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12 tháng 10 2021

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

28 tháng 11 2021
Lol .ngudoots
24 tháng 9 2021

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

6 tháng 11 2021

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

19 tháng 11 2021

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

19 tháng 11 2021

\(=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1=\left(x^2+3x\right)\left(x^2+3x+1\right)+1\)

Đặt \(x^2+3x=t\)

\(\left(x^2+3x\right)\left(x^2+3x+2\right)+1=t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)

NV
24 tháng 11 2021

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

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

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

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

24 tháng 11 2021

x(x+1)(x+2)(x+3)+1

= [x(x+3)][(x+1)(x+2)]+1

=(x2+3x)(x2+3x+2)+1

Đặt x2+3x+1=y, ta có: 

(y-1)(y+1)+1

=y2-1+1

=y2

Thay y=x2+3x+1, lại có: 

(x2+3x+1)2

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

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

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

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

16 tháng 9 2023

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)

2 tháng 8 2016
[x + (3+√5)/2]^2[x + (3-√5)/2]^2
2 tháng 8 2016

Còn không ghi là (x^2 + 3x + 1)^2