`a,(2x-5)(12+5x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{12}{5}\end{matrix}\right.\)

`b, (x-3)(x-4)-2(x-3)=0`

`<=>(x-3)(x-4-2)=0`

`<=>(x-3)(x-6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

`c, x(x-1)(x+1)=0`

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

`d, (2x)/3 +(2x-1)/6=0`

`<=> (4x)/6 +(2x-1)/6=0`

`<=> (4x+2x-1)/6=0`

`<=> (6x-1)/6=0`

`<=> 6x-1=0`

`<=> 6x=1`

`<=>x=1/6` ( đề là vậy à bạn )

 a) \(\left(2x-5\right)\left(12+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-2,4\end{matrix}\right.\)

b) \(\left(x-3\right)\left(x-4\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(x-4\right)-2\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

c) \(x\left(x-1\right)\left(x+1\right)=0\)

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

d) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=0\)

\(\Leftrightarrow\dfrac{4x+2x-1}{6}=0\)

\(\Leftrightarrow6x-1=0\)

\(\Leftrightarrow6x=1\Leftrightarrow x=\dfrac{1}{6}\)

a) \(\Leftrightarrow36+3-11x=0\)

\(\Leftrightarrow-11x=-39\)

\(\Leftrightarrow x=\frac{39}{11}\)

b) \(x^2-2x\frac{1}{4}+\frac{1}{16}-\frac{81}{16}=0\)

\(\left(x-\frac{1}{4}\right)^2=\frac{81}{16}\)

\(x-\frac{1}{4}=\frac{9}{4}\)

\(x=\frac{10}{4}=\frac{5}{2}\)

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

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

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

x = 2 hoặc x = - 2 hoặc x = 3

a) \(\frac{8}{2}\)

b) \(\frac{5}{2}\)

c) x=2 hoạc x=-2 hoặc x=3

a) \(2x-12-x=0\)

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)

Mọi người giúp mình nhanh nhé ! Mình đang cần gấp

a) x = 15

b) x = 11

c) x = 68

d) x = 310

(x4)3=x6x19

a) x = 15.              b)      x = 11.        c) x = 68.              d) x = 310

Tìm xx biết:    \left(x^{4}\right)^{3}=\dfrac{x^{19}}{x^{6}}(x4)3=x6x19​

Trả lời: x=x= 

a) \(9x^2-6x+3=0\)

\(\Leftrightarrow\left(3x\right)^2-2.3x.1+1^2+2=0\)

\(\Leftrightarrow\left(3x-1\right)^2=-2\) ( vô lí )

b) \(x^2-7x+12=0\)

\(\Leftrightarrow x^2-2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2=\frac{1}{4}=\left(-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{2}=\frac{1}{2}\\x-\frac{7}{2}=-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}\)

Vậy : \(x\in\left\{3,4\right\}\)

c) \(x^2-8x+6=0\)

\(\Leftrightarrow x^2-2.x.4+4^2-10=0\)

\(\Leftrightarrow\left(x-4\right)^2=10\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=\sqrt{10}\\x-4=-\sqrt{10}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{10}+4\\x=-\sqrt{10}+4\end{cases}}\)