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 \(\left(3.x-1\right)^3=-\dfrac{8}{27}\Rightarrow3.x-1=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{9}\)

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 = \(\dfrac{-18}{11}\)

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= \(\dfrac{1}{6}\)

a. \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

=> \(\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)

=> x + \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\) hoặc \(x+\dfrac{1}{2}=\dfrac{-1}{4}\)

=> x = \(\dfrac{-1}{4}\) hoặc \(x=\dfrac{-3}{4}\)

\(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x-\frac{1}{2}=0\)

<=> \(x=\frac{1}{2}\)

\(\left(x-2\right)^2=1\)

<=> \(\hept{\begin{cases}x-2=1\\x-2=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\x=1\end{cases}}\)

\(\left(2x+3\right)^2=\frac{9}{121}\)

<=-> \(\hept{\begin{cases}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{cases}}\)

<=> \(\hept{\begin{cases}2x=\frac{-30}{11}\\2x=\frac{-36}{11}\end{cases}}\)

\(2x^{10}=25x^8\)

<=> \(2x^{10}-25x^8=0\)

<=> \(x^8.\left(2x^2-25\right)=0\)

<=> \(\hept{\begin{cases}x^8=0\\2x^2-25=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=0\\x^2=\frac{25}{2}\end{cases}}\)

<=> \(\hept{\begin{cases}x=0\\x=\sqrt{\frac{25}{2}}\\x=-\sqrt{\frac{25}{2}}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{cases}}\)

`@` `\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`

`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)

\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)

\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)

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

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 \(x\in\left\{0;3;-11;11\right\}\)

`P <= 1` là `P` ở đâu cậu nhỉ cộng `A` với `B` lại với nhau à?