$P\le \sqrt{a^2+2\sqrt{a}ab+2b^2}+\sqrt{b^2+2\sqrt{2}bc+2c^2}+\sqrt{c^2+2\sqrt{2}ca+2a^2}$

$P\le \sqrt{{\left(a+\sqrt{2}b\right)}^2}+\sqrt{{\left(b+\sqrt{2}c\right)}^2}+\sqrt{{\left(c+\sqrt{2}a\right)}^2}$

$P\le \left(1+\sqrt{2}\right)\left(a+b+c\right)=1+\sqrt{2}$

Dấu "=" xảy ra khi $\left(a;b;c\right)=\left(0;0;1\right)$ và các hoán vị

Giải nhanh dùm mem đi

phan h nhan vo la duoc

Bài 3: 

a: Ta có: \(\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\)

\(=y^2+8y-5y-40-y^2+y-4y+4\)

=-36

b: Ta có: \(y^4-\left(y^2-1\right)\left(y^2+1\right)\)

\(=y^4-y^4+1\)

=1

Bài 2: 

a: \(\left(2a-b\right)\left(4a+b\right)+2a\left(b-3a\right)\)

\(=8a^2+2ab-4ab-b^2+2ab-6a^2\)

\(=2a^2-b^2\)

b: \(\left(3a-2b\right)\left(2a-3b\right)-6a\left(a-b\right)\)

\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)

\(=6b^2-7ab\)

c: \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)

\(=10bx-5b^2-16bx+8b^2+2x^2-xb\)

\(=3b^2-7xb+2x^2\)

Xét hiệu \(2a^2+2b^2-\left(a^3+ab^2\right)=\left(2a^2-a^3\right)+\left(2b^2-ab^2\right)\)

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

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

Do \(a^2+b^2\ge0;\forall a;b\) nên:

\(2a^2+2b^2>a^3+ab^2\) khi \(\left\{{}\begin{matrix}a^2+b^2\ne0\\2-a>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2\ne0\\a< 2\end{matrix}\right.\)

\(2a^2+2b^2=a^3+ab^2\) khi \(\left[{}\begin{matrix}a^2+b^2=0\\2-a=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=b=0\\a=2\end{matrix}\right.\)

\(2a^2+2b^2< a^3+ab^2\) khi \(\left\{{}\begin{matrix}a^2+b^2\ne0\\a>2\end{matrix}\right.\) \(\Rightarrow a>2\)

\(2a^2+2b^2\ge a^3+ab^2\) khi \(2-a\ge0\Leftrightarrow a\le2\)

\(a,=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4-x-x^2+x^3-x^4+x^5+1+x-x^2+x^3-x^4\\ =2x-2x^2+2x^3-2x^4\)

\(5a^2-14ab-3b^2\\ =5a^2-15ab-ab-3b^2\\ =5a\left(a-3b\right)-b\left(a-3b\right)\\ =\left(5a-b\right)\left(a-3b\right)\)

\(5a^2-14ab-3b^2\)

\(=5a^2-15ab+ab-3b^2\)

\(=5a\left(a-3b\right)+b\left(a-3b\right)\)

\(=\left(a-3b\right)\left(5a+b\right)\)

a:b:c=3:4:5⇒a/3=b/4=c/5=k

⇒a=3k, b=4k, c=5k

2a²+2b²-3c²=-100

⇔2.(3k)²+2.(4k)²-3.(5k)²=-100

⇔2.9k²+2.16k²-3.25k²=-100

⇔18k²+32k²-75k²=-100

⇔ -25k²=-100

⇔k²=4

⇔k=+-2

k=-2⇔a/3=-2⇔a=-6

           b/4=-2⇔b=-8

           c/5=-2⇔c=-10

k=2⇔a/3=2⇔a=6

           b/4=2⇔b=8

           c/5=2⇔c=10

Ta có: 

a:b:c=3:4:5 => \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\)=> a=3k; b=4k; c=5k

=>\(2a^2=\left(6k\right)^2\text{​​};2b^2=\left(8k\right)^2;3c^2=\left(15k\right)^2\)

mà theo bài ra ta có: 2a²+2b²-3c²=-100 

=> \(6k^2+8k^2-15k^2=-100\)

=> \(\left(6+8-15\right)k^2=-100\)

=>\(\left(-1\right)k^2=-100\)

=>\(k^2=\dfrac{-100}{-1}=100\)

=> k= 10 hoặc k=-10

TH1: a=3.10=30 

         b=4.10=40

         c=5.10=50

TH2: a=3.(-10)=-30 

         b=4.(-10)=-40

         c=5.(-10)=-50