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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
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Tham khảo!
x5+x-1 x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1
= x3(x2-x+1)+x2(x2-x+1)-(x2-x+1)
= (x2-x+1)(x3+x2-1)
x5+x-1 x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1
= x3(x2-x+1)+x2(x2-x+1)-(x2-x+1)
= x2-x+1x3+x2-1
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^5+x^4+1\\ =x^5-x^3+x^2+x^4-x^2+x+x^3-x+1\\ =x^2\left(x^3-x+1\right)+x\left(x^3-x+1\right)+\left(x^3-x+1\right)\\ =\left(x^3-x+1\right)\left(x^2+x+1\right)\)
\(x^5+x^4+1\)
\(=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình bổ sung nhé:
\(=\left(x+1\right)\left(x^4+x^3+x^2-x^3+1\right)\)
\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x^3-1\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
=x^3(x^2+x+1)+(x^2+x+1)
=(x^2+x+1)(x^3+1)
=(x^2+x+1)(x+1)(x^2-x+1)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)=\left(x^3-1\right)\left(x^2+1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x^2+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^5+x+1=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^3-x\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
Chúc bạn học tốt.
x5+x+1
=x(x4+1)
+1 nữa mà