Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x^3.(x^2-7)^2-36x
=x(x^6-14x^4+49x^2-36)
=x.[x^4(x^2-1)-13x^2(x^2-1)+36(x^2-1)
=x(x-1)(x+1)(x^4-13X^2+36)
=x(x-1)(x+1)[x^2(x^2-4)-9(x^2-4)]
=x(x-1)(x+1)(x-2)(x+2)(x-3)(x+3)
Ta có : x3 . ( x2 - 7 )2 - 36x
=> x ( x6 - 14x4 + 49x2 - 36 )
=> x [ x4 ( x2 - 1 ) - 13x2 ( x2 - 1 ) + 36 ( x2 - 1 )
=> x ( x - 1 ) ( x + 1 ) ( x4 - 13x2 + 36 )
=> x ( x - 1 ) ( x + 1 ) [ x2 ( x2 - 4 ) - 9 ( x2 - 4 ) ]
=> x ( x - 1 ) ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 3 ) ( x + 3 )
bài 2 nè
a+b+c = 0
=>(a+b+c)^3 = 0
a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c) = 0
vì a+b = -c
a+c = -b
b+c = -a
thay vào => a^3 + b^3 + c^3 - 3abc = 0
=> a^3 + b^3 + c^3 = 3abc
\(A=x^4-14x^3+71x^2-154x+120\)
\(=x^3\left(x-2\right)-12x^2\left(x-2\right)+47x\left(x-2\right)-60\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-12x^2+47x-60\right)\)
\(=\left(x-2\right)\left[x^2\left(x-3\right)-9x\left(x-3\right)+20\left(x-3\right)\right]\)
\(=\left(x-2\right)\left(x-3\right)\left(x^2-9x+20\right)=\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\)
b, Vì A là tích của 4 số nguyên liên tiếp nên A chia hết cho 24
A = x.[x^2.(x^2-7)^2-36]
= x.[(x^3-7x)^2-6^2]
= x.(x^3-7x-6).(x^3-7x+6)
= x.[(x^3+1)-(7x+7)].[(x^3-x)-(6x-6)]
= x.(x+1).(x^2-x-7).(x-1).(x^2+x-6)
= x.(x+1).(x-1).(x-2).(x+3).(x^2-x-7)
Tk mk nha
x3(x2−7)2−36x=x3(x4−14x2+49)−36xx3(x2−7)2−36x=x3(x4−14x2+49)−36x
=x7−14x5+49x3−36xx7−14x5+49x3−36x
=x7−x6+x6−x5−13x5+13x4−13x4+13x3+36x3−36xx7−x6+x6−x5−13x5+13x4−13x4+13x3+36x3−36x
=x6(x−1)+x5(x−1)−13x4(x−1)−13x3(x−1)+36x(x2−1)x6(x−1)+x5(x−1)−13x4(x−1)−13x3(x−1)+36x(x2−1)
=x(x−1)(x5+x4−13x3−13x2+36x+36)x(x−1)(x5+x4−13x3−13x2+36x+36)
=x(x−1)[x4(x+1)−13x2(x+1)+36(x+1)]x(x−1)[x4(x+1)−13x2(x+1)+36(x+1)]
=x(x−1)(x+1)(x4−13x2+36)x(x−1)(x+1)(x4−13x2+36)
đặt x^2 =a (a>=0) thì xét đa thức x4−13x2+36=a2−13a+36x4−13x2+36=a2−13a+36
xét Δ=b2−4ac=169−4.36=25Δ=b2−4ac=169−4.36=25
Δ>0Δ>0→phương trình có 2 nghiệm riêng biệt là ⎡⎣a1=−b+Δ√2a=13+52=9a2=−b−Δ√2a=13−52=4[a1=−b+Δ2a=13+52=9a2=−b−Δ2a=13−52=4(t/m a>=0)
vậy bt ban đầu :x(x−1)(x+1)(x2−4)(x2−9)x(x−1)(x+1)(x2−4)(x2−9)
=(x−3)(x−2)(x−1)x(x+1)(x+2)(x+3)