viết dưới dạng tichj theo hàng đẳng thưc
a ,8x^3 - 64y^3
9x^2-30xy+25y^2
4x^2+16x+7
-5+18y-9y^2
K
Khách
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NC
0
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
3 tháng 12 2021
a) \(39x-39y=39\left(x-y\right)\)
b) \(3x^2\left(x-3y\right)-5y\left(3y-x\right)=3x^2\left(x-3y\right)+5y\left(x-3y\right)\)
\(=\left(3x^2+5x\right)\left(x-3y\right)=x\left(3x+5\right)\left(x-3y\right)\)
c) \(16x^2+24xy+9y^2=\left(4x\right)^2+4x.3y.2+\left(3y\right)^2=\left(4x+3y\right)^2\)
d) \(25x^2-\frac{1}{25y^2}=\left(5x\right)^2-\left(\frac{1}{5y}\right)^2=\left(5x-\frac{1}{5y}\right)\left(5x+\frac{1}{5y}\right)\)
e) \(7x^2-7xy+5x-5y=7x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(7x+5\right)\)
f) \(5x^2-45y^2-30y-5=5\left(x^2-9y^2-6y-1\right)=5\left[x^2-\left(9y^2+6y+1\right)\right]\)
\(=5\left[x^2-\left(3y+1\right)^2\right]=5\left(x-3y-1\right)\left(x+3y+1\right)\)
g) \(x^2+2x+1-y^2-4y-1=\left(x^2+2x+1\right)-\left(y^2+2y+1\right)\) ( Chắc đề vậy :v )
\(=\left(x+1\right)^2-\left(y+1\right)^2=\left(x+1-y-1\right)\left(x+1+y+1\right)=\left(x-y\right)\left(x+y+2\right)\)
h) \(4x^2+8x-5=4x^2-2x+10x-5=2x\left(2x-1\right)+5\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x+5\right)\)
20 tháng 7 2023
`a, 4x^2 - 25y^2 = (2x-5y)(2x+5y)`.
`b, 8x^3 +27 = (2x+3)(4x^2 - 6x + 9)`.
`c, 125x^3 - 64y^3 = (5x)^3 - (4y)^3 = (5x-4y)(25x^2 + 20xy + 16y^2)`.
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
20 tháng 7 2023
\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
HP
24 tháng 9 2021
a. x2 + 6x + 9 = (x + 3)2
b. 25 + 10x + x2 = (5 + x)2
c. x2 + 8x + 16 = (x + 4)2
d. x2 + 14x + 49 = (x + 7)2
e. 4x2 + 12x + 9 = (2x + 3)2
f. 9x2 + 12x + 4 = (3x + 2)2
h. 16x2 + 8 + 1 = (4x + 1)2
i. 4x2 + 12xy + 9y2 = (2x + 3y)2
k. 25x2 + 20xy + 4y2 = (5x + 2y)2
24 tháng 9 2021
a) \(=\left(x+3\right)^2\)
b) \(=\left(x+5\right)^2\)
c) \(=\left(x+4\right)^2\)
d) \(=\left(x+7\right)^2\)
e) \(=\left(2x+3\right)^2\)
f) \(=\left(3x+2\right)^2\)
h) \(=\left(4x+1\right)^2\)
i) \(=\left(2x+3y\right)^2\)
k) \(=\left(5x+2y\right)^2\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 10 2019
\(=-16x^2+24xy+6xy-9y^2\)
\(=-8x\left(2x-3y\right)+3y\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(3y-8x\right)\)
4 tháng 10 2019
k
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5 tháng 10 2019
\(30xy-16x^2-9y^2\)
\(=-\left(16x^2-24xy+9y^2\right)+6xy\)
\(=-\left(4x-3y\right)^2+6xy\)
\(=-\left[\left(4x-3y\right)^2-6xy\right]\)
\(=-\left(4x-3y-\sqrt{6xy}\right)\left(4x-4y+\sqrt{6xy}\right)\)
ND
Nguyễn Đức Trí
VIP
25 tháng 8 2023
a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+8x+16=\left(x+4\right)^2\)
c) \(x^2+6x+9=\left(x+3\right)^2\)
d) \(4x^2+4x+1=\left(2x+1\right)^2\)
e) \(36+x^2-12x=x^2-12x+36=\left(x-6\right)^2\)
f) \(4x^2+12x+9=\left(2x+3\right)^2\)
g) \(x^4+81+18x^2=x^4+18x^2+81=\left(x^2+9\right)^2\)
