Question

Các bạn giúp mình vs

Answer 1

Bài T5 nhé

Answer 2

ĐKXĐ : \(2-x^4\ge0\)

Áp dụng Cô si

\(\sqrt[4]{2-x^4}=\sqrt[4]{\left(2-x^4\right).1.1.1}\le\dfrac{2-x^4+1+1+1}{4}=\dfrac{5-x^4}{4}\)

\(VT\le\dfrac{x^2\left(5-x^4\right)}{4}-x^4+x^3-1=\dfrac{-\left(x-1\right)^2\left(\left(x^2+x\right)^2+6\left(x+\dfrac{2}{5}\right)^2\right)}{4}\le0=VP\)

Dấu "=" \(x=1\)

Vậy S = {1}

Answer 3

Khá căn bản ta dự đoán x=1.

`x^2\root{4}{2-x^4}-x^4+x^3-1=0`
`đk:-root{4}{2}<=x<root{4}{2}`
`<=>x^3-1+x^2\root{4}{2-x^4}-x^4=0`
`<=>(x-1)(x^2+x+1)+x^2(1-root{4}{2-x^4})=0`
`<=>(x-1)(x^2+x+1)+x^2((1-2+x^4)/(1+root{4}{2-x^4}))=0`
`<=>(x-1)(x^2+x+1)+(x^2(x-1)(x^3+x^2+x+1))/(1+root{4}{2-x^4})=0`
`<=>(x-1)(x^2+x+1+(x^2(x^3+x^2+x+1))/(1+root{4}{2-x^4}))=0`
`<=>x-1=0` do `x^2+x+1+(x^2(x^3+x^2+x+1))/(1+root{4}{2-x^4})>0`
`<=>x=1`