\(4x^3=x\)

\(\Rightarrow4x^3-x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\4x^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\Rightarrow x=\frac{1}{2}\end{cases}}\)

KL....

\(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=\left(x+2\right)\left[\left(x+2\right)\left(x-2\right)\right]\)

\(=\left(x+2\right)\left(x^2-4\right)=x^3-4x+2x^2-8\)

=X^2 + 4X + 4 - (X^2 - 4)
= 4X + 8 = 4(X+2)

Thực hiện phép tính

a,  6x³y⁵z : 3xy³z=2x²y²

b,  \(\frac{3x+6}{x+2}+\frac{2x+4}{x+2}\)

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

=3+2=5

a)\(3.3x^3+9x\)

\(=9x.\left(x^2+1\right)\)

b) \(xy+y^2-3x-3y\)

\(y.\left(x+y\right)-3.\left(x+y\right)=\left(x+y\right).\left(y-3\right)\)

\(3x^33+9x=9x\left(x^2+1\right)\)

\(xy+y^2-3x-3y=y\left(x+y\right)-3\left(x+y\right)=\left(y-3\right)\left(x+y\right)\)

\(a,\left(x+2\right)-\left(x-2\right).\left(x+2\right)\)

\(=\left(x+2\right).\left(1-x+2\right)=\left(x+2\right).\left(-x+3\right)\)

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

\(b,3x^2-6x=0\)

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

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

cau b co sai thi dung k nha

3x²-6x=0

=>3x(x-2)=0

=>3x=o hoac x-2=0

=>x=0 hoac x=2

dap so: x = 2 hoac 0

dung thi k nha

\(-3x^2-7x+10\)

\(=3x-3x^2+10-10x\)

\(=3x.\left(1-x\right)+10.\left(1-x\right)=\left(3x+10\right).\left(1-x\right)\)

\(-3x^2+3y^2-4xz-4yz\)

\(=3\left(y^2-x^2\right)-4z\left(x+y\right)\)

\(=3\left(y-x\right)\left(x+y\right)-4z\left(x+y\right)\)

\(=\left(x+y\right)\left(3y-3x-4z\right)\)

r3t4yjytuky

ai luot wa xinn co tam tra loi ho

\(B1,a,A=x^2-6x+11\)

               \(=\left(x^2-6x+9\right)+2\)

                \(=\left(x-3\right)^2+2\ge2\)

Dấu "=" <=> x=3

Vậy ..........

\(b,B=x^2-20x+101\)

        \(=\left(x^2-20x+100\right)+1\)

         \(=\left(x-10\right)^2+1\ge1\)

Dấu "=" <=> x = 10

Vậy .

\(2,a,A=4x-x^2+3\)

            \(=7-\left(x^2-4x+4\right)\)'

             \(=7-\left(x-2\right)^2\le7\)

Dấu ''='' <=> x = 2

Vậy .

\(b,B=-x^2+6x-11\)

       \(=-2-\left(x^2-6x+9\right)\)

        \(=-2-\left(x-3\right)^2\le-2\)

Dấu ""=" <=> x = 3

Vậy..

\(\left(3x+12\right)-\left(x^2-16\right)=3\left(x+4\right)-\left(x+4\right)\left(x-4\right)=\left(x+4\right)\left(3-x+4\right)\)

=(x+4)(7-x)