(x+3)(5x+10)=0

x=-3 hoặc x=-2 

Vậy \(x\in\left\{-3;-2\right\}\)

Làm lại :V

(x-3)(5x+10)=0

x=3 hoặc x=-2 

Vậy \(x\in\left\{3;-2\right\}\)

Bạn thử xem lại đề nhé, giữa 3 số này là dấu cộng hay dấu nhân.

Nếu là dấu cộng thì ta có:

Nếu là dấu nhân thì ta có:

a) \(5x\left(x-7\right)-30\cdot\left(x-7\right)=0\)

   \(\Rightarrow\left(5x-30\right)\left(x-7\right)=0\)

   \(\Rightarrow\left[{}\begin{matrix}5x-30=0\\x-7=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=7\end{matrix}\right.\)

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

   \(\Rightarrow4x^2-16-x^2-3x=3x^2+15x\)

   \(\Rightarrow-16=18x\Rightarrow x=-\dfrac{8}{9}\)

Answer:

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

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

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

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

\(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(5x\left(x-3\right)-x+3=0\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

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

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

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

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

\(\Rightarrow x^2+3x-2x-6=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

\(5x\left(6-x\right)+\left(5x-3\right)x=27\)

\(30x-5x^2+5x^2-3x=27\)

\(27x=27\)

\(x=1\)

Vậy \(x=1\)

\(3\left(2x-1\right)-5\left(x-3\right)+6.\left(3x-4\right)=83\)

\(6x-3-5x+15+18x-24=83\)

\(19x-12=83\)

\(19x=95\)

\(x=\frac{95}{19}\)

\(x=5\)

Vậy \(x=5\)

\(5x.\left(6-x\right)+\left(5x-3\right)x=27\)

\(\Rightarrow30x-5x^2+5x^2-3x=27\)

\(\Rightarrow\left(30x-3x\right)-\left(5x^2-5x^2\right)=27\)

\(\Rightarrow27x=27\)

\(\Rightarrow x=\)................................................(XIn lỗi em không biết chia :D)

\(a,5x\left(6-x\right)+\left(5x-3\right)x=27\)

\(\Leftrightarrow30x-5x^2+5x^2-3x=27\)

\(\Leftrightarrow27x=27\)

\(\Leftrightarrow x=1\)

\(b,3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)=83\)

\(\Leftrightarrow6x-3-5x+15+18x-24=83\)

\(\Leftrightarrow19x-12=83\)

\(\Leftrightarrow19x=95\)

\(\Leftrightarrow x=5\)

a: =>x=3/7+3/5=15/35+21/35=36/35

b: =>x/35=4/5-5/7=28/35-25/35=3/35

=>x=3

c: =>x<3/4+8/4=11/4

=>\(x\in\left\{0;1;2;3\right\}\)

d: =>5/3<x<5/6+24/6=29/6

=>\(x\in\left\{2;3;4\right\}\)

e: =>x<10/12-9/12=1/12

=>x=0

f: =>2/3<x<12/6-5/6=7/6

=>x=1

x= 5³:5

x=5²

25

x=25

nhấn vào đúng 0 sẽ có câu trả lời

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)