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Nguyễn Hoàng Minh
12 tháng 10 2021 lúc 15:39

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

Trịnh Đình Thi
28 tháng 11 2021 lúc 10:48
Lol .ngudoots
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Bánh cá nướng :33
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Nguyễn Hoàng Minh
24 tháng 9 2021 lúc 7:50

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

Kiệt
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Midoriya Izuku
6 tháng 10 2023 lúc 21:06

1. x^2 -x-y^2-y= ( x^2-y^2) - ( x+y)= (x+y).(x-y) - ( x+y)= (x+y). ( x-y-1)

2. x^2-y^2+x-y= (x-y).(x+y) + (x-y)= (x-y).(x+y+1)

3. 3x-3y+x^2-y^2= 3.(x-y) + (x-y).(x+y)= (x-y).(3+x+y)

Hà Quang Minh
7 tháng 10 2023 lúc 8:32

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

Bơ Ngố
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ṡú✿ṡıȗ✿ṅɢầȗ
2 tháng 1 2022 lúc 16:13

\(\dfrac{1}{x-y}-\dfrac{1}{x+y}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y}{\left(x-y\right)\left(x+y\right)}-\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y-x+y+2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2x+2y}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2}{x-y}\)

Kudo Shinichi
2 tháng 1 2022 lúc 16:13

\(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{4x}{\left(x-y\right)\left(x+y\right)}\)

Kudo Shinichi
2 tháng 1 2022 lúc 16:17

\(\dfrac{1}{x-y}-\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}\\ =\dfrac{x+y}{x-y}-\dfrac{x-y}{x+y}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{2x+2y}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{2}{x-y}\)

Vũ Nhật Long
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sdsdfdfdf
21 tháng 10 2021 lúc 18:59

a) \(x^2-xy+x-y\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x+1\right)\left(x-y\right)\)

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sdsdfdfdf
21 tháng 10 2021 lúc 19:00

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

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

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sdsdfdfdf
21 tháng 10 2021 lúc 19:01

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

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Hoàng văn tiến
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Nguyễn Lê Phước Thịnh
4 tháng 12 2023 lúc 20:05

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

\(=x^2-x+x^2+2x+1\)

\(=2x^2+x+1\)

Đa thức này ko phân tích được nha bạn

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

\(=\left(x+1\right)\cdot\left(x+1\right)-\left(x+1\right)\cdot3\)

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

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

3: \(2x\cdot\left(x-2\right)-\left(x-2\right)^2\)

\(=2x\left(x-2\right)-\left(x-2\right)\cdot\left(x-2\right)\)

\(=\left(x-2\right)\left(2x-x+2\right)\)

\(=\left(x-2\right)\left(x+2\right)\)

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

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

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

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

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

5: \(3x\left(x+2\right)-5\left(x+2\right)^2\)

\(=\left(x+2\right)\cdot3x-\left(x+2\right)\cdot\left(5x+10\right)\)

\(=\left(x+2\right)\left(3x-5x-10\right)\)

\(=\left(-2x-10\right)\left(x+2\right)\)

\(=-2\left(x+5\right)\left(x+2\right)\)

6: \(4x\left(x-y\right)+3\left(y-x\right)^2\)

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

\(=\left(x-y\right)\cdot4x+\left(x-y\right)\left(3x-3y\right)\)

\(=\left(x-y\right)\cdot\left(4x+3x-3y\right)\)

\(=\left(x-y\right)\left(7x-3y\right)\)

Nguyễn Hương Giang
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HT.Phong (9A5)
12 tháng 8 2023 lúc 14:51

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

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

\(=2\cdot6x\)

\(=12x\)

_________

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y\)

\(=4xy\)

HT.Phong (9A5)
12 tháng 8 2023 lúc 14:59

\(\left(x+y\right)^3+\left(x-y\right)^3\)

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

\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)

\(=2x\cdot\left(x^2+3y^2\right)\)

______

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)

Tuyết Ngọc
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HT.Phong (9A5)
15 tháng 8 2023 lúc 9:45

\(x\left(y-1\right)-y\left(y-1\right)\)

\(=\left(y-1\right)\left(x-y\right)\)

_________________

\(5x\left(x-2\right)-x+2\)

\(=5x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(5x-1\right)\)

_________________

\(5x\left(x+1\right)-8x-8\)

\(=5x\left(x+1\right)-\left(8x+8\right)\)

\(=5x\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(5x-8\right)\)

Vy trần
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Nguyễn Lê Phước Thịnh
10 tháng 10 2021 lúc 23:51

g: \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(5x+3\right)\)

f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)\)

\(=6x^2\left(x+1\right)\)