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a) Ta có: \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
b) Ta có: \(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(-5x+1\right)\)
c) Ta có: \(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
d) Ta có: \(2x^2+3x-5\)
\(=2x^2+5x-2x-5\)
\(=x\left(2x+5\right)-\left(2x+5\right)\)
\(=\left(2x+5\right)\left(x-1\right)\)
e) Ta có: \(x^3-3x^2+1-3x\)
\(=\left(x+1\right)\cdot\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
f) Ta có: \(x^2-4x-5\)
\(=x^2-4x+4-9\)
\(=\left(x-2\right)^2-3^2\)
\(=\left(x-2-3\right)\left(x-2+3\right)\)
\(=\left(x-5\right)\left(x+1\right)\)
g) Ta có: \(\left(a^2+1\right)^2-4a^2\)
\(=\left(a^2+1\right)^2-\left(2a\right)^2\)
\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)
h) Ta có: \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
k) Ta có: \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
m) Ta có: \(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
My Nguyễn ơi,bạn truy cập vào đường link này để tìm câu hỏi tương tự của câu a/Bài 1 nhé
https://vn.answers.yahoo.com/question/index?qid=20110206184834AAokV5m&sort=N
a, \(x^3-x^2y-xy^2+y^3\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x^2-y^2\right)\left(x-y\right)\)
\(=\left(x-y\right)^2\left(x+y\right)\)
b, \(x^3+x^2-4x-4\)
\(=x^2\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x^2-4\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\left(x+1\right)\)
c, \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
d, \(\left(7x+3\right)^2-\left(2x-1\right)^2\)
\(=\left(7x+3-2x+1\right)\left(7x+3+2x-1\right)\)
\(=\left(5x+4\right)\left(9x+2\right)\)
e, \(x^3-3x^2-3x+1\) sai đề
f, \(x^2-2x-3\)
\(=x^2-3x+x-3=x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x+1\right)\left(x-3\right)\)
g, \(x^2-2x-8\)
\(=x^2-4x+2x-8=x\left(x-4\right)+2\left(x-8\right)\)
\(=\left(x+2\right)\left(x-8\right)\)
h, \(x^2-10x+21\)
\(=x^2-7x-3x+21\)
\(=x\left(x-7\right)-3\left(x-7\right)=\left(x-3\right)\left(x-7\right)\)
i, \(x^2-4xy+3y^2\)
\(=x^2-4xy+4y^2-y^2\)
\(=\left(x-2y\right)^2-y^2\)
\(=\left(x-2y-y\right)\left(x-2y+y\right)\)
\(=\left(x-3y\right)\left(x-y\right)\)
a) \(x^3 - x^2y - xy^2 + y^3\)
\(=\left(x^3-x^2y\right)-\left(xy^2-y^3\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)^2\left(x+y\right)\)
b) \(x^3 + x^2 - 4x - 4\)
\(=\left(x^3+x^2\right)-\left(4x+4\right)\)
\(=x^2\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4\right)\)
\(=\left(x+1\right)\left(x^2-2^2\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x-2\right)\)
d) mk chỉnh lại đề
\(8xy^2-5xyz-24y+15z\)
\(=xy\left(8y-5z\right)-3\left(8y-5z\right)\)
\(=\left(8y-5z\right)\left(xy-3\right)\)
e) \(x^4-x^3-x+1=\left(x-1\right)^2\left(x^2+x+1\right)\)
f) \(x^4+x^2y^2+y^4=\left(x^2-xy+y^2\right)\left(x^2+xy-y^2\right)\)
g) \(x^3+3x-4=\left(x-1\right)\left(x^2+x+4\right)\)
h) \(x^3-3x^2+2=\left(x-1\right)\left(x^2-2x-2\right)\)
i) \(2x^3+x^2-4x-12=\left(x-2\right)\left(2x^2+5x+6\right)\)
k) \(25x^2\left(x-5\right)-x+y=\left(1-5x\right)\left(1+5x\right)\left(y-x\right)\)
c) thay x=1 vào đa thức f(x) ta có: f(1)=4.1^3-1^2+2.1-5
=4-2+2-5
=- 1
vậy 1 k phải là nghiệm của đa thức f(x)
MÌNH CHỈ LÀM ĐƯỢC C THÔI HOK TỐT
làm sai nha chỗ nào là 1 thì thay bằng -1 nha kq sẽ ra nha