a: \(\left|a-2b+3\right|^{2023}>=0\forall a,b\)

\(\left(b-1\right)^{2024}>=0\forall b\)

Do đó: \(\left|a-2b+3\right|^{2023}+\left(b-1\right)^{2024}>=0\forall a,b\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}a-2b+3=0\\b-1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=1\\a=2b-3=2\cdot1-3=-1\end{matrix}\right.\)

Thay a=-1 và b=1 vào P, ta được:

\(P=\left(-1\right)^{2023}\cdot1^{2024}+2024=2024-1=2023\)

a^3+b^3+c^3=3abc

=>(a+b)^3+c^3-3ab(a+b)-3bac=0

=>(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)=0

=>(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0

=>a^2+b^2+c^2-ab-bc-ac=0

=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0

=>(a-c)^2+(a-b)^2+(b-c)^2=0

=>a=b=c

=>A=(1+b/b)(1+b/b)(1+c/c)

=2*2*2=8

\(x=\sqrt{\dfrac{2\sqrt{3}+2-6\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2-4\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}\) ko tồn tại vì 2-4căn 3<0

Sửa đề: \(5x^2+5y^2+8xy-2x+2y+2=0\)

=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

=>\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

=>\(\left\{{}\begin{matrix}2x+2y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(M=\left(x-y\right)^{2023}-\left(x-2\right)^{2024}+\left(y+1\right)^{2023}\)

\(=\left(1+1\right)^{2023}-\left(1-2\right)^{2024}+\left(-1+1\right)^{2023}\)

\(=2^{2023}-1\)

\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)

Dấu bằng xảy ra \(\Leftrightarrow a=b=c\)

ta có : \(a^3+b^3+c^3=3abc\Rightarrow a=b=c\)

\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=2.2.2=8\)

o0o I am a studious person o0o: Theo em thì: \(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\orbr{\begin{cases}a=b=c\\a+b+c=0\end{cases}}\) chứ ạ?

Có Thực Dương đâu mak BĐT Cô-si???

Lời giải:

\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)

Lời giải:

\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)