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AH
Akai Haruma
Giáo viên
28 tháng 12 2023

Lời giải:
\(x^3(x^2-7)^2-36x=x[x^2(x^2-7)^2-36]\\ =x[(x^3-7x)^2-6^2]=x(x^3-7x-6)(x^3-7x+6)\\ =x[x^2(x-3)+3x(x-3)+2(x-3)][x^2(x-2)+2x(x-2)-3(x-2)]\\ =x(x-3)(x^2+3x+2)(x-2)(x^2+2x-3)\\ =x(x-3)(x+1)(x+2)(x-2)(x-1)(x+3)\)

13 tháng 2 2016

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 )

21 tháng 10 2016

\(x^3\left(x^2-7\right)^2-36x=x^3\left(x^4-14x^2+49\right)-36x\)

=\(x^7-14x^5+49x^3-36x\)

=\(x^7-x^6+x^6-x^5-13x^5+13x^4-13x^4+13x^3+36x^3-36x\)

=\(x^6\left(x-1\right)+x^5\left(x-1\right)-13x^4\left(x-1\right)-13x^3\left(x-1\right)+36x\left(x^2-1\right)\)

=\(x\left(x-1\right)\left(x^5+x^4-13x^3-13x^2+36x+36\right)\)

=\(x\left(x-1\right)\left[x^4\left(x+1\right)-13x^2\left(x+1\right)+36\left(x+1\right)\right]\)

=\(x\left(x-1\right)\left(x+1\right)\left(x^4-13x^2+36\right)\)

đặt x^2 =a (a>=0) thì xét đa thức \(x^4-13x^2+36=a^2-13a+36\)

xét \(\Delta=b^2-4ac=169-4.36=25\)

\(\Delta>0\)→phương trình có 2 nghiệm riêng biệt là \(\left[\begin{array}{nghiempt}a_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{13+5}{2}=9\\a_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{13-5}{2}=4\end{array}\right.\)(t/m a>=0)

vậy bt ban đầu :\(x\left(x-1\right)\left(x+1\right)\left(x^2-4\right)\left(x^2-9\right)\)

=\(\left(x-3\right)\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

13 tháng 7 2016

a) \(x^3\left(x^2-7\right)^2-36x=x\left[\left(x^3-7x\right)^2-6^2\right]\)

\(=x\left(x^3-7x-6\right)\left(x^3-7x+6\right)\)

\(x\left[\left(x-3\right)\left(x+1\right)\left(x+2\right)\right].\left[\left(x+3\right)\left(x-2\right)\left(x-1\right)\right]\)

\(=\left(x-3\right)\left(x-2\right)\left(x-1\right).x.\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

b) Không pt được.

c) Không pt được.

25 tháng 2 2018

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

31 tháng 7 2018

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)

22 tháng 8 2023

2) \(x^4-5x^2+4\)

\(=x^4-x^2-4x^2+4\)

\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2-4\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)

22 tháng 8 2023

\(8x^3+36x^2y+54xy^2+27y^3\\ =\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\\ =\left(2x+3y\right)^3\\ =\left(2x+3y\right)\left(2x+3y\right)\left(2x+3y\right)\)

\(\left(x-y\right)^3-\left(x+y\right)^3\\ =\left(x-y-x-y\right)\left(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2\right)\\ =-2y\left(3x^2+y^2\right)\)

\(\left(x+1\right)^3+\left(x-1\right)^3\\ =\left(x+1+x-1\right)\left(x^2+2x+1-x^2+1+x^2-2x+1\right)\\ =2x\left(x^2+3\right)\)

\(\left(x-1\right)^2-\left(x+1\right)^2\\ =\left(x-1-x-1\right)\left(x-1+x+1\right)\\ =-2.2x=-4x\)

a: =(2x)^3+3*(2x)^2*3y+3*2x*(3y)^2+(3y)^3

=(2x+3y)^3

b: (x-y)^3-(x+y)^3

=(x-y-x-y)[(x-y)^2+(x-y)(x+y)+(x+y)^2]

=-2y*[x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2]

=-2y(3x^2+y^2)

c: (x+1)^3+(x-1)^3

=(x+1+x-1)[(x+1)^2-(x+1)(x-1)+(x-1)^2]

=2x*[x^2+2x+1-x^2+1+x^2-2x+1]

=2x(x^2+3)

d: =(x-1-x-1)(x-1+x+1)

=2x*(-2)=-4x