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\(a,4x^2+28x+49=\left(2x\right)^2+2.2x.7+7^2=\left(2x+7\right)^2\\ b,16y^2-8y+1=\left(4y\right)^2-2.4y.1+1^2=\left(4y-1\right)^2=\left(1-4y\right)^2\\ 4a^2+20ab+25b^2=\left(2a\right)^2+2.2a.5b+\left(5b\right)^2=\left(2a+5b\right)^2\\ d,9x^2-6xy+y^2=\left(3x\right)^2-2.3x.y+y^2=\left(3x-y\right)^2=\left(y-3x\right)^2\)
a) \(7x^2-63y^2\)
\(=7\cdot\left(x^2-9y^2\right)\)
\(=7\left[x^2-\left(3y\right)^2\right]\)
\(=7\left(x-3y\right)\left(x+3y\right)\)
b) \(36-4a^2+20ab-25b^2\)
\(=36-\left(4a^2-200ab+25b^2\right)\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
c) \(2x-2y-x^2+2xy-y^2\)
\(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
a: =7(x^2-9y^2)
=7(x-3y)(x+3y)
b: =36-(4a^2-20ab+25b^2)
=36-(2a-5b)^2
=(6-2a+5b)(6+2a-5b)
c: =2(x-y)-(x-y)^2
=(x-y)(2-x+y)
a)7x2 + 6x - 13
= 7x2 - 7x + 13x - 13
= (7x2 - 7x) + (13x - 13)
= 7x(x - 1) + 13(x - 1)
= (7x + 13)(x - 1)
b) 15(x - y) - 25x + 25y
= 15(x - y) - (25x - 25y)
= 15(x - y) - 25(x - y)
= (15 - 25)(x - y)
= -10(x - y)
c) 36 - 4a2 + 20ab - 25b2
= 36 - (4a2 - 20ab + 25b2)
= 62 - (2a - 5b)2
= (6 - 2a + 5b)(6 + 2a - 5b)
= (5b - 2a + 6)(2a - 5b + 6)
nhớ tik mik nha. Ak có j ko hiểu thì cứ bình luận ở dưới mik giải cho
a, \(7x^2+6x-13\)
=\(7x^2-7x+13x-13\)
=\(\left(7x^2-7x\right)+\left(13x-13\right)\)
=\(7x\left(x-1\right)+13\left(x-1\right)\)
=\(\left(x-1\right)\left(7x+13\right)\)
a)\(36-4a^2+20ab-25b^2=6^2-\left(4a^2-20ab+25b^2\right)\)
\(=6^2-\left[\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
b)\(a^3+3a^2+3a+1-27b^3=\left(a+1\right)^3-\left(3b\right)^3\)(chỗ này mình sửa 27b2 thành 27b3 vì mình nghĩ nhầm đề)
\(=\left(a+1-3b\right)\left[\left(a+1\right)^2+\left(a+1\right)3b+\left(3b\right)^2\right]\)
\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)
c)\(x^3+3x^2+3x+1-3x^2-3x=\left(x+1\right)^3-3x\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\)
Bài 1:
a) x3 - 3x2 + 3x - 1 + 2(x2 - x)
= (x - 1)3 + 2x(x - 1)
= (x - 1)[(x - 1)2 + 2x]
= (x - 1)(x2 - 2x + 1 + 2x)
= (x - 1)(x2 + 1)
b) 36 - 4a2 + 20ab - 25b2
= 36 - (2a - 5b)2
= (6 - 2a + 5b)(6 + 2a - 5b)
c) 5a3 - 10a2b + 5ab2 - 10a + 10b
= 5(a3 - 2a2b + ab2 - 2a + 2b)
= 5[a(a2 - 2ab + b2) - 2(a - b)]
= 5[a(a - b)2 - 2(a - b)]
= 5(a - b)(a2 - ab - 2)
\(a,36-4a^2+20ab-25b^2\)
\(=6^2-\left(2a-5b\right)^2=\left(6-2a+5b\right)\left(6+2a-5b\right)\)\(b,x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
\(d,5a^2-10a^2b+5ab^2-10a+10b\)
\(=5a^2-5a^2b-5a^2b+5ab^2-10a+10b\)
\(=5a\left(a-b\right)-5ab\left(a-b\right)-10\left(a-b\right)\)
\(=\left(a-b\right)\left(5a-5ab-10\right)\)
a) \(3xy^2-12xy+12x\)
\(=3x\left(y-4y+4\right)\)
b) \(3x^3y-6x^2y-3xy^3-6axy^2-3a^2xy+3xy\)
\(=3xy\left(x^2-2x-y^2-2ay-a^2+1\right)\)
\(=3xy\left[\left(x^2-2\cdot x\cdot1+1^2\right)-\left(y^2+2\cdot y\cdot a+a^2\right)\right]\)
\(=3xy\left[\left(x-1\right)^2-\left(y+a\right)^2\right]\)
\(=3xy\left(x-1-y-a\right)\left(x-1+y+a\right)\)
c) \(36-4a^2+20ab-25b^2\)
\(=6^2-\left[\left(2a\right)^2-2\cdot2a\cdot5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
d) \(5a^3-10a^2b+5ab^2-10a+10b\)
\(=5a\left(a^2-2ab+b^2\right)-10\left(a-b\right)\)
\(=5a\left(a-b\right)^2-10\left(a-b\right)\)
\(=\left(a-b\right)\left[5a\left(a-b\right)-10\right]\)
\(=5\left(a-b\right)\left[a\left(a-b\right)-2\right]\)
\(=5\left(a-b\right)\left(a^2-ab-2\right)\)
a. 3xy2-12xy+12x
= 3x(y2-4y+4)
= 3x(y-2)2
b. 3x3y-6x2y-3xy3-6axy2-3a2xy+3xy
= 3xy( x2-2x-y2-2ay-a2+1)
= 3xy ((x2-2x+1)-(a2-2ay-y2))
=3xy((x-1)2-(a-y)2)
= 3xy((x-1+a-y)(x-1-(a-y))
=3xy(x-1+a-y)(x-1-a+y)
d. =( 5a(a2-2ab+b2))-(10(a+b))
= 5a(a-b)2-10(a-b)
=5a(a-b)(a-b)-10(a-b)
=(a-b)(5a(a-b)-10)
Hình như mik làm sai hết rồi
\(=81-\left(2a-5b\right)^2=\left(9-2a+5b\right)\left(9+2a-5b\right)\)
:v