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

\(=ab-ac-ba-bc+ca-cb=-2bc\)

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

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

\(=ab-ac-ba-bc+ca-cb\)

\(=\left(ab-ba\right)+\left(-ac+ca\right)+\left(-bc-cb\right)\)

\(=0+0-2bc\)

\(=-2bc\)

Vậy \(a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)=-2bc\).

Học tốt

a/

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

\(=ab-ac-ab-bc+ac-bc=-2bc\)

b/

\(a\left(1-b\right)+a\left(a^2-1\right)=\)

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

c/

\(a\left(b-x\right)+x\left(a+b\right)=ab-ax+ax+bx=\)

\(=ab+bx=b\left(a+x\right)\)

a) VT: a(b - c) - b(a + c) + c(a - b)

= ab - ac - ab - bc + ac - bc

= -2bc

Vậy a(b - c) - b(a + c) + c(a - b) = -2bc.

b) VT: a(1 - b) + a(a² - 1)

= a - ab + a³ - a

= a³ - ab

= a(a² - b)

Vậy a(1 - b) + a(a² - 1) = a(a² - b).

3 số thực dương nhé.

Áp dụng bất đẳng thức Cauchy Schwarz dạng Engel có :

\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{\left(a^2+2bc\right)+\left(b^2+2ca\right)+\left(c^2+2ab\right)}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1^2}=9\)

Dấu bằng xảy ra \(\Leftrightarrow\frac{1}{a^2+2bc}=\frac{1}{b^2+2ca}=\frac{1}{c^2+2ab}\)và \(a+b+c=1\)

\(\Leftrightarrow a^2+2bc=b^2+2ca=c^2+2ab\)

Mong có ai giúp mình từ đẳng thức trên giải ra a=b=c.

a=b=c ket hop voi a+b+c=<1 =>a=b=c=1/3 nhe

\(a^2+2bc=b^2+2ca=c^2+2ab\)

Ta có: \(\hept{\begin{cases}a^2+2bc=b^2+2ca\\b^2+2ca=c^2+2ab\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)\left(a+b\right)-2c\left(a-b\right)=0\\\left(b-c\right)\left(b+c\right)-2a\left(b-c\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)\left(a+b-2c\right)=0\\\left(b-c\right)\left(b+c-2a\right)=0\end{cases}}\)

Tới đây thì suy được ra là \(a=b=c\) rồi nhé Trần Thùy Dung - Trang của Trần Thùy Dung - Học toán với OnlineMath

\(2bc+b^2+c^2-a^2\)

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

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

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

\(=2p\left(2p-2a\right)=4p\left(p-a\right)\)

biến đổi vế phải ta được:

4p(p -a ) = 4p\(^2\)-4pa

=(2p)\(^2\)-2p.2a

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

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

=\(2bc+b^2+c^2-a^2\)=vế trái (đpcm)

a+b+c = 2p => 4p = 2(a+b+c); p=(a+b+c)/2

VP = 4p(p-a) = 2(a+b+c)(\(\frac{a+b+c}{2}-a\))

= \(2\left(a+b+c\right)\left(\frac{a+b+c-2a}{2}\right)\)

=\(2\left(a+b+c\right)\cdot\frac{b+c-a}{2}=\left(a+b+c\right)\left(b+c-a\right)\)

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

\(=2bc+b^2+c^2-a^2\) = VT (đpcm)

\(2bc+b^2+c^2-a^2.\)'

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

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

Theo đề ta có \(a+b+c=2p\)

\(\Rightarrow b+c=2p-a\)

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

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

\(=\left(2p-a+a\right)\left(2p-a-a\right)\)

\(=2p\left(2p-2a\right)\)

\(=2p\cdot2\left(p-a\right)=4p\left(p-a\right)\)

\(\Rightarrow2bc+b^2+c^2-a^2=4p\left(p-a\right)\)(đpcm)

2bc + b² + c² - a²

= ( b² + 2ab + c² ) - a²

= ( b + c )² - a²

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

Từ gt a + b + c = 2p => b + c = 2p - a

Thế vào (*) ta được

( 2p - a - a )( 2p - a + a )

= ( 2p - 2a )2p

= 4p² - 4pa

= 4p( p - a ) ( đpcm )