TQ
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
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
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
CG
2
11 tháng 3 2021
Ta có:
\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)
\(A=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)
\(A=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)
\(A=\left(2c+1\right)\left(4ab+2a+2b+1\right)\)
\(A=\left(2c+1\right)\left[2a\left(2b+1\right)+\left(2b+1\right)\right]\)
\(A=\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)\)
NC
11 tháng 3 2021
Ta có:\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)
\(=8abc+4ab+4bc+4ca+2a+2b+2c+1\)
\(=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)
\(=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)
\(=\left(2c+1\right)\left(4ab+2b+2a+1\right)\)
\(=\left(2c+1\right)\left[2b\left(2a+1\right)+\left(2a+1\right)\right]\)
\(=\left(2c+1\right)\left(2b+1\right)\left(2a+1\right)\)
5 tháng 12 2016
\(a^4+a^3+a^2+a+1-a^3-a=\left(a^4-a^3+a^2\right)+\left(a^3-a^2+a\right)+\left(a^2-a+1\right)\)
=\(a^2\left(a^2-a+1\right)+a\left(a^2-a+1\right)+\left(a^2-a+1\right)\)
=\(\left(a^2+a+1\right)\left(a^2-a+1\right)\)
Dùng phương pháp thêm bớt hạng tử
DK
2
29 tháng 8 2021
\(a^4+a^3+a^2+a\)
\(=a^3\left(a+1\right)+a\left(a+1\right)\)
\(=\left(a+1\right)\left(a^3+a\right)\)
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
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
ta có a^4 + a^2 + 1 = a ^ 4 + 2a^2 + 1 - a^2 = (a^2+1)^2 - a^2 = (a^2- a + 1)(a^2 + a + 1)
\(a^4+a^2+1\)
\(=a^4+2a^2+1-a^2\)
\(=\left(a^2+1\right)^2-a^2\)
\(=\left(a^2+1-a\right)\left(a^2+a+1\right)\)