ko pit

2x^2-50x=0

2x(x-25)=0

+2x=0=>x=0

+x-25=0=>x=25

      2x^2 - 50x=0

<=>x.(2x-50)=0

<=>2x-50=0:x=0

<=>2x=50 => x=25 và 0 nhá :D

c)(x-4).(2x+6)=0

=>(x-4)=0 hoặc (2x+6)=0

với x-4 = 0

      x    =0+4

      x    =4

với 2x+6=0

      2x    =0-6

      2x    =-6

      x      =-6:2

      x      =-3

3x=(-7)+1

3x=(-6)

x=(-6):3

x=(-2)

=(-5)=(-5).(-2) là sao vậy bạn

a) (x + 3) . (x - \(\frac{1}{2}\)) = 0

=> \(\hept{\begin{cases}x+3=0\\x-\frac{1}{2}=0\end{cases}}\)

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

ok nha!! 5756758769723414657765887805674765756568678568

a, 1 - 2x < 7

=> -2x < 6

=> x < -3

=> x thuộc {-4; -5; -6; ...}

b, \(\left(x-1\right)\left(x-2\right)>0\)

th1 :

\(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 1\\x< 2\end{cases}\Rightarrow}x< 1\Rightarrow x\in\left\{0;-1;-2;...\right\}}\)

th2 :

\(\hept{\begin{cases}x-1>0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>2\end{cases}\Rightarrow}x>2\Rightarrow x\in\left\{3;4;5;...\right\}}\)

vậy_

c tương tự b

\(a.1-2x< 7\Leftrightarrow2x< 7+1=8\Leftrightarrow x< 8:2\Leftrightarrow x< 4\)

Vậy x < 4

\(b.\left(x-1\right)\left(x-2\right)>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1>0;x-2>0\\x-1< 0;x-2< 0\end{cases}}\)

\(TH1\Leftrightarrow\orbr{\begin{cases}x-1>0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x>0+1=1\\x>0+2=2\end{cases}\Rightarrow x>2}}\)

\(TH2\Leftrightarrow\orbr{\begin{cases}x-1< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 0+1=1\\x< 0+2=2\end{cases}\Rightarrow}}x< 2\)

Vậy \(x\ne2\)

Sửa lại câu a,

\(1-2x< 7\Leftrightarrow2x< 1-7=-6\Leftrightarrow x< -6:2\Leftrightarrow x< -3.\)

c, \(\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)

\(TH1\hept{\begin{cases}\left(x-2\right)^2< 0\\\left(x+1\right)>0\\\left(x-4\right)>0\end{cases}}\)không xảy ra

\(TH2\hept{\begin{cases}\left(x-2\right)^2>0\\x+1< 0\\x-4>0\end{cases}\Leftrightarrow\hept{\begin{cases}\text{mọi x}\\x< -1\\x>-4\end{cases}}}\Rightarrow-4< x< -1\)

\(TH3\hept{\begin{cases}\left(x-2\right)^2>0\\x+1>0\\x-4< 0\end{cases}\Leftrightarrow\hept{\begin{cases}\text{mọi x}\\x>-1\\x< -4\end{cases}}}\left(loại\right)\)

Vậy -4 < x < -1 hay x thuộc { -2; -3 }

`@` `\text {Ans}`

`\downarrow`

`a,`

`(2x - 1)^2 - 25 = 0`

`<=> (2x - 1)^2 = 25`

`<=> (2x - 1)^2 = (+-5)^2`

`<=>`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy, `S = {-2; 3}`

`b,`

`8x^3 - 50x = 0`

`<=> x(8x^2 - 50) = 0`

`<=>`\(\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\8x^2=50\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=\pm\dfrac{5}{2}\end{matrix}\right.\)

Vậy, `S = {-5/2; 0; 5/2}.`

a) (2x - 1)² - 25 = 0

(2x - 1)² - 5² = 0

(2x - 1 - 5)(2x - 1 + 5) = 0

(2x - 6)(2x + 4) = 0

2x - 6 = 0 hoặc 2x + 4 = 0

*) 2x - 6 = 0

2x = 6

x = 3

*) 2x + 4 = 0

2x = -4

x = -2

Vậy x = -2; x = 3

b) 8x³ - 50x = 0

2x(4x² - 25) = 0

2x[(2x)² - 5²] = 0

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

2x = 0 hoặc 2x - 5 = 0 hoặc 2x + 5 = 0

*) 2x = 0

x = 0

*) 2x - 5 = 0

2x = 5

x = 5/2

*) 2x + 5 = 0

2x = -5

x = -5/2

Vậy x = -5/2; x = 0; x = 5/2

1) Ta có: |x+3| \(\ge\)0; |2x+y-4| \(\ge\)0

\(\Rightarrow\) |x + 3| + |2x + y - 4| \(\ge\) 0

Dấu = xảy ra khi x+3=0 và 2x+y-4 = 0 \(\Rightarrow\)x=-3; y=10

1)  |x + 3| + |2x + y - 4| = 0

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\2x+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\-6+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=10\end{cases}}\)

2) Ta có:  |x + 3| + |x + y - 6| + |7 - 1| = 0

\(\Leftrightarrow\) |x + 3| + |x + y - 6| + 6= 0

\(\Leftrightarrow\)|x + 3| + |x + y - 6| = -6