Phân tích thành nhân tử: 2 x 2 + 3 x + 5
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6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
HQ
Hà Quang Minh
Giáo viên
1 tháng 8 2023
\(\left(x-5\right)^3-2y\left(5-x\right)^2\\ =\left(x-5\right)^2\left(x-5-2y\right)\)
AH
Akai Haruma
Giáo viên
22 tháng 11 2021
Biểu thức này không phân tích được thành nhân tử nữa bạn nhé.
MN
2
28 tháng 8 2021
\(\left(x-3\right)^2-5\left(x-2\right)+5=0\\ \Leftrightarrow x^2-6x+9-5x+10+5=0\\ \Leftrightarrow x^2-11x+24=0\\ \Leftrightarrow\left(x-8\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\)
\(\left(2x-1\right)^2-3\left(x-2\right)\left(x+2\right)-25=0\\ \Leftrightarrow4x^2-4x+1-3\left(x^2-4\right)-25=0\\ \Leftrightarrow4x^2-4x-24-3x^2+12=0\\ \Leftrightarrow x^2-4x-12=0\\ \Leftrightarrow\left(x-6\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
29 tháng 8 2021
a: Ta có: \(\left(x-3\right)^2-5\left(x-2\right)+5=0\)
\(\Leftrightarrow x^2-6x+9-5x+10+5=0\)
\(\Leftrightarrow x^2-11x+24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\)
b: Ta có: \(\left(2x-1\right)^2-3\left(x-2\right)\left(x+2\right)-25=0\)
\(\Leftrightarrow4x^2-4x+1-3x^2+12-25=0\)
\(\Leftrightarrow x^2-4x-12=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
T
16 tháng 9 2023
Mình bổ sung nhé:
\(=\left(x+1\right)\left(x^4+x^3+x^2-x^3+1\right)\)
\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x^3-1\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
16 tháng 9 2023
=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)
8 tháng 10 2021
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)=\left(x^3-1\right)\left(x^2+1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x^2+1\right)\)
15 tháng 9 2015
(x+2).(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)-24
=(x2+7x+10)(x2+7x+12)-24
=(x2+7x+11-1)(x2+7x+11+1)-24
Đặt x2+7x+11=a thì
=(a-1)(a+1)-24
=a2-1-24=a2-25=a2-52
=(a+5)(a-5)
=(x2+7x+16)(x2+7x+6)
N
3
25 tháng 10 2021
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
2 x 2 + 3 x + 5 = 2 x 2 - 2 x + 5 x - 5 = 2 x 2 - 2 x + 5 x - 5 = 2 x x - 1 + 5 x - 1 = x - 1 2 x + 5