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a. Đề phải là thế này:
x3 + y3 + z3 - 3xyz
= (x3 + 3x2y + 3xy2 + y3) + z3 - (3x2y + 3xy2 + 3xyz)
= (x + y)3 + z3 - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - z(x + y) + z2] - 3xy(x + y + z)
= ( x + y + z)(x2 + 2xy + y2 - xz - zy + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
b. (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x - 1)(x - 4)(x - 2)(x - 3) - 3
= (x2 - 5x + 4)(x2 - 5x + 6) - 3
Đặt t = x2 - 5x + 4
=> Đa thức
<=> t.(t + 2) - 3
= t2 + 2t - 3
= t2 + 3t - t - 3
= t.(t + 3) - (t + 3)
= (t + 3)(t + 1) (1)
Thay t = x2 - 5x + 4 vào (1):
=> (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x2 - 5x + 4 + 3)(x2 - 5x + 4 + 1)
= (x2 - 5x + 7)(x2 - 5x + 5)
a) (x+y)2-(x-y)2
=(x+y)(x-y)
b)(3x+1)2-(x+1)2
=[(3x+1)+(x+1)].[(3x+1)-(x+1)]
=(3x+1+x+1)(3x+1-x-1)
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=2x.2\left(2x+1\right)=4x\left(2x+1\right)\)
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)
b) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy+xz+yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=\frac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2}\)
a/ \(=x^4+x^3+x^2+5x^2+5x+5\)
\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)=\left(x^2+5\right)\left(x^2+x+1\right)\)
b/ \(=x^3+x^2+2x-x^2-x-2\)
\(=x\left(x^2+x+2\right)-\left(x^2+x+2\right)=\left(x-1\right)\left(x^2+x+2\right)\)
c/ \(=x^3+4x^2+4x-x^2-4x-4\)
\(=x\left(x^2+4x+4\right)-\left(x^2+4x+4\right)=\left(x-1\right)\left(x+2\right)^2\)
câu d khó quá , mk lm k nổi , sr nha ^^
a) x4 + x3 + 6x2 + 5x + 5
= x4 + x3 + x2 + 5x2 + 5x + 5
= x2 ( x2 + x + 1) + 5 (x2 + x + 1)
= (x2 + x + 1) (x2 + 5)
b) x3 + x - 2
= x3 + x2 + 2x - x2 - x - 2
= x (x2 + x + 2) - (x2 + x + 2)
= (x2 + x + 2) (x - 1)
c) x3 + 3x2 - 4
= x3 + 4x2 + 4x - x2 - 4x - 4
= x (x2 + 4x + 4) - (x2 + 4x + 4)
= (x2 + 4x + 4) (x - 1)
= (x + 2)2 (x - 1)
d) xy(x + y) + yz(y + z) + xz(x + z) + 3xyz
= xy(x + y) + xyz + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + y + z)
= (x + y + z) (xy + yz + xz)
2
a
\(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^3=\left(-z\right)^3\)
\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)
\(\Rightarrow x^3+y^3+z^3=3xy\left(x+y\right)=3xyz\)
b
Đặt \(a-b=x;b-c=y;c-a=z\Rightarrow x+y+z=0\)
Ta có bài toán mới:Cho \(x+y+z=0\).Phân tích đa thức thành nhân tử:\(x^3+y^3+z^3\)
Áp dụng kết quả câu a ta được:
\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
a) \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left[\left(x-2\right)\left(x-5\right)\right]\left[\left(x-3\right)\left(x-4\right)\right]+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)
Đặt: \(x^2-7x+11=t\)
\(\Rightarrow\hept{\begin{cases}x^2-7x+10=t-1\\x^2-7x+12=t+1\end{cases}}\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)
\(=\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1\)
\(=t^2\)
Vậy: \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left(x^2-7x+11\right)^2\)
a) Đề bài phải là : \(\left(x+y\right)^2-\left(x-y\right)^2\)thì mới phân tích được.
Nếu đề bài như trên ta có:
\(\left(x+y\right)^2-\left(x-y\right)^2=\)\(\left(x+y-x+y\right)\left(x+y+x-y\right)=2x\cdot2y=4xy\)
b) Ta có: \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
= \(2x\cdot\left(4x+2\right)=2x\cdot2\cdot\left(2x+1\right)=4x\cdot\left(2x+1\right)\)
c) Ta có : \(x^3+y^3+z^3-3xyz\)
= \(\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xy\)
=\(\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y+z\right)\)
=\(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
=\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
x3 + y3 + z3 - 3xyz
= (x3 + 3x2y + 3xy2 + y3) + z3 - (3x2y + 3xy2 + 3xyz)
= (x + y)3 + z3 - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - z(x + y) + z2] - 3xy(x + y + z)
= ( x + y + z)(x2 + 2xy + y2 - xz - zy + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
b. (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x - 1)(x - 4)(x - 2)(x - 3) - 3
= (x2 - 5x + 4)(x2 - 5x + 6) - 3
Đặt t = x2 - 5x + 4
=> Đa thức
<=> t.(t + 2) - 3
= t2 + 2t - 3
= t2 + 3t - t - 3
= t.(t + 3) - (t + 3)
= (t + 3)(t + 1) (1)
Thay t = x2 - 5x + 4 vào (1):
=> (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x2 - 5x + 4 + 3)(x2 - 5x + 4 + 1)
= (x2 - 5x + 7)(x2 - 5x + 5)