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Cr746
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Dương Lam Hàng
13 tháng 8 2018 lúc 10:03

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

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

            \(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)

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

4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)

                            \(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)

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

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

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

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

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Cr746
14 tháng 8 2018 lúc 9:54

Phân tích các đa thức sau thành nhân tử :

1) x^3 + x^2y - 4x - 4y

2) x^3 - 3x^2 +1 - 3x

3) 3x^2 - 6xy + 3y^2 - 12z^2

4) x^2 - 2x - 15

5) 2x^2 +3x - 5

6) 2x^2 - 18

7) x^2 - 7xy + 10y^2

8) x^3 - 2x^2 + x - xy^2

Làm nhanh giúp mình với nhé .....mình đang cần gấp[[[[

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Thuy_Vy_89
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Phương Anh Đỗ
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Nguyễn Lê Phước Thịnh
21 tháng 10 2021 lúc 21:08

a: \(x^4+3x^3+x^2+3x\)

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

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

c: \(x^2-xy-x+y\)

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

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

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Uyên Thảo
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💋Bevis💋
17 tháng 7 2019 lúc 12:18

\(a,xy+1-x-y\)

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

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

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

\(b,ax+ay-3x-3y\)

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

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

\(c,x^3-2x^2+2x-4\)

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

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

\(d,x^2+ab+ax+bx\)

\(=\left(x^2+ax\right)+\left(ab+bx\right)\)

\(=x\left(a+x\right)+b\left(a+x\right)\)

\(=\left(a+x\right)\left(b+x\right)\)

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

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

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

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

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💋Bevis💋
17 tháng 7 2019 lúc 12:22

\(f,ax^2+ax-bx^2-bx-a+b\)

\(=\left(ax^2-bx^2\right)+\left(ax-bx\right)-\left(a-b\right)\)

\(=x^2\left(a-b\right)+x\left(a-b\right)-\left(a-b\right)\)

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

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Uyên Thảo
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Trung Art
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Akai Haruma
7 tháng 7 2019 lúc 22:54

5.

\(4x^5y^2+8x^4y^3+4x^3y^4=4x^3y^2(x^2+2xy+y^2)\)

\(=4x^3y^2(x+y)^2\)

9.

\(4x^5y^2+16x^4y^2-6x^3y^2=2x^3y^2(2x^2+4x-3)\)

13.

\(-3x^4y+6x^3y-3x^2y=-3x^2y(x^2-2x+1)=-3x^2y(x-1)^2\)

17.

\(8x^3-8x^2y+2xy^2=2x(4x^2-4xy+y^2)\)

\(=2x[(2x)^2-2.2x.y+y^2]=2x(2x-y)^2\)

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Akai Haruma
7 tháng 7 2019 lúc 22:57

21.

\((a^2+4)^2-16a^2b^2=(a^2+4)^2-(4ab)^2\)

\(=(a^2+4-4ab)(a^2+4+4ab)\)

25.

\(100a^2-(a^2+25)^2=(10a)^2-(a^2+25)^2\)

\(=(10a-a^2-25)(10a+a^2+25)\)

\(=-(a^2-10a+25)(a^2+10a+25)=-(a-5)^2(a+5)^2\)

29.

\(25a^2b^2-4x^2+4x-1=25a^2b^2-(4x^2-4x+1)\)

\(=(5ab)^2-(2x-1)^2=(5ab-2x+1)(5ab+2x-1)\)

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Akai Haruma
7 tháng 7 2019 lúc 23:00

33.

\(1-2m+m^2-x^2-4x-4=(m^2-2m+1)-(x^2+4x+4)\)

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

\(=(m-x-3)(m+x+1)\)

37.

\(ax^2+bx^2+2xy(a+b)+ay^2+by^2\)

\(=x^2(a+b)+2xy(a+b)+y^2(a+b)\)

\(=(a+b)(x^2+2xy+y^2)=(a+b)(x+y)^2\)

41.

\(5a^2-5=5(a^2-1)=5(a^2-1^2)=5(a-1)(a+1)\)

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Uyên Thảo
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Ngân Vũ Thị
17 tháng 7 2019 lúc 10:51

Hỏi đáp Toán

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Trần Thanh Phương
17 tháng 7 2019 lúc 10:52

a) \(xy+1-x-y\)

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

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

b) \(ax+ay-3x-3y\)

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

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

c) \(x^3-2x^2+2x-4\)

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

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

d) \(x^2+ab+ax+bx\)

\(=x\left(b+x\right)+a\left(b+x\right)\)

\(=\left(b+x\right)\left(a+x\right)\)

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

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

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

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

f) \(ax^2+ax-bx^2-bx-a+b\)

\(=\left(ax^2+ax-a\right)-\left(bx^2+bx-b\right)\)

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

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

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Nguyễn Thị Diễm Quỳnh
17 tháng 7 2019 lúc 10:58

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

b) \(ax+ay-3x-3y=a\left(x+y\right)-3\left(x-y\right)=\left(a-3\right)\left(x+y\right)\)

