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A=(a+b)(b+c)(c+a)+abcA=(a+b)(b+c)(c+a)+abc

=a2b+ab2+a2c+ac2+b2c+bc2+2abc+abc=a2b+ab2+a2c+ac2+b2c+bc2+2abc+abc

=ab(a+b+c)+bc(a+b+c)+ca(a+b+c)=ab(a+b+c)+bc(a+b+c)+ca(a+b+c)

=(a+b+c)(ab+bc+ca)=(a+b+c)(ab+bc+ca)

Vậy....

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

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

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

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

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

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

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

giúp mình làm  bài này đi rrooiif mình giúp cho

cho tam giac abc . co canh bc=12cm, duong cao ah=8cm

a> tinh s tam giac abc

b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách

c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame

a) a(b-c)+c(a-b)=ab-ac+ca-cb=ab-cb=b(a-c)

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

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

d) a(b-c)-a(b+d)=ab-ac-ab-ad=-ac-ad=-a(a+d)

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

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

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

d) a(b - c) - a(b + d) = ab - ac - ab - ad = -ac - ad = -a(c + d)

a) a(b-c)+c (a-b)=ab-ac+ca-cb=cb=b(a-c)

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

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

d) a(b-c)-a(b+d)=ab-ac-ab-ad=ac-ab-ad=ac-ad=a(a+d)

nhanh lên với ak

Ta có :

a^3+b^3+c^3-3abc

=(a+b)^3+c^3-3ab(a+b) - 3abc

=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)

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

=> 2(a^3+b^3+c^3-3abc)= (a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)

=(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]

a)-b+c

d)-2a-2c

e)2b-2c

b)-2a+b

c)-a+c

f)a

-a-b+a+c=-b+c

-a+b-c-a-b-c=-2a-2c

a+b-a-b+a-c-a-c=-2c

-a-c+a-b-c=-2c+b

b-b-a+c=-a+c

a+b-c+a-b+c-b+c-a-a+b+c=2c

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

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

\(\Rightarrow VT=VP\left(đpcm\right)\)

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

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

\(\Rightarrow VT=VP\left(đpcm\right)\)

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

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

\(\Rightarrowđpcm\)

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

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

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

\(\Rightarrow VT=VP\left(đpcm\right)\)

e, \(-\left(-a+b+c\right)+\left(b+c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)

\(a-b-c+b+c-1=b-c+6-7+a-b+c\)

\(a-1=-1+a\Rightarrow a-1=a+\left(-1\right)\Rightarrow a-1=a-1\)

\(\Rightarrow VT=VP\left(đpcm\right)\)

\( a)\left( {a + b - c} \right) + \left( {b + c - a} \right) + \left( {a + c - b} \right)\\ = a + b - c + b + c - a + a + c - b\\ = a + b + c\\ b)\left( {a - b} \right) + \left( {b - c + a} \right) + \left( {c - b} \right)\\ = a - b + b - c + a + c - b\\ = 2a + b\\ c)\left( {2a - b + c} \right) + \left( {b - c + a} \right) + \left( {c - 2a + b} \right)\\ = 2a - b + c + b - c + a + c - 2a + b\\ = a + b + c\\ d)\left( {a - c + b} \right) + \left( {b - c - a} \right) - a - b - c\\ = a - c + b + b - c - a - a - b - c\\ = - a - b - 3c \)

a) (a + b - c) + (b + c - a) + (a +​c - b)

= a + b - c + b + c - a + a + c - b

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

= a + b + c

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

= a - b + b - c + a + c - b

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

= 2a - b

c) (2a - b + c) + (b - c + a) + (c - 2a + b)

= 2a - b + c + b - c + a + c - 2a + b

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

= b + c

d) (a - c + b) + (b - c - a) - a - b - c

= a - c + b + b - c - a - a - b - c

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

= b - 2a - 3c

Chúc bạn học tốt@@

a/ \((a + b - c) + (b + c - a) + (a +​c - b)\)

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

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

\(=a+b+c\)

b/ \(\text{ (a - b) + (b - c + a) + (c - b)}\)

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

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

\(=2a-b\)

c/ \(\text{(2a - b + c) + (b - c + a) + (c - 2a + b)}\)

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

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

\(=a+b+c\)

d) \(\text{(a - c + b) + (b - c - a) - a - b - c}\)

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

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

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

Đáp án: C

A ∩  B = {b; d}; A ∩  C = {a; b}; B ∩ C = {b; e}

A \ B = {a; c}; A \ C = {c; d}; B \ C = {d}

A ∪  B = {a; b; c; d; e}; A ∪  C = {a; b; c; d; e}

A ∩  (B \ C) = {d}. (A ∩  B) \ (A ∩  C) =  {d}.

A \ (B ∩ C) = {a; c; d}. (A \ B) ∪  (A \ C) = {a; c; d}.

(A \ B) ∩  (A \ C) = {c}.

a. A ∩  (B \ C) = (A ∩  B) \ (A ∩  C) ={d} ⇒ a đúng.

b. A \ (B ∩ C)= {a; c; d}  (A \ B) ∩  (A \ C)={c} ⇒ b sai.

c. A ∩  (B \ C) ={d}  (A \ B) ∩  (A \ C)={c}  ⇒ c sai

d. A \ (B ∩C) = (A \ B) ∪ (A \ C)= {a; c; d} ⇒ d đúng.