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1) \(\left(x+1\right)^2-16y^2\)
\(=\left(x+1\right)^2-\left(4y\right)^2\)
\(=\left(x+1-4y\right).\left(x+1+4y\right)\)
2) \(2x^2-8\)
\(=\left[\left(\sqrt{2x}\right)^2\right]^2-\left(\sqrt{8}\right)^2\)
\(=\left[\left(\sqrt{2x}\right)^2-\sqrt{8}\right].\left[\left(\sqrt{2x}\right)^2+\sqrt{8}\right]\)
6) \(x^2+1-y^2+2x\)
\(=x^2+2x+1-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right).\left(x+1+y\right)\)
Chúc bạn học tốt!
a) =(x-y)*(x+y)-(5*(x+y))
=(x+y)*(x-y-5)
Mấy bài còn lại cũng tương tự nha bạn = cách đặt nhân tử chung
bai nao khong hieu thi pan nhan tin vào nick minh minh se giai đùm ban
a) (x2 - y2) - 5(x + y)
= (x - y)(x + y) - 5 (x + y)
= (x + y) (x - y -5)
b) 5x3 - 5x2y - 10x2 + 10 xy
= 5[(x3 - x2y) - (2x2 - 2 xy)]
=5[x2(x - y) - 2x(x - y)]
=5x(x-y)(x - 2)
c) 2x2 - 5x = x(2x - 5)
d) x3 - 3x2 +1 - 3x
= (x3 + 1) - (3x2 + 3x)
= (x + 1)(x2 - x + 1) - 3x(x + 1)
= (x + 1) [x2 - x + 1 - 3x]
= (x + 1)[x2 - 4x + 1]
= (x + 1)[x2 - 2.x.2 + 22 - 22 + 1]
= (x + 1)[(x - 2)2 - 3]
= \(\left(x+1\right)\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\)
e) 3x2 - 6xy + 3y2 - 12z2
= 3[ x2 - 2xy + y2 - 4z2]
= 3[ (x - y)2 - (2z)2]
= 3(x - y + 2z)(x - y - 2z)
f) 3x2 - 7x - 10
= 3x2 - 7x - 7 - 3
= (3x2 -3) - (7x + 7)
= 3(x2 - 1) - 7(x + 1)
= 3 (x + 1)(x - 1) - 7(x + 1)
= (x + 1)[3(x - 1) - 7]
= (x +1)(3x - 8)
g) x4 + 1 - 2x2 = (x2)2 - 2.x2 + 1 = (x2 - 1)2
= (x + 1)2(x - 1)2
h) 3x2 - 3y2 - 12x + 12y
= 3(x2 - y2) - 12(x - y)
= 3(x - y)(x + y) - 12(x -y)
= (x - y) [3(x + y) - 12]
= (x - y). 3. (x+y - 4)
j) x2 - 3x + 2 = x2 - x - 2x +2
= x(x - 1) - 2(x -1)
=(x - 1)(x - 2)
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
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) \(4x^2+8x-5=4x^2-2x+10x-5=2x\left(2x-1\right)+5\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x+5\right)\)
\(5x^2\left(3y-1\right)-\left[3x^2\left(5y+2\right)-2x\left(3x^2-7x\right)\right]=15x^2y-5x^2-\left(15x^2y+6x^2-6x^3+14x^2\right)=15x^2y-5x^2-15x^2y-6x^2+6x^3-14x^2=6x^3-25x^2\)
bạn ơi mik nghĩ là khi bạn viết thì nên xuống dòng