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12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
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
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
TT
1
26 tháng 8 2021
a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)
Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}=\left(x+2\right)\left(x+3\right)\)
b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)
Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)
c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)
Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)
d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)
Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)
HG
1
29 tháng 11 2023
a: \(x^4-2x^3+x^2-2x\)
\(=\left(x^4-2x^3\right)+\left(x^2-2x\right)\)
\(=x^3\left(x-2\right)+x\left(x-2\right)\)
\(=x\left(x-2\right)\left(x^2+1\right)\)
b: \(x^4+x^3-8x-8\)
\(=\left(x^4+x^3\right)-\left(8x+8\right)\)
\(=x^3\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-8\right)\)
\(=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
DN
1
MT
5 tháng 7 2015
(x+2)(x+3)(x+4)(x+5) - 8
=(x+2)(x+5)(x+3)(x+4)-8
=(x2+7x+10)(x2+7x+12)-8
đặt t=x2+7x+10 ta được:
t(t+2)-8=t2+2t-8
=t2-2t+4t-8
=t(t-2)+4(t-2)
=(t-2)(t+4)
thay t=x2+7x+10 ta được:
(x2+7x+8)(x2+7x+14)
vậy (x+2)(x+3)(x+4)(x+5) - 8=(x2+7x+8)(x2+7x+14)
(x+1)(x+2)(x+3)(x+4)-8
=[(x+1).(x+4)].[(x+2).(x+3)]-8
=(x2+5x+4).(x2+5x+6)-8
Đặt (x2+5x+4)=t =>(x2+5x+6)=t+2
Thay vào biểu thức ta có:
(x2+5x+4).(x2+5x+6)-8
t.(t+2)-8
=t2+2t+1-9
=(t+1)2-32
=(x2+5x+4+1)-32
=(x2+5x+5+3).(x2+5x+5-3)
=(x2+5x+8).(x2+5x+2)
=
ta làm như sau :
\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-8.\)
\(\Rightarrow\left(x^2+5X+4\right)\left(x^2+5x+6\right)-8\)
Đặt \(x^2+5x+4=t\)
\(\Leftrightarrow t\left(t+2\right)-8\)
\(\Leftrightarrow t^2+2t-8\Leftrightarrow t^2+2t+1-9\)
\(\Leftrightarrow\left(t+1\right)^2-3^2\)
\(\Leftrightarrow\left(t-2\right)\left(t+4\right)\)
\(\Leftrightarrow\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)