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30 tháng 6 2021

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

30 tháng 6 2021

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

a: Ta có: \(2x\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)

a) Ta có: \(\left(x-3\right)^2-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)

b) Ta có: \(x:0.25+x:0.2+x:0.1+x=34\)

\(\Leftrightarrow4x+5x+x+x=34\)

\(\Leftrightarrow11x=34\)

hay \(x=\dfrac{34}{11}\)

16 tháng 11 2021

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)

7 tháng 9 2021

a) \(x^2-64=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

b) \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

c) \(9-6x+x^2=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

a: Ta có: \(x^2-64=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

b: Ta có: \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

hay \(x=\dfrac{1}{2}\)

c: ta có: \(x^2-6x+9=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

hay x=3

21 tháng 9 2021

a. 9x2 - 6x - 3 = 0

<=> 3(3x2 - 2x - 1) = 0

<=> 3(3x2 - 3x + x - 1) = 0

<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)

<=> 3(3x + 1)(x - 1) = 0

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

b. (2x + 1)2 - 4(x + 2)2 = 9

<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)

<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9

<=> -3(4x + 5) = 9

<=> 4x + 5 = -3

<=> 5 + 3 = -4x

<=> -4x = 8

<=> -x = 2

<=> x = -2

21 tháng 9 2021

a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)

\(\Leftrightarrow\left(3x-1\right)^2-4=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)

c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)