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23 tháng 10 2021
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
24 tháng 10 2021
a: \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=x^2y^2+a^2b^2+x^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(x^2+a^2\right)\)
NC
26 tháng 8 2020
Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
G
26 tháng 8 2020
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
NH
2
3 tháng 8 2015
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
7 tháng 11 2021
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
7 tháng 11 2021
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
TL
2
Lời giải:
$N=p^{m+2}q-pq^{m+3}-p^{m+3}q^{n+4}$
$=pq(p^{m+1}-q^{m+2}-p^{m+2}q^{n+3})$