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

1/ (x+12)(x-3) = 0 nên x+12 = 0 hoặc x-3 = 0 => x = 0 - 12 = -12 hoặc x = 0 + 3 = 3.Vậy x \(\in\left\{-12;3\right\}\)

2/ (-x+5)(3-x) = 0 nên -x+5 = 0 hoặc 3-x = 0 => x = -(0-5) = -(-5) = 5 hoặc x = 3 - 0 = 3.Vậy x \(\in\left\{3;5\right\}\)

1)  - x + 5 = 0 hoặc 3 - x = 0 => x = 5 hoặc x = 3

2)   x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0 =. x = 0 hoặc x = - 2 hoặc x = 7

3)   x - 1 = 0 hoặc x + 2 = 0 hoặc - x - 3 = 0 => x = - hoặc x = - 2 hoặc x = - 3

1, => x + 12 = 0 => x = -12

x - 3 = 0 => x = 3 

=> x \(\in\) { -12; 3 } 

1; (\(x\) + 12)(\(x\) - 3) = 0

    \(\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)

    Vậy \(x\) \(\in\) { -12; 3}

a) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=\frac{-1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{-1}{6}\end{cases}}}\)

Vậy x= 5/6 hoặc -1/6

b) - Nếu x=0 thì \(5^y=2^0+624=1+624=625=5^4\Rightarrow y=4\left(y\in N\right)\)

- Nếu x \(\ne\) 0 thì vế trái là số chẵn , vế phải là số lẻ \(\forall x;y\inℕ\) ( vô lí)

Vậy x=0, y=4

thank you bạn.

a: \(\Leftrightarrow\left(2-x\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)=-4x\)

\(\Leftrightarrow2x-6-x^2+3x+x^2+3x-x-3=-4x\)

=>7x-9=-4x

=>11x=9

hay x=9/11

b: \(\Leftrightarrow\left(5-x\right)\left(x-4\right)+\left(x+2\right)\left(x+4\right)=-3x\)

\(\Leftrightarrow5x-20-x^2+4x+x^2+6x+8=-3x\)

=>15x-12=-3x

=>18x=12

hay x=2/3

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

_Tần vũ_

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{18}\)

_Tần Vũ_

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{3^2}{5^2}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}:3\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1^3}{3^3}\right)\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)

\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)

\(\Rightarrow3x=\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{6}:3\)

\(\Rightarrow x=\frac{1}{18}\)

a. \(1-2x< 7\)

mà: \(1-n\le1\)với mọi n

\(\Rightarrow2x=n\Rightarrow x=\frac{n}{2}\)với mọi n

b.để: (x-1).(x-2)>0

=> x-1>0hoặc x-2<0

=>x>1hoặc x<2

(mik chỉ làm 2 câu mẫu thôi, bạn cố gắng tự làm nha, rất vui được kết bạn với bạn)