Ta có: 3x³ + xy=3

=>x( 3x² + y) =3

Xét các trường hợp:

* Nếu x= 1 thì y=0

* Nếu x= -1 thì y= -6

* Nếu x= 3 thì y= -26

* Nếu x=-3 thì y= -28

Vay...

 Nhows k cho mình nhá

\(PT< =>8x\left(8x-1\right)^2\left(8x-2\right)=72\)

\(< =>8x\left(8x-2\right)\left(64x^2-16x+1\right)=72\)

\(< =>\left(64x^2-16x\right)\left(64x^2-16x+1\right)=72\)

Đặt \(64x^2-16x+\frac{1}{2}=t\)

\(PT< =>\left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right)=72\)

\(< =>t^2=\frac{289}{4}\)

\(< =>\orbr{\begin{cases}t=\frac{17}{2}\\t=\frac{-17}{2}\end{cases}}\)

\(TH1:t=\frac{17}{2}\)

\(PT< =>64x^2-16x+\frac{1}{2}=\frac{17}{2}\)

\(< =>\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{4}\end{cases}}\)

\(TH2:t=\frac{-17}{2}\)

\(PT< =>64x^2-16x+\frac{1}{2}=\frac{-17}{2}\)

\(< =>64x^2-16x+9=0\)

\(< =>\left(8x-1\right)^2+8=0\left(VL\right)\)

Vậy S={1/2;-1/4}

\(\left(x-6\right)^4+\left(x-8\right)^4=16.\)

\(\left(x-6\right)^4+\left(x-8\right)^4=2^4.\)

\(\left(x-6\right)^4=2^4-\left(x-8\right)^4.\)

\(\left(x-6\right)^4=\left(2+x-8\right)^2\left(2-x-8\right)^2\)(hđt)

\(\left(x-6\right)^4=\left(10+x\right)^2\left(-6-x\right)^2\)

Từ đây xét nghiệm PT => kết quả cuối

= 2049

= 2055

Hy vs nhé!!

29+1+2019=2049

24+12+2018=2054