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1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)
2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)
4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
5, \(2x^3+3x^2+2x+3\)
\(=x^2\left(2x+3\right)+2x+3\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
6, \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xy^2\)
\(=xz\left(x^2-xz\right)+xz\left(xy-yz\right)\)
\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)
\(=xz\left(x+y\right)\left(x-z\right)\)
8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)
9, \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
10, \(x^2-8x+12\)
\(=x^2-6x-2x+12\)
\(=x\left(x-6\right)-2\left(x-6\right)\)
\(=\left(x-2\right)\left(x-6\right)\)
Chỗ còn lại mai làm nốt nha.
Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha
11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)
\(=ab\left(a-b\right)-ac\left(a-b\right)-bc\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
12, \(x^3-7x-6\)
\(=x^3-3x^2+3x^2-9x+2x-6\)
\(=x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+2\right)\)
\(=\left(x-3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)
13, \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
14, \(a^4+64\)
\(=a^4+16a^2+64-16a^2\)
\(=\left(a^2+8\right)^2-16a^2\)
\(=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)
15, \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)
16, \(x^5+x-1\)
\(=x^5-x^4+x^3+x^4-x^3+x^2-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x^2\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x^2-1\right)\)
17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)
19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)
Đặt \(x^2+8x+7=a\) ta có:
(*) \(\Leftrightarrow a\left(a+8\right)+15\)
\(\Leftrightarrow a^2+8a+15\)
\(\Leftrightarrow a^2+3a+5a+15\)
\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)
\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)
Đặt \(x^2+3x+1=a\) ta có:
(*) \(\Leftrightarrow a\left(a+1\right)-6\)
\(\Leftrightarrow a^2+a-6\)
\(\Leftrightarrow a^2+3a-2a-6\)
\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)
1.=(x-y)(5x+1)
2.=(x+3)(2x+1)
3.=(3x-2y)2(1+1)=2(3x-2y)2
4.bạn chép sai hay sao ý
5.=(2x-3y)2
6. = -(x+y)2
7. = -(a-5)2
1. 5x(x-y)-(y-x)
= 5x(x-y)+(x-y)
= (x-y)(5x+1)
2. 2x(x+3)+(3+x)
= (x+3)(2x+1)
3. (3x-2y)2-(2x-3y)2
= (3x-2y-2x+3y)(3x-2y+2x-3y)
=(x+y)(5x-5y)
=5(x+y)(x-y)
4. 4-(a-b)2
= 22-(a-b)2
= (2-a+b)(2+a-b)
5. 4x2-12xy+9y2
= (2x-3y)2
6. -x2-2xy-y2
= -(x+y)2
7. 10a-a2-25
= -a2+10a-25
= -(a-5)2
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(\Rightarrow A=\left(x^3+8\right)-\left(x^3-2\right)\)
\(\Rightarrow A=x^3+8-x^3+2\)
\(\Rightarrow A=\left(x^3-x^3\right)+\left(8+2\right)\)
\(\Rightarrow A=10\)
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(=x^3+8-x^3+2\)
\(=10\)
\(B=\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3+8\right)\left(x^3-8\right)\)
\(=x^6-64\)
\(C=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
\(=\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x+1-3x+1\right)^2\)
