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
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+64x^4\)
\(=\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+100x^4-36x^4\)
\(=\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^4\)
\(=\left(x^4-4x^3+18x^2-8x+4\right)^2-\left(6x^2\right)^2\)
\(=\left(x^4-4x^3+24x^2-8x+4\right)\left(x^4-4x^3+12x^2-8x+4\right)\)
dat x 2 − 2x + 2 = y ta co pt y 4 + 20x 2 y 2 + 64x 4 = 8x 2 2 + 2.8x 2 . 8 10 y 2 + 8 10 y 2 2 − 64 36 y 4 = 8x 2 + 8 10 y 2 2 − 8 6 y 2 2 = 8x 2 + 2 y 2 8x 2 + 2y 2 hi kết bạn nha
sau chữ x,y là mũ hai nhé mình không biết gi mũ
phân tích đa thức thành nhân tử: ( x2-2x+2)4 + 20x^2(x2 -2x+2)2+64x4
( x2-2x+2)4 + 20x^2(x2 -2x+2)2+64x4
=(x^2−2x+2)^4+20x^2(x^2−2x+2)^2+100x^4−36x^4
=[(x^2−2x+2)^2+10x2^]^2−36x^4
=(x64−4x^3+18x^2−8x+4)^2−(6x2)^2
=(x^4−4x^3+24x^2−8x+4)(x^4−4x^3+12x^2−8x+4)
