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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,hđt số 3 = \(\left(a^2+2a\right)^2-9\)
b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)
a) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)
\(=\left(a^2+2a\right)^2+3.\left(-3\right)\)
\(=\left(a^2+2a\right)^2-9\)
b) \(\left(x-y+6\right)\left(x+y-6\right)\)
\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)
\(=x^2-\left(y-6\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(x^{20}+x+1\)
\(=\left(x^{20}-x^2\right)+\left(x^2+x+1\right)\)
Đặt \(x^2+x+1=A\)
\(\Rightarrow x^{20}+x+1=x^2\left(x^{18}-1\right)+A\)
\(=x^2\left(x^9+1\right)\left(x^9-1\right)+A\)
\(=\left(x^{11}+x^2\right)\left[\left(x^3\right)^3-1^3\right]+A\)
\(=\left(x^{11}+x^2\right)\left(x^6+1+x^3\right)\left(x^3-1\right)+A\)
\(=\left(x^{17}+x^{14}+x^{11}+x^8+x^5+x^2\right)\left(x-1\right)\left(x^2+x+1\right)+A\)
\(=A.\left(x^{18}-x^{17}+x^{15}-x^{14}+x^{12}-x^{11}+x^9-x^8+x^6-x^5+x^3-x^2\right)+A\)
\(=A.\left(x^{18}-x^{17}+x^{15}-x^{14}+x^{12}-x^{11}+x^9-x^8+x^6-x^5+x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^{18}-x^{17}+x^{15}-x^{14}+x^{12}-x^{11}+x^9-x^8+x^6-x^5+x^3-x^2+1\right)\)
bạn nào giải dùm mình bài này với 16x4+y16 phân tích đa thức thành nhân tử .... ai giải đc là ghê nề
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(16x^4+y^{16}+16x^2y^8\right)-16x^2y^8\)
\(=\left(4x^2+y^8\right)^2-\left(4xy^4\right)^2\)
\(=\left(4x^2+y^8-4xy^4\right)\left(4x^2+y^8+4xy^4\right)\)
\(=\left(2x-y^4\right)^2\left(2x+y^4\right)^2\)
16x4+y16=(4x2)2+(y8)2=(4x2+y8)2-8x2y8=(4x2+y8)2-(√8x2y8)2=(4x2+y8-√8x2y8)(4x2+y8+√8x2y8)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn tham khảo nhé :
Ta có :
\(\frac{x-3}{2012}+\frac{x-2}{2013}=\frac{x-2013}{2}+\frac{x-2012}{3}\)
\(\Leftrightarrow\)\(\left(\frac{x-3}{2012}-1\right)+\left(\frac{x-2}{2013}-1\right)=\left(\frac{x-2013}{2}-1\right)+\left(\frac{x-2012}{3}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}=\frac{x-2015}{2}+\frac{x-2015}{3}\)
\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}-\frac{x-2015}{2}-\frac{x-2015}{3}=0\)
\(\Leftrightarrow\)\(\left(x-2015\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\ne0\)
\(\Rightarrow\)\(x-2015=0\)
\(\Rightarrow\)\(x=2015\)
Vậy \(x=2015\)
Chsuc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-y^2+6y-9=x^2-\left(y^2-6y+9\right)\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x+y-3\right)\left(x-y+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=xy\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(xy-7\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x2+x-3
=2x2+3x-2x-3
=2x(2x+3)-(2x+3)
=(2x+3)(2x-1)
Chúc bạn học tốt
2x2+x-3
=2𝑥2+3𝑥−2𝑥−3
=𝑥(2𝑥+3)−1(2𝑥+3)
=(𝑥−1)(2𝑥+3)
ez mà bro :D
62.64-43.(36-1)
= 66-64.728
= 66-46592
= 0
tui nha
lp 8 hả
=66-43(729-1)
=66-43728
=46656-46592
=64
=82