tìm x
( 2x + 3 )2 = \(\dfrac{9}{121}\)
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10 tháng 6 2017
a) Vì \(\left(2.x+3\right)^2=\dfrac{9}{121}\Rightarrow\left\{{}\begin{matrix}2.x+3=\dfrac{3}{11}\\2.x+3=-\dfrac{3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{15}{11}\\x=-\dfrac{18}{11}\end{matrix}\right.\)
b) Vì \(\left(3.x-1\right)^3=-\dfrac{8}{27}\Rightarrow3.x-1=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{9}\)
VN
22 tháng 7 2018
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
VN
22 tháng 7 2018
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
NH
1
TN
24 tháng 7 2018
\(\left(x-\frac{1}{2}\right)^2=0\)
<=> \(x-\frac{1}{2}=0\)
<=> \(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
<=> \(\hept{\begin{cases}x-2=1\\x-2=-1\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\x=1\end{cases}}\)
\(\left(2x+3\right)^2=\frac{9}{121}\)
<=-> \(\hept{\begin{cases}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{cases}}\)
<=> \(\hept{\begin{cases}2x=\frac{-30}{11}\\2x=\frac{-36}{11}\end{cases}}\)
\(2x^{10}=25x^8\)
<=> \(2x^{10}-25x^8=0\)
<=> \(x^8.\left(2x^2-25\right)=0\)
<=> \(\hept{\begin{cases}x^8=0\\2x^2-25=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x^2=\frac{25}{2}\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x=\sqrt{\frac{25}{2}}\\x=-\sqrt{\frac{25}{2}}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{cases}}\)
15 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
`=> (x-3)5 = (2x+1)3`
`=> 5x-15 = 6x+3`
`=> 5x-6x = 15+3`
`=> -x=18`
`=> x=-18`
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
`=> (x+1)x = 22*6`
`=> (x+1)x = 132`
`=> x^2 + x = 132`
`=> x^2+x-132=0`
`=> (x-11)(x+12)=0`
`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
`=> (2x-1)x = 2*5`
`=> 2x^2 - x =10`
`=> 2x^2 - x - 10 =0`
`=> 2x^2 + 4x - 5x - 10 =0`
`=> (2x^2 + 4x) - (5x+10)=0`
`=> 2x(x+2) - 5(x+2)=0`
`=> (2x-5)(x+2)=0`
`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
`=> (2x-1)(2x+1)=21*3`
`=> 4x^2 + 2x - 2x - 1 = 63`
`=> 4x^2 - 1=63`
`=> 4x^2 - 1 - 63=0`
`=> 4x^2 - 64 = 0`
`=> 4(x^2 - 16)=0`
`=> 4(x^2 + 4x - 4x - 16)=0`
`=> 4[(x^2+4x)-(4x+16)]=0`
`=> 4[x(x+4)-4(x+4)]=0`
`=> 4(x-4)(x+4)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
`=> (2x+1)(x+1) = 9*5`
`=> (2x+1)(x+1)=45`
`=> 2x^2 + 2x + x + 1 = 45`
`=> 2x^2 + 3x + 1 =45`
`=> 2x^2 + 3x + 1 - 45 =0`
`=> 2x^2+3x-44=0`
`=> 2x^2 + 11x - 8x - 44=0`
`=> (2x^2 +11x) - (8x+44)=0`
`=> x(2x+11) - 4(2x+11)=0`
`=> (x-4)(2x+11)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)
BD
15 tháng 6 2023
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)
6 tháng 1 2022
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
6 tháng 1 2022
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)
\(\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\left(2x+3\right)=\dfrac{3}{11}\)
\(2x=\dfrac{3}{11}-3\)
\(2x=-\dfrac{14}{11}\)
\(x=-\dfrac{7}{11}\)
\(\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\Rightarrow2x+3=\dfrac{3}{11}\)
=> 2x = \(\dfrac{3}{11}-3\)
=> 2x = \(-\dfrac{30}{11}\)
=> x = \(-\dfrac{30}{11}\div2=-\dfrac{15}{11}\)