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

x⁴+8x³+15x²-4x-2

= (x⁴+4x³+x²)+(4x³+16x²+4x)-(2x²+8x+2)

= x².(x²+4x+1)+4x.(x²+4x+1) -2(x²+4x+1)

= (x²+4x+1).(x²+4x-2)

ai h minh minh h lai cho

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

a) 15x²-5x³=5x²(3-x)

a: \(15x^2-5x^3=5x^2\left(3-x\right)\)

b: \(8x^3-y^3+4x^2y-2xy^2\)

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

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

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

c: Ta có: \(x^8+64y^4\)

\(=x^8+16x^4y^2+64y^4-16x^4y^2\)

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

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

b mk thấy nó sai đề sao ý 

c) \(C=\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)

\(=\left(x^2+x+4\right)^2+2.4x.\left(x^2+x+4\right)+16x^2-x^2\)

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

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

\(=\left(x^2+5x+4-x\right)\left(x^2+5x+4+x\right)=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)

Nhập vào cốc cốc để hệ thống máy tính làm cho