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Ta có : (x3 + 3x + 1)(x3 + 3x + 2) - 6
= (x3 + 3x + 1,5 - 0,5)(x3 + 3x + 1,5 + 0,5) - 6
= (x3 + 3x + 1,5)2 - 0,52 - 6
= (x3 + 3x + 1,5)2 - 6,25
= (x3 + 3x + 1,5 - 2,5) (x3 + 3x + 1,5 + 2,5)
= (x3 + 3x - 1) (x3 + 3x + 3)
\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\)\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\)\(\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\)\(\left(3x-2\right)\left(3x-6\right)\)
\(=\)\(3\left(x-2\right)\left(3x-2\right)\)
Chúc bạn học tốt ~
c) Đặt \(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1,5=a\)
\(\Rightarrow A=\left(a-0,5\right)\left(a+0,5\right)-6\)
\(\Rightarrow A=a^2-0,25-6\)
\(\Rightarrow A=a^2-\frac{25}{4}\)
\(\Rightarrow A=\left(a-\frac{5}{2}\right)\left(a+\frac{5}{2}\right)\)
Thay \(a=x^2+3x+0,5\)vào A ta có :
\(A=\left(x^2+3x+0,5-\frac{5}{2}\right)\left(x^2+3x+0,5+\frac{5}{2}\right)\)
\(A=\left(x^2+3x-2\right)\left(x^2+3x+3\right)\)
c, Đặt \(x^2+3x+2=a\)
Ta có : \(\left(a-1\right)a-6=a^2-a-6=\left(a^2-3a\right)+\left(2a-6\right)\)
\(=a\left(a-3\right)+2\left(a-3\right)\)
\(=\left(a+2\right)\left(a-3\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
Câu d làm tương tự .
Gợi ý : (x+3)(x+5) = x2 + 8x + 15
đặt bằng a rồi giải tiếp
a)(x+y)2-(x-y)2
=(x+y-x+y)(x+y+x-y)
=2y.2x=4xy
b)(3x+1)2-(x+1)2
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
=4x(2x+1)
c) x3+y3+z3-3xyz
= (x+y)3- 3xy(x+y) +z3-3xyz
=(x+y+z)( x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)
=(x+y+z)(x2+y2+z2-xy-xz-yz)
Phân tích đa thức sau thành nhân tử :
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
b) \(x^3+y^3+z^3-3xyz\)
x3+y(1-3x2)+x(3y2-1)-y3
= x3-3x2y+3xy2-y3+y-x
=(x-y)3 -(x-y)
=(x-y)(x2-2xy+y2-1)
cái chỗ kia giải thích dùm mìh đy : \(x^3-3x^2y+3xy^2-y^3+y-x\)
\(x^3+y\left(1-3x^2\right)+x\left(3y^2-1\right)-y^3\)
\(=x^3-3x^2y+3xy^2-y^3+y-x\)
\(=\left(x-y\right)^3-\left(x-y\right)\)
phân tích đa thức thành nhân tử cơ mà
=(x-y)3-(x-y)
=(x-y)[(x-y)2-1]
a) \(3^2\left(y-x\right)+6x^2\left(x-y\right)^2\)
\(=3\left(y-x\right)\left[3+2x^2\left(y-x\right)\right]\)
\(=3\left(y-x\right)\left(3+2x^2y-2x^3\right)\)
b) \(x^4-3x^3+3x-1\)
\(=\left(x^4+x^3\right)-\left(4x^3+4x^2\right)+\left(4x^2+4x\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-4x^2+4x-1\right)\)
\(=\left(x+1\right)\left[\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(x-1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(x^2-3x+1\right)\)
d)\(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2+2xy\right)\left(x^2+y^2-z^2-2xy\right)\)
\(=\left[\left(x^2+2xy+y^2\right)-z^2\right]\left[\left(x^2-2xy+y^2\right)-z^2\right]\)
\(=\left[\left(x+y\right)^2-z^2\right]\left[\left(x-y\right)^2-z^2\right]\)
\(=\left(x+y-z\right)\left(x+y+z\right)\left(x-y-z\right)\left(x-y+z\right)\)
e)Đặt \(x^2+3x=a\)
Có: \(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
\(=\left(a+1\right)\left(a-3\right)-5\)
\(=a^2-3a+a-3-5\)
\(=a^2-2a-8\)
\(=a^2+2x-4x-8\)
\(=a\left(a+2\right)-4\left(a+2\right)\)
\(=\left(a+2\right)\left(a-4\right)\)
\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)
\(=\left(x^2+x+2x+2\right)\left(x^2-x+4x-4\right)\)
\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x-1\right)+4\left(x-1\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x-1\right)\left(x+4\right)\)
\(d,\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-z^2\right]\left[\left(x^2+2xy+y^2\right)-z^z\right]\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
\(e,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\left(1\right)\)
\(\text{Đặt }x^2+3x+\frac{1-3}{2}=t\)
\(\text{hay }x^2+3x-2=t\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left(t+3\right)\left(t-1\right)-5\)
\(\Rightarrow t^2-t+3t-3-5\)
\(=t^2+2t-8\)
\(=t^2-2t+4t-8\)
\(=t\left(t-2\right)+4\left(t-2\right)\)
\(=\left(t-2\right)\left(t+4\right)\left(3\right)\)
\(\text{Thay (2) vào (3),ta được:}\)
\(\left(x^2+3x-2-2\right)\left(x^2+3x-2+4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x^2-x+4x-4\right)\left(x^2+x+2x+2\right)\)
\(=\left[x\left(x-1\right)+4\left(x-1\right)\right]\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)