h) \(9x^2+30xy+25y^2=\left(3x+5y\right)^2\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 8 2023
a, \(x^2\) + 2\(x\) + 1 = (\(x\) + 1)2
b, \(x^2\) + 8\(x\) + 16 = (\(x\) + 4)2
c, \(x^2\) + 6\(x\) + 9 = (\(x\) + 3)2
d, 4\(x^2\) + 4\(x\) + 1 = (2\(x\) + 1)2
1 tháng 8 2017
\(1.\)
\(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xyz^2\)
\(=x^2z\left(x-z\right)-xyz\left(x-z\right)\)
\(=\left(x^2z-xyz\right)\left(x-z\right)\)
\(=xz\left(x-y\right)\left(x-z\right)\)
\(2.\)
\(x^2-\left(a+b\right)xy+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x^2-bxy-axy+aby^2\)
\(=x\left(x-by\right)-ay\left(x-by\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
\(3.\)
\(ab\left(x^2+y^2\right)+xy\left(x^2+y^2\right)\)
\(=abx^2+aby^2+a^2xy+b^2xy\)
\(=abx^2+b^2xy+a^2xy+aby^2\)
\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)
\(=\left(ax+by\right)\left(bx+ay\right)\)
\(4.\)
\(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2b^2+a^2y^2+b^2x^2\)
\(=x^2y^2+b^2x^2+a^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(a^2+x^2\right)\left(b^2+y^2\right)\)
\(5.\)
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)
\(=a^2b-ab^2-a^2c-b^2c+ac^2-bc^2\)
\(=ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left(ab-bc-ac+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a-c\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(6.\)
\(16x^2-40xy+2y^2\)
\(=\left(4x\right)^2-2\cdot4\cdot5xy+\left(5y\right)^2\)
\(=\left(4x-5y\right)^2\)
\(7.\)
\(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2\cdot5x^2y+y^2\)
\(=\left(5x^2+y\right)^2\)
\(8.\)
\(-16x^4y^6-24x^5y^5-9x^6y^4\)
\(=-\left(4^2x^4y^6+2\cdot4\cdot3x^5y^5+3^2x^6y^4\right)\)
\(=-\left[\left(4x^2y^3\right)^2+2\left(4x^2y^3\right)\left(3x^3y^2\right)+\left(3x^3y^2\right)^2\right]\)
\(=\left(4x^2y^3+3x^3y^2\right)^2\)
\(9.\)
\(16x^2-4y^2-8x+1\)
\(=\left(4x\right)^2-\left(2y\right)^2-8x+1\)
\(=\left(4x\right)^2-8x+1-\left(2y\right)^2\)
\(=\left(4x+1\right)^2-\left(2y\right)^2\)
\(=\left(4x-2y+1\right)\left(4x+2y+1\right)\)
\(10.\)
\(49x^2-25+42xy+9y^2\)
\(=\left(7x\right)^2-5^2+2\cdot7\cdot3xy+\left(3y\right)^2\)
\(=\left(7x\right)^2+2\cdot7\cdot3xy+\left(3y\right)^2-5^2\)
\(=\left(7x+3y\right)^2-5^2\)
\(=\left(7x+5y+5\right)\left(7x+3y-5\right)\)
3 tháng 9 2016
2) (a-1)2+(b-2)2+(2c-1)2=0
do (a-1)2, (b-2)2 và (2c-1)2 lớn hơn hoặc bằng 0 nên để thỏa mãn biểu thức trên thì (a-1)2, (b-2)2 và (2c-1)2 đồng thời bằng 0
suy ra a=1, b=2, c=1/2
\(8x^3-64y^3=\left(2x\right)^3-\left(4y\right)^3=\left(2x-4y\right)\left(4x^2+8xy+16y^2\right)\)
\(9x^2-30xy+25y^2=\left(3x\right)^2-2\cdot3x\cdot5y+\left(5y\right)^2=\left(3x-5y\right)^2\)
\(4x^2+16x+7=\left(2x^2\right)+2\cdot2x\cdot4+4^2-9=\left(2x+4\right)^2-3^2=\left(2x+1\right)\left(2x+7\right)\)
\(-5+18y-9y^2=-\left[\left(3y\right)^2-2\cdot3y\cdot3+3^2-4\right]=-\left[\left(3y-3\right)^2-2^2\right]=-\left(3y-5\right)\left(3y-1\right)\)
đến đó giải tiếp hộ mik nx ddi đc k