Lời giải:

$2x=3y\Leftrightarrow \frac{x}{3}=\frac{y}{2}\Leftrightarrow \frac{x}{6}=\frac{y}{4}$

$5y=4z\Leftrightarrow \frac{y}{4}=\frac{z}{5}$

Vậy:

$\frac{x}{6}=\frac{y}{4}=\frac{z}{5}$

$\Rightarrow (\frac{x}{6})^3=(\frac{y}{4})^3=(\frac{z}{5})^3=\frac{xyz}{6.4.5}=\frac{120}{120}=1$

$\Rightarrow \frac{x}{6}=\frac{y}{4}=\frac{z}{5}=1$

$\Rightarrow x=6; y=4; z=5$

Ta có: 2x=3y

nên \(\dfrac{x}{3}=\dfrac{y}{2}\)

hay \(\dfrac{x}{6}=\dfrac{y}{4}\)(1)

Ta có: 5y=4z

nên \(\dfrac{y}{4}=\dfrac{z}{5}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=4k\\z=5k\end{matrix}\right.\)

Ta có: xyz=120

nên \(120k^3=120\)

\(\Leftrightarrow k^3=1\)

hay k=1

Suy ra: x=6; y=4; z=5

(x-5)(2x-4)=0

\(\Rightarrow\hept{\begin{cases}x-5=0\\2x-4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=2\end{cases}}\)

Vậy x=2 hoặc x=5

(x - 5) . (2x - 4) = 0

x - 5=0  => x=5

2x - 4=0  => 2x= 4  => x=2

vậy x=5 hoặc x=2

xog rùi nka

\(\left(x-5\right)\left(2x-4\right)\)

\(\Rightarrow\hept{\begin{cases}x-5=0\\2x-4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=2\end{cases}}\)

vậy x=5 hoặc x=2

Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x² - 10x - 2x² + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x² + 2x + x + 2 - x² - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x² - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x² - 4x + 5 - 4x² + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x² = 4x - 5

<=> 2x² + x + 6x + 3 - 2x² - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0

<=> - 8x + 32 + 8x² + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

x=-15/6

a)\(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

    \(4x^2-20x-\left(4x^2-7x+3\right)=5\)

    \(4x^2-20x-4x^2+7x-3=5\)

     \(-13x=8\)

      \(x=-\frac{8}{13}\)

b)\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

     \(48x^2-32x+5+3x-48x^2-7+112x=81\)

     \(83x-2=81\)

     \(x=1\)

\(\frac{2x-1}{2}=\frac{8}{2x-1}\)\(\Rightarrow\)\(\left(2x-1\right)^2=16\)

\(\Rightarrow2x-1=\pm4\)

\(\Rightarrow\hept{\begin{cases}2x-1=4\\2x-1=-4\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\x=\hept{\begin{cases}-3\\2\end{cases}}\end{cases}}}\)

x=0,25 nha ban

ĐKXĐ: \(x\ne\frac{1}{2}\)

Ta có:

\(\frac{2x-1}{2}=\frac{8}{2x-1}\)

\(\Rightarrow\left(2x-1\right)^2=16=4^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=4\\2x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\left(TMĐKXĐ\right)\\x=-\frac{3}{2}\left(TMĐKXĐ\right)\end{cases}}}\)