a, \(\dfrac{a}{b}\)  = \(\dfrac{3}{5}\) ⇒ a = \(\dfrac{3}{5}\)b;  \(\dfrac{b}{c}\) = \(\dfrac{4}{5}\) ⇒ c = b : \(\dfrac{4}{5}\) = \(\dfrac{5}{4}\)b

⇒ a.c =  \(\dfrac{3}{5}\)b. \(\dfrac{5}{4}\)b = \(\dfrac{3}{4}\) ⇒ b².\(\dfrac{3}{4}\)  = \(\dfrac{3}{4}\) ⇒ b² = 1 ⇒ \(\left[{}\begin{matrix}b=1\\b=-1\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}a=\dfrac{3}{5}\\a=-\dfrac{3}{5}\end{matrix}\right.\); \(\left[{}\begin{matrix}c=\dfrac{5}{4}\\c=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy các cặp số a;b;c thỏa mãn đề bài là:

(a; b; c) = (-\(\dfrac{3}{5}\); -1; - \(\dfrac{5}{4}\)) ; (\(\dfrac{3}{5}\); 1; \(\dfrac{5}{4}\))

b, a.(a+b+c) = -12; b.(a+b+c) =18; c.(a+b+c) = 30

     ⇒a.(a+b+c) - b.(a+b+c) + c.(a+b+c) = -12 + 18 + 30

    ⇒ (a +b+c)(a-b+c) = 0

     ⇒ a - b + c = 0 ⇒ a + c  =b

Thay a + c  =  b vào biểu thức: b.(a+b+c) =18 ta có:

            b.(b + b) = 18

             2b.b = 18

              b² = 18: 2

              b² = 9 ⇒ \(\left[{}\begin{matrix}b=-3\\b=3\end{matrix}\right.\)

Thay a + c = b vào biểu thức c.(a + b + c) = 30 ta có:

        c.(b+b) = 30 ⇒ 2bc = 30 ⇒ bc = 30: 2 = 15 ⇒ c = \(\dfrac{15}{b}\)

Thay a + c = b vào biểu thức a.(a+b+c) = -12 ta có:

     a.(b + b) = -12 ⇒2ab = -12 ⇒ ab = -12 : 2 = - 6 ⇒ a = - \(\dfrac{6}{b}\)

Lập bảng ta có: 

Vậy các cặp số a; b; c thỏa mãn đề bài là:

(a; b; c) = (2; -3; -5); (-2; 3; 5)

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

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

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

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

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

mình nhầm.câu hỏi 2=-1

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

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

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

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

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

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

\(=\left(a-b\right)\left(b-c\right)\left[ac\left(a-c\right)+b\left(a-c\right)\left(a+c\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ac+ab+bc\right)\)

b, \(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=a^4\left(b-a+a-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=a^4\left(b-a\right)+a^4\left(a-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=\left(a-b\right)\left(c^4-a^4\right)+\left(a-c\right)\left(a^4-b^4\right)\)

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

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

\(=\left(a-b\right)\left(a-c\right)\left[a^3+ab^2+a^2b+b^3-c^3-a^2c-ac^2-a^3\right]\)

\(=\left(a-b\right)\left(a-c\right)\left[a^2\left(b-c\right)+a\left(b^2-c^2\right)+\left(b^3-c^3\right)\right]\)

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

\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left[a^2+b^2+c^2+ab+bc+ca\right]\)

Bổ đề : Chứng minh (a + b)² + (a - b)² = 2(a² + b²)

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

Áp dụng vào bài toán,ta có :

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

= 2[(b + c)² + a²] + 2[a² + (b - c)²] = 2[2a² + (b + c)² + (b - c)²] = 2[2a² + 2(b² + c²)] = 4(a² + b² + c²)

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

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

= 2[2(a² + b²) + 2(c² + d²)] = 4(a² + b² + c² + d²)

câu a) cái khúc =2[(b+c)^2 +a^2] +2[a^2 +(b-c)^2] là răng 

ghi rõ ra dùm

(a + b + c)² + (b + c - a)² = [(b + c) + a]² + [(b + c) - a]² = 2[(b + c)² + a²]

(c + a - b)² + (a + b - c)² = [a - (b - c)]² + [a + (b - c)]² = 2[a² + (b - c)²]

b) F=3.(b-c)-(a+c)+5.(a-b-c)

   F=3b-3c-a-c+5a-5b-5c

   F=(-a+5a)+(3b-5b)+(-3c-c-5c)

  F= 4a+(-2b)+(-9c)

  F=4a-2b-9c

a) E=2.(a+b)+3.(a-c)-4(b+c)

   E=2a+2b+3a-3c-4b-4c

   E=(2a+3a)+(2b-4b)+(-3c-4c)

   E=5a+(-2b)+(-7c)=5a-2b-7c

c)G=7.(a-b-c)+4.(b+c)-6(b-c-a)

G=7a-7b-7c+4b+4c-6b+6c+6a

G=(7a+6a)+(-7b+4b-6b)+(-7c+4c+6c)

G=13a+(-9b)+3c

G=13a-9b+3c

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

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

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

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

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

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

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)\)

Giúp tôi ! Làm ơn đi.....Help me@@

Đinh mệnh:))) Ko giúp rồi còn cho sai