=(a+b)(a²-ab+b²)+3ab(a²+b²+2ab-2ab)+6a²b²(a+b)

=(a+b)((a+b)²-3ab)+3ab((a+b)²-2ab)+6a²b²(a+b)

Thay a+b=1

=> 1(1-3ab)+3ab(1-2ab)+6a²b²= 1-3ab +3ab-6a²b²+6a²b²=1

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

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

=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

Câu hỏi của Jungkookie - Toán lớp 7 - Học toán với OnlineMath