a) \(x^2+4y^2+z^2-2x-6z+8y+15\)

\(=\left(x^2-2x+1\right)+\left[\left(2y\right)^2+2\cdot2y\cdot2+2^2\right]+\left(z^2-6z+9\right)+1\)

\(=\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\ge1\forall x;y;z\)

\(\Rightarrowđpcm\)

b) tương tự

 \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left(a-b\right)-b^2\left(b-c\right)+c^2\left(a-b\right)\)

\(=\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

       \(ab^2-ac^2-b^3+bc^2\)

\(=b^2\left(a-b\right)-c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)

Vậy \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)

\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{a-c}{b+c}\)

Có a²(b-c) + b²(c-a) + c²(a-b)

= a²(b-c) - b²(a-c) + c²(a-b)

= a²(b-c) - b²(b-c+a-b) + c²(a-b)

= a²(b-c) - b²(b-c) - b²(a-b) + c²(a-b)

=[a²(b-c) - b²(b-c)] - [b²(a-b) - c²(a-b)]

=(b-c)(a²-b²) - (a-b)(b²-c²)

=(b-c)(a-b)(a+b) - (a-b)(b-c)(b+c)

=(b-c)(a-b)[(a+b)-(b+c)]

=(b-c)(a-b)(a-c)

 Có ab² - ac² - b³ + bc²

   = (ab²-ac²) - (b³-bc²)

   =a(b²-c²) - b(b²-c²)

=(b²-c²)(a-b)

=(b-c)(b+c)(a-b)

Có  a²(b-c) + b²(c-a) + c²(a-b)   /   ab² - ac² - b³ + bc²

  = (b-c)(a-b)(a-c) / (b-c)(b+c)(a-b)

= (a-c) / (b+c)

Quy đồng tí là ra.. :>> 

\(A=\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(A=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(A=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(A=\frac{4+4x^4+4-4^2x}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}\)

\(A=\frac{8}{1-x^8}+\frac{8}{1+x^8}=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}=\frac{16}{1-x^{16}}\)

Chúc bạn học tốt ~ 

mình đồng ý với bài trên

\(4y-2y^2\ge0\)

\(\Leftrightarrow2y^2-4y\le0\)

\(\Leftrightarrow2y\left(y-2\right)\le0\)

Vì y > y - 2

\(\Rightarrow\hept{\begin{cases}y\ge0\\y-2\le0\end{cases}\Rightarrow}\hept{\begin{cases}y\ge0\\y\le2\end{cases}\Rightarrow0\le y\le2}\)

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ko bt thì  ko nói nha mình đang cần gấp lém xin đừng trêu

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