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

d) \(x^2+ab+ax+bx=\left(x^2+ax\right)+\left(ab+bx\right)=x\left(x+a\right)+b\left(a+x\right)=\left(x+b\right)\left(x+a\right)\)

e) \(16-x^2+2xy-y^2=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

f) \(ax^2+ax-bx^2-bx-a+b=\left(a-b\right)x^2+\left(a-b\right)x-a+b=\left(a-b\right)\left(x^2+x-1\right)\)

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Trần Trung Kiên
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Khôi Bùi
10 tháng 9 2018 lúc 12:23

1 ) \(x\left(a-b\right)+a-b=\left(x+1\right)\left(a-b\right)\)

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

3 ) \(-2x-2y+ax+ay=-2\left(x+y\right)+a\left(x+y\right)=\left(a-2\right)\left(x+y\right)\)

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

5 ) \(5x^2y+5xy^2+a^2x+a^2y\)

\(=5xy\left(x+y\right)+a^2\left(x+y\right)\)

\(=\left(5xy+a^2\right)\left(x+y\right)\)

6 ) \(2x^2-6xy+5x-15y\)

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

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

7 ) \(ax^2-3axy+bx-3by\)

\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)

\(=x\left(ax+b\right)-3y\left(ax+b\right)\)

\(=\left(x-3y\right)\left(ax+b\right)\)

8 ) \(x^2+4x-5x-20=0\)

\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

9 ) \(x^2+10x-2x-20=0\)

\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

10 ) \(x^2-6x-4x+24=0\)

\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

:D

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Alex Ich
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Trần Quốc Lộc
19 tháng 10 2017 lúc 15:21

\(\text{a) }ax-bx+ab-x^2\\ \\=\left(ax+ab\right)-\left(x^2+bx\right)\\ \\=a\left(x+b\right)-x\left(x+b\right)\\ \\=\left(a-x\right)\left(x+b\right)\\ \)

\(\text{b) }x^2-y^2+4x+4\\ \\ =\left(x^2+4x+4\right)-y^2\\ \\ =\left(x+2\right)^2-y^2\\ \\ =\left(x+2+y\right)\left(x+2-y\right)\\ \)

\(\text{c) }ax+ay-3x-3y\\ \\=\left(ax+ay\right)-\left(3x+3y\right)\\ \\ =a\left(x+y\right)-3\left(x+y\right)\\ \\=\left(a-3\right)\left(x+y\right)\\ \)

\(\text{d) }x^3+x^2+x+1\\ \\=\left(x^3+x^2\right)+\left(x+1\right)\\ \\=x^2\left(x+1\right)+\left(x+1\right)\\ \\=\left(x^2+1\right)\left(x+1\right)\\ \)

\(\text{e) }x^3-3x^2+3x-9\\ \\=\left(x^3-3x^2\right)+\left(3x-9\right)\\ \\ =x^2\left(x-3\right)+3\left(x-3\right)\\ \\=\left(x^2+3\right)\left(x-3\right)\\ \)

\(\text{f) }x^2+ab+ax+bx\\ \\=\left(x^2+ax\right)+\left(bx+ab\right)\\ \\ =x\left(x+a\right)+b\left(x+a\right)\\ \\=\left(x+b\right)\left(x+a\right)\\ \)

\(\text{g) }xy+1+x+y\\ \\=\left(xy+x\right)+\left(y+1\right)\\ \\=x\left(y+1\right)+\left(y+1\right)\\ \\=\left(x+1\right)\left(y+1\right)\)

\(\text{h) }9-x^2-2xy-y^2\\ \\=9-\left(x^2+2xy+y^2\right)\\ \\=3^2-\left(x+y\right)^2\\ \\=\left(3-x-y\right)\left(3+x+y\right)\\ \)

\(\text{i) }x^2-2xy+y^2-1\\ \\=\left(x^2-2xy+y^2\right)-1\\ \\=\left(x-y\right)^2-1^2\\ \\=\left(x-y-1\right)\left(x-y+1\right)\\ \)

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Gia Hân Ngô
19 tháng 10 2017 lúc 15:40

d) x3 + x2 + x + 1

= x2(x + 1) + (x + 1)

= (x + 1)(x2 + 1)

e) x3 - 3x2 + 3x - 9

= x2(x - 3) + 3(x - 3)

= (x - 3)(x2 + 3)

f) x2 + ab + ax + bx

= x2 + bx + ab + ax

= x(x + b) + a(b + x)

= (b + x)(x + a)

g) xy + 1 + x + y

= xy + x + y + 1

= x(y + 1) + (y + 1)

= (y + 1)(x + 1)

h) 9 - x2 - 2xy - y2

= 9 - (x2 + 2xy + y2)

= 32 - (x + y)2

= (3 - x - y)(3 + x + y)

i) x2 - 2xy + y2 - 1

= (x - y)2 - 1

= (x - y - 1)(x - y + 1)

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