Đặt \(A=\left(x+3\right)^4+\left(x+1\right)^4-16\)

\(=\left\lbrack\left(x+2\right)+1\right\rbrack^4+\left\lbrack\left(x+2\right)-1\right\rbrack^4-16\)

Đặt b=x+2

=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)

\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)

\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)

\(=2\left(b^2+1\right)^2+8b^2-16\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)

\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)

\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)

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

Đặt \(A=\left(x+3\right)^4+\left(x+1\right)^4-16\)

\(=\left\lbrack\left(x+2\right)+1\right\rbrack^4+\left\lbrack\left(x+2\right)-1\right\rbrack^4-16\)

Đặt b=x+2

=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)

\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)

\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)

\(=2\left(b^2+1\right)^2+8b^2-16\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)

\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)

\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)

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

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

k mk nha

bạn ơi bạn chưa bớt 2x^2 kìa

x⁵-x⁴-1=x⁵-x³-x²-x⁴+x²+x+x³-x-1

=x².(x³-x-1)-x.(x³-x-1)+(x³-x-1)

=(x³-x-1)(x²-x+1)

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

k mk nha

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

\(x^4+3x^3+12x-16\)

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

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

\(=\left(x-1\right)\left(x^3+4x^2+4x+16\right)\)

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

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