\(=\left(x^2+2\right)^2\)
\(D=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1+3x\right)\left(3x^3+1-3x\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)
\(=-9x^2\)
\(E=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)
\(A=\left(x+1\right)^2+\left(x+2\right)^2=\left(x+1\right)^2+\left(-2-x\right)^2\ge\frac{1}{2}\left(x+1-2-x\right)^2=\frac{1}{2}.1^2=\frac{1}{2}\Rightarrow A_{min}=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}\)
\(B=-2x^2-4\le0-4=-4\Rightarrow B_{max}=-4\Leftrightarrow x=0\)
\(C=-5x^2+10x-7=-5x^2+10x-5-2=-5\left(x-1\right)^2-2\le0-2=-2\Rightarrow C_{min}=-2\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
Bài 5:
`a)x^3+8`
`=x^3+2^3`
`=(x+2)(x^2-2x+4)`
`b)x^3-8`
`=x^3-2^3`
`=(x-2)(x^2+2x+4)`
`c)x^3+1`
`=x^3+1^3`
`=(x+1)(x^2-x+1)`
`d)x^3-1`
`=x^3-1^3`
`=(x-1)(x^2+x+1)`
`e)8x^3-y^3`
`=(2x)^3-y^3`
`=(2x-y)(4x^2+2xy+y^2)`
`f)x^3-27y^3`
`=x^3-(3y)^3`
`=(x-3y)(x^2+3xy+9y^2)`
`g)8x^3+y^3`
`=(2x)^3+y^3`
`=(2x+y)(4x^2-2xy+y^2)`
`i)8x^3+64y^3`
`=(2x)^3+(4y)^3`
`=(2x+4y)(4x^2-8xy+16y^2)`
`m)8x^3-64y^3`
`=(2x)^3-(4y)^3`
`=(2x-4y)(4x^2+8xy+16y^2)`
`n)x^2-9`
`=x^2-3^2`
`=(x+3)(x-3)`
`p)x^2-1`
`=x^2-1^2`
`=(x+1)(x-1)`
`q)9x^2-1`
`=(3x)^2-1^2`
`=(3x-1)(3x+1)`
`k)1-4x^2`
`=1^2-(2x)^2`
`=(1-2x)(1+2x)`
`t)16-25y^2`
`=4^2-(5y)^2`
`=(4-5y)(4+5y)`
`i)4x^2-9y^2`
`=(2x)^2-(3y)^2`
`=(2x-3y)(2x+3y)`
`h)x^4-y^4`
`=(x^2)^2-(y^2)^2`
`=(x^2-y^2)(x^2+y^2)`
`=(x+y)(x-y)(x^2+y^2)`
`j)x^2-7`
`=x^2-(\sqrt{7})^2`
`=(x-\sqrt{7})(x+\sqrt{7})`
Bài 4:
a: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3-x^2+x+x^2-x+1=x^3+1\)
b: \(\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3+x^2+x-x^2-x-1\)
\(=x^3-1\)
c: \(\left(x+3\right)\left(x^2-3x+9\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27\)
\(=x^3+27\)
d: \(\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+3x^2+9x-3x^2-9x-27\)
\(=x^3-27\)
e: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3\)
\(=8x^3+y^3\)
g: \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3=8x^3-y^3\)
m: \(\left(x+1\right)^3=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\)
\(=x^3+3x^2+3x+1\)
n: \(\left(x-1\right)^3=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=x^3-3x^2+3x-1\)
p: \(\left(2x+1\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3\)
\(=8x^3+12x^2+6x+1\)
q: \(\left(2x-1\right)^3=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)
\(=8x^3-3\cdot4x^2+3\cdot2x-1=8x^3-12x^2+6x-1\)
k: \(\left(x+2y\right)^3\)
\(=x^3+3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2+\left(2y\right)^3\)
\(=x^3+6x^2y+12xy^2+8y^3\)
h: \(\left(x-2y\right)^3\)
\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=x^3-6x^2y+12xy^2-8y^3\)
t: \(\left(3x+4y\right)^3=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot4y+3\cdot3x\cdot\left(4y\right)^2+\left(4y\right)^3\)
\(=27x^3+3\cdot9x^2\cdot4y+3\cdot3x\cdot16y^2+64y^3\)
\(=27x^3+108x^2y+144xy^2+64y^3\)
i: \(\left(3x-4y\right)^3=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot4y+3\cdot3x\cdot\left(4y\right)^2-\left(4y\right)^3\)
\(=27x^3-3\cdot9x^2\cdot4y+3\cdot3x\cdot16y^2-64y^3\)
\(=27x^3-108x^2y+144xy^2-64y^3\)