{3}{7}\)
\(x\) \(=\) \(\dfrac{1}{9}\)
NH
15 tháng 7 2023
`3x-15/(5*8)-15/(8*11)-15/(11*14)-...-15/(47*50)=2 1/10`
`3x-(15/(5*8)+15/(8*11)+15/(11*14)+...+15/(47*50))=21/10`
`3x-5(3/(5*8)+3/(8*11)+3/(11*14)+...+3/(47*50))=21/10`
`3x-5(1/5-1/8+1/8-1/11+1/11-1/14+...+1/47-1/50)=21/10`
`3x-5(1/5-1/50)=21/10`
`3x-5*9/50=21/10`
`3x-9/10=21/10`
`3x=21/10+9/10`
`3x=3`
`x=1`
14 tháng 5 2017
\(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+\dfrac{3x}{8.11}+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
\(\Rightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)
\(\Rightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Rightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Rightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\)
\(\Rightarrow x=\dfrac{1}{21}.\dfrac{7}{3}\)
\(\Rightarrow x=\dfrac{1}{9}\)
Vậy \(x=\dfrac{1}{9}\)
5 tháng 1 2022
\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)
\(b,\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x+3}{x^2-1}\\ =\dfrac{3x^2+4x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+4x+1-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2+2x-3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{2x^2+6x-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x+3}{\left(x-1\right)^2}\)
MB
4
12 tháng 5 2015
3x/2.5 + 3x/5.8+3x/8.11+3x/11.14 = 1/21
x(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14) = 1/21
x(1/2-1/14) = 1/21
x . 3/7 = 1/21
=> x = 1/21 : 3/7
=> x = 1/9
Hihi mình giải zầy mk hk bik đúng hay sai
1 tháng 4 2023
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
24 tháng 7 2017
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
TV
3
12 tháng 5 2015
Ta có : \(\frac{3x}{2\times5}+\frac{3x}{5\times8}+\frac{3x}{8\times11}+\frac{3x}{11\times14}=\frac{1}{21}\)
\(\Rightarrow x\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}\right)=\frac{1}{21}\)
\(\Rightarrow x\times\left(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}\right)=\frac{1}{21}\)
\(x\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(x\times\left(\frac{1}{2}-\frac{1}{14}\right)\) \(=\frac{1}{21}\)
\(x\times\frac{3}{7}\) \(=\frac{1}{21}\)
\(x\) \(=\frac{1}{21}\div\frac{3}{7}=\frac{1}{21}\times\frac{7}{3}\)
\(\Rightarrow x=\frac{1}{9}\)
6 tháng 5 2018
Ta có 3x/2.5+3x/5.8+3x/8.11+3x/11.14=1/21
=>x(3/2.5+3/5.8+3/8.11+3/11.14)=1/21
=>3x(1/2.5+1/5.8+1/8.11+1/11.14)=1/21
=>3x(1/2-1/14)=1/21
=>3x.3/7=1/21
=>3x=1/21:3/7
=>3x=1
=>x=1:3=1/3
11 tháng 7 2023
a: =>2(2x-3)-9=5-3x-2
=>4x-6-9=-3x+3
=>4x-15=-3x+3
=>7x=18
=>x=18/7
b: =>\(\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\dfrac{21}{3x}+2\)
=>\(\dfrac{23}{3x}=\dfrac{4}{5}+2+\dfrac{1}{4}=\dfrac{61}{20}\)
=>3x=460/61
=>x=460/183
a, dễ, tự làm
b, \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+.........+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.....+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\)
\(\Leftrightarrow x=\dfrac{1}{9}\)
Vậy ...
a) (x-2)3 = (x-2)2
<=> (x-2)3-(x-2)2 = 0
<=> (x-2)2(x-2-1) = 0
<=> \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x-3=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
<=> \(x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)
<=> \(x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
<=> \(x=\dfrac{1}{21}:\dfrac{3}{7}\)
<=> \(x=\dfrac{1}{9}\)