Giải:

Ta có: a + b + c = 0 nên suy ra: b = – (a + c) thay vào biểu thức:

ab + 2bc + 3ca = -a.(a + c) – 2c.(a + c) + 3ac = -a² – ac – 2ac – 2c² + 3ac = – (a² + 2c²) ≤ 0 (đpcm). 

Trả lời

Theo đề ra ta có:

a+b+c=0

\(\Rightarrow\)ab+2ab+3ac=-a(a+c)-2c(a+c)+3ac

          =\(-a^2-ac-2ac-2ac^2+3ac\)

           \(=-\left(a^2+2c^2\right)\le0\)

Vậy nếu a+b+c=0 thì \(ab+2bc+3ac\le0\left(đpcm\right)\)

Ta có : a + b + c = 0

\( \implies\) b + c = - a ; a + b = - c 

Ta có : ab + 2bc + 3ca 

= ab + 2bc + ca + 2ca 

= ( ab + ca ) + ( 2bc + 2ca )

= a ( b + c ) + 2c ( a + b )

= a ( - a ) + 2c ( - c ) 

= - a² - 2c² 

= - ( a² + 2c² ) ( * )

Mà : a² \(\geq\)  0 ; 2c² \(\geq\)  0 

\( \implies\)  a² + 2c² \(\geq\)  0 ( ** )

Từ ( * ) ; ( ** ) 

\( \implies\)  - ( a² + 2c² )  \(\leq\)  0 

\( \implies\) ab + 2bc + 3ca  \(\leq\)  0 

\(ab+2bc+3ac\\ =\left(ab+ac\right)+\left(2bc+2ac\right)\\ =a\left(b+c\right)+2c\left(a+b\right)\\ =a.\left(-a\right)+2c\left(-c\right)\\ =-a^2-2c^2\\ =-\left(a^2+2c^2\right)\le0\)

Ta có: a + b + c = 0 nên suy ra: b = – (a + c) thay vào biểu thức:

ab + 2bc + 3ca = -a.(a + c) – 2c.(a + c) + 3ac = -a² – ac – 2ac – 2c² + 3ac = – (a² + 2c²) ≤ 0 (đpcm).

hok tôts

vì a+b+c=0 nên a,b,c lớn nhất chỉ có thể bằng ko,nên ab+2bc+3ca chỉ có thể < hoặc bằng 0

Ta có: 

\(\left(3a-2b+c\right)^2=9a^2+4b^2+c^2+2\left(3ac-6ab-2bc\right)\)

\(\Rightarrow b^2=9a^2+4b^2+c^2\)

(vì \(3a-3b+c=0\Leftrightarrow3a-2b+c=-b\), \(6ab+2bc-3ac=0\))

\(\Leftrightarrow9a^2+3b^2+c^2=0\)

\(\Leftrightarrow a=b=c=0\). 

Khi đó: \(P=\left(-1\right)^{2019}+\left(-1\right)^{2020}+\left(-1\right)^{2021}=-1\)

Ta có: 

(3a−2b+c)2=9a2+4b2+c2+2(3ac−6ab−2bc)

⇒b2=9a2+4b2+c2

(vì 3a−3b+c=0⇔3a−2b+c=−b, 6ab+2bc−3ac=0)

⇔9a2+3b2+c2=0

⇔a=b=c=0. 

Khi đó: P=(−1)2019+(−1)2020+(−1)2021=−1

Lời giải:

$N=a(b+3c)+5bc=(1-b-c)(b+3c)+5bc$

$=b+3c-b^2-3c^2+bc$

$-N=b^2+3c^2-bc-b-3c$

$-2N=2b^2+6c^2-2bc-2b-6c$

$\geq b^2+5c^2-2b-6c$

$=(b+c-1)^2+(2c-1)^2-2bc-2$

$\geq -2(bc+1)$

Mà $bc\leq \frac{(b+c)^2}{4}\leq \frac{1}{4}$

$\Rightarrow bc+1\leq \frac{5}{4}$

$\Rightarrow -2(bc+1)\geq \frac{-10}{4}$
$\Rightarrow -2N\geq \frac{-10}{4}$

$\Rightarrow N\leq \frac{5}{4}$

Vậy $N_{\max}=\frac{5}{4}$ khi $(a,b,c)=(0,\frac{1}{2}, \frac{1}{2})$

`(2bc-2016)/(3c-2bc+2016)`

`=(-(3c-2bc+2016)+3c)/(3c-2bc+2016)`

`=-1+(3c)/(3c-2bc+2016)`

`(2b)/(3-2b+ab)

`=(2bc)/(3c-2bc+abc)`

`=(2bc)/(3c-2bc+2016)`

`(4032-3ac)/(3ac-4032+2016a)`

`=(-(3ac-4032+2016a)+2016a)/(3ac-4032+2016a)`

`=-1+(2016a)/(3ac-2abc+2016a)`

`=-1+(2016)/(3c-2bc+2016)`

`=>M=-1+(3c)/(3c-2bc+2016)-(2bc)/(3c-2bc+2016)-1+(2016)/(3c-2bc+2016)

`=>M=-2+(3c-2bc+2016)/(3c-2bc+2016)`

`=>M=-2+1`

`=>M=-1`

`(2bc-2016)/(3c-2bc+2016)`

`=(-(3c-2bc+2016)+3c)/(3c-2bc+2016)`

`=-1+(3c)/(3c-2bc+2016)`

`(2b)/(3-2b+ab)`

`=(2bc)/(3c-2bc+abc)`

`=(2bc)/(3c-2bc+2016)`

`(4032-3ac)/(3ac-4032+2016a)`

`=(-(3ac-4032+2016a)+2016a)/(3ac-4032+2016a)`

`=-1+(2016a)/(3ac-2abc+2016a)`

`=-1+(2016)/(3c-2bc+2016)`

`=>M=-1+(3c)/(3c-2bc+2016)-(2bc)/(3c-2bc+2016)-1+(2016)/(3c-2bc+2016)`

`=>M=-2+(3c-2bc+2016)/(3c-2bc+2016)`

`=>M=-2+1`

`=>M=-1`

Nãy thiếu latex ạ sorry~~