a, 2x + 12 = 434

2x = 434 - 12 = 422

x = 422 : 2 = 211

b. 25 + 27 ( x - 8 ) = 106

27 ( x - 8 ) = 106 - 25 = 81

x - 8 = 81 : 27 = 3

x = 3 + 8 = 11

Bài làm

a) Ta có: x( x + 7 ) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)

Vậy x = { 0; -7 }

b) Ta có: ( x + 12 )( x - 3 ) = 0

\(\Rightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)

Vậy x = { -12; 3 }

c) \(\frac{2}{x}=\frac{-8}{12}\)

\(\Rightarrow x.\left(-8\right)=2.12\)

\(\Rightarrow x.\left(-8\right)=24\)

\(\Rightarrow x=24:\left(-8\right)=-3\)

Vậy x = -3

a, x(x+7)=0                b,(x+12).(x-3)=0                       c,2/x= -8/12

 x+7=0.x                       x+12=0 hoặc x-3=0                   2/x= -2/3

x+7=0                           x=0-12 hoặc x=0+3                  2/x= 2/-3

x=0                                x=-12 hoặc x=3                     =>x=-3

a, x(x + 7) = 0

<=> x = 0 hoặc x + 7 = 0

<=> x = 0 hoặc x = -7

b, (x + 12)(x - 3) = 0

<=> x + 12 = 0 hoặc x - 3 = 0

<=> x = -12 hoặc x = 3

c, \(\frac{2}{x}=\frac{-8}{12}\)

\(\Leftrightarrow-8x=2.12\)

\(\Leftrightarrow-8x=24\)

\(\Leftrightarrow x=-3\)

`7xx8xx12/6xx21xx16`

`=7xx8xx2xxx21xx16`

`=56xx2xx21xx16`

`=112xx21xx16`

`=2352xx16`

`=37632`

\(\dfrac{7\times8\times12}{6\times21\times16}=\dfrac{7\times8\times3\times2\times2}{2\times3\times3\times7\times8\times2}=\dfrac{1}{3}\)

Cảm ơn bạn nha không là mình chết mất

      \(8\left(x-\frac{1}{2}\right)\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)-4x\left(1-x+2x^2\right)+2=0\)

\(\Leftrightarrow8\left[x^3-\left(\frac{1}{2}\right)^3\right]-4x+4x^2-8x^3+2=0\)

\(\Leftrightarrow8x^3-1-4x+4x^2-8x^3+2=0\)

\(\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\)

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

8(x-1/2)(x^2+1/2x+1/4)  -  4x(1-x+2x^2)+2=0

=>   8𝑥^3 − 1  −  8𝑥^3 + 4𝑥2 − 4𝑥 + 2  =  0

=>   4𝑥2 − 4𝑥 + 1 = 0

=>  ( 2x - 1 )^2  = 0

=>  2x - 1 = 0

=>  2x  =  1

=>  x  = 1/2

<=> 8( x³ - 1/8 ) - 4x + 4x² - 8x³ + 2 = 0

<=> 8x³ - 1 - 4x + 4x² - 8x³ + 2 = 0

<=> 4x² - 4x + 1 = 0

<=> ( 2x - 1 )² = 0 <=> x = 1/2

x+ 2 + x+ 4 + x +6 + x + 8 + x + 10 + x + 12 = 192

6x +  2+ 4 + 6 + 8 + 10 + 12  = 192

6x+ 42 = 192 

6x= 192 - 42

6x= 150

x = 150 : 6

x= 25

x+2+x+4+x+6+x+8+x+10+x+12= 192

<=> 6x + 42 = 192

<=> 6x = 150

<=> x = 25

x+ 2 + x+ 4 + x +6 + x + 8 + x + 10 + x + 12 = 192

6x +  2+ 4 + 6 + 8 + 10 + 12  = 192

6x+ 42 = 192 

6x= 192 - 42

6x= 150

x = 150 : 6

x= 25

\(a)\)\(\left[\left(8.x-12\right)\div4\right].3^3=3^6\)

\(\left[\left(8.x-12\right)\div4\right]=3^6\div3^3\)

\(\left[\left(8.x-12\right)\div4\right]=3^3\)

\(\left(8.x-12\right)\div4=27\)

\(\left(8.x-12\right)=27.4\)

\(8.x-12=108\)

\(8.x=108+12\)

\(8.x=120\)

\(x=120\div8\)

\(x=15\)

\(b)\)\(3^{2.x-4}-x^0=8\)

\(3^{2.x-4}-1=8\)

\(3^{2.x-4}=8+1\)

\(3^{2.x-4}=9\)

\(3^{2.x-4}=3^2\)

\(2.x-4=2\)

\(2.x=2+4\)

\(2.x=6\)

\(x=3\)

Sửa đề

\(c)\)\(65-4^{x+2}=2014^0\)

\(65-4^{x+2}=1\)

\(4^{x+2}=65-1\)

\(4^{x+2}=64\)

\(4^{x+2}=4^3\)

\(x+2=3\)

\(x=3-2\)

\(x=1\)

1) 5|x-3| = \(\frac{5}{7}\)

=> |x-3| = \(\frac{1}{7}\)

\(\Rightarrow\orbr{\begin{cases}x-3=\frac{1}{7}\\x-3=\frac{-1}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{22}{7}\\x=\frac{20}{7}\end{cases}}\)

Vậy x \(\in\){\(\frac{22}{7};\frac{20}{7}\)}

2) |x-11/7| = 0

=> x-11/7 = 0

=> x = 11/7

- Cía chỗ bài tìm x thứ 2 mình viết nhầm số 11/12 thành 11/7

2) |x-11/12|=0

=> x-11/12 = 0

=> x = 11/12