\(A=4x^2+12xy+9y^2\)

\(B=25x^2-10xy+y^2\)

\(C=8x^3+12x^2y^2+6xy^4+y^6\)

\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)

\(E=x^3-27y^3\)

\(F=x^6-27\)

bài 1: x.(x+7) = 0

Th1:x=0              Th2:x+7=0

                          =>x=-7

bài 2 (x+12).(x-3)= 0

Th1:x+12=0                                         Th2:x-3=0

=>x=-12                                                =>x=3

bài 3 (-x+5).(3-x)=0

Th1 (-x)+5=0                                          Th2:3-x=0

=>-x=-5                                                  =>x=3

bài 4 x.(2+x).(7-x)=0

Th1:x=0                                               Th3:7-x=0

Th2:2+x=0                                             =>x=7

=>x=-2

bài 5 (x-1).(x+2).(-x-3)=0

Th1:x-1=0                                               Th2:x+2=0

=>x=1                                                   =>x=-2

Th3:-x-3=0

=>-x=-3

\(x+\frac{1}{2}\times x=\frac{4}{5}\)

\(x\times\left(1+\frac{1}{2}\right)=\frac{4}{5}\)

\(x\times\frac{3}{2}=\frac{4}{5}\)

\(x=\frac{4}{5}:\frac{3}{2}\)

\(x=\frac{8}{15}\)

Vậy đáp án đúng là C

1) \(x.\left(x+7\right)=0\)
\(=>\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.=>\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
2) \(\left(x+12\right).\left(x-3\right)=0\)
\(=>\left[\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.=>\left[\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
3) \(\left(-x+5\right).\left(3-x\right)=0\)
\(=>\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.=>\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)
4) \(x.\left(2+x\right).\left(7-x\right)=0\)
\(=>\left[\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.=>\left[\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
5) \(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)
\(=>\left[\begin{matrix}x-1=0\\x+2=0\\-x-3=0\end{matrix}\right.=>\left[\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)

Tổng quát: tích ab = 0 thì a=0 hoặc b=0

\(a\cdot b\cdot c=0\Leftrightarrow\left[\begin{matrix}a=0\\b=0\\c=0\end{matrix}\right.\)(nếu có 3 thừa số)

giúp mình với, đi mà T.T

Sao đến bài này là ko ai giúp mình vậy T.T

\(\left(2x+1\right)\left(x^2-x\right)+x\left(5+x-2x^2\right)=3x+7\)

\(2x^3-2x^2+x^2-x+5x+x^2-2x^3=3x+7\)

\(5x-x=3x+7\)

\(4x-3x=7\)

\(x=7\)

(2x+1)(x^2-x)+x(-2x^2+x+5)=3x+7

=>2x^3-2x^2+x^2-x-2x^3+x^2+5x=3x+7

=>-x^2-x+x^2+5x=3x+7

=>4x=3x+7

=>x=7

\(a,\dfrac{3}{4}:\dfrac{6}{x}=\dfrac{5}{4}\\ \dfrac{6}{x}=\dfrac{3}{4}:\dfrac{5}{4}\\ \dfrac{6}{x}=\dfrac{3}{5}\\ x=6:\dfrac{3}{5}\\ x=10\\ b,\left(x+\dfrac{7}{4}\right)\times\dfrac{2}{3}=6\\ x+\dfrac{7}{4}=6:\dfrac{2}{3}\\x+\dfrac{7}{4}=9\\ x=9-\dfrac{7}{4}\\ x=\dfrac{29}{4}\)

giúp mình đi, sao dạo này ít ngừi giúp qué zại