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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 )
\(36x^2-\left(3x-2\right)^2\)
\(=\left(6x\right)^2-\left(3x-2\right)^2\)
\(=\left(6x+3x-2\right)\left(6x-3x+2\right)\)
\(=\left(9x-2\right)\left(3x+2\right)\)
\(-49x^2+9\)
\(=3^2-\left(7x\right)^2\)
\(=\left(3-7x\right)\left(3+7x\right)\)
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)
\(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)
\(\Leftrightarrow\left(1+x^2\right)^2+4x\left(1+x^2\right)\)
\(\Leftrightarrow\left(1+x^2\right)\times\left[\left(1+x^2\right)+4\right]\)
( 1+x2 )2 -4x( 1- x2 )
=x4+2x2+1-4x+4x3
=x3+2x2-x+2x3+4x2-2x-x2-2x+1
=x(x2+2x-1)+2x(x2+2x-1)-(x2+2x-1)
=(x2+2x-1)(x2+2x-1)
=(x2+2x-1)2
\(\left(1+x\right)^2-4x\left(1-x^2\right)\)
\(=\left(1+x\right)^2-4x\left(1-x\right)\left(1+x\right)\)
\(=\left(1+x\right)\left(1+x-4\left(1-x\right)\right)\)
\(=\left(1+x\right)\left(1+x-4+4x\right)\)
\(=\left(1+x\right)\left(5x-3\right)\)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
dat \(x^2-2x+2=y\)
ta co pt
\(y^4+20x^2y^2+64x^4\)
\(=\left(8x^2\right)^2+2.8x^2.\frac{10}{8}y^2+\left(\frac{10^{ }}{8^{ }}y^2\right)^2-\frac{36}{64}y^4\)
\(=\left(8x^2+\frac{10}{8}y^2\right)^2-\left(\frac{6}{8}y^2\right)^2\)
\(=\left(8x^2+\frac{y^2}{2}\right)\left(8x^2+2y^2\right)\)
bạn thay y nữa là xong
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+64x^4\)
\(=\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+100x^4-36x^4\)
\(=\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^4\)
\(=\left(x^4-4x^3+18x^2-8x+4\right)^2-\left(6x^2\right)^2\)
\(=\left(x^4-4x^3+24x^2-8x+4\right)\left(x^4-4x^3+12x^2-8x+4\right)\)
\(=\left(x+3\right)^2\cdot\left(x-3\right)^2\)
cảm ơn bạn