Đặt \(\dfrac{a}{b^2}=\dfrac{b^2}{c^3}=\dfrac{c^3}{a^4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=k.b^2\\b^2=k.c^3\\c^3=k.a^4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=k.k.c^3=k^2c^3\\c^3=k.a^4\end{matrix}\right.\)

\(\Rightarrow a=k^2.k.a^4\)

\(\Rightarrow a=k^3a^4\)

\(\Rightarrow\left(ka\right)^3=1\)

\(\Rightarrow ka=1\)

\(\Rightarrow a=\dfrac{1}{k}\) (1)

Thế vào \(c^3=k.a^4\Rightarrow c^3=k.\dfrac{1}{k^4}=\dfrac{1}{k^3}\)

\(\Rightarrow c=\dfrac{1}{k}\) (2)

Thế vào \(b^2=kc^3\Rightarrow b^2=k.\dfrac{1}{k^3}=\dfrac{1}{k^2}\)

\(\Rightarrow b=\dfrac{1}{k}\) hoặc \(b=-\dfrac{1}{k}\) (3)

(1);(2);(3) \(\Rightarrow\left[{}\begin{matrix}a=b=c\\a=c=-b\end{matrix}\right.\)

TH1: \(a=b=c\)

\(\Rightarrow P=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)=2.2.2=8\)

Th2: \(a=c=-b\)

\(\Rightarrow P=\left(1+\dfrac{-b}{b}\right)\left(1+\dfrac{b}{-b}\right)\left(1+\dfrac{-b}{-b}\right)=0.0.2=0\)

Hình như thiếu đề bn ơi

1)(a+b-c)(a-b+c)-(a+b+c)= a+b-c-a+b-c-a-b-c= -a+b-3c

2)-a(b-c)-b(a+c)-c(b+a)-ab-ac= -ab+ac-ab-bc-bc-ac-ab-ac= -3ab-ac-2bc

D = (a + c) - (b + d) = a + c - b - d = (a - d) + (c - b) = C

=> D = C

Chúc bạn học tốt.

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

\(=a+c-b-d\)(1)

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

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

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

từ (1) và (2) => đpcm

D = (a + c) - (b + d)

D = a + c - b - d

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

mà C = (a - d) + (c - d)

Vậy D = C

a)    \(\left(A+B\right)^2=\left(A+B\right)\left(A+B\right)=A^2+AB+AB+B^2=A^2+2AB+B^2\)

b)  \(\left(A+B\right)^3=\left(A+B\right)^2\left(A+B\right)=\left(A^2+2AB+B^2\right)\left(A+B\right)\)( NHÂN  ra nốt hộ mk nha ) :D !

c)\(\left(A+B\right)\left(A-B\right)=A^2+AB-AB-B^2=A^2-B^2\)

ý d tương tự nha :D !

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

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

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

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

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

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

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

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

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

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

ta có (a+b+c)^2 = (a+b+c).(a+b+c) =a^2+ab+ac+ab+b^2+bc+ac+bc+c^2 = a^2+b^2+c^2+2ab+2ac+2bc
   và   (a-b-c)^2 = (a-b-c)(a-b-c)     = a^2-ab-ac-(ab-b^2-bc)-(ac-cb-c^2)   =a^2-ab-ac-ab+b^2+bc-ac+cb+c^2=a^2 -2ab-2ac+bc+b^2+c^2

Ta có\(a>b-c\)

Mà a;b;c là độ dài 3 cạnh của 1 tam giác nên a;b;c>0

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

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

\(\Leftrightarrow a^2-b^2-c^2+2bc>0\)

Vậy \(a^2-b^2-c^2+2bc>0\)

a^2 - b^2 + ab - 4   (1)

thay a = 2 , b= 4 vào (1) ta có 

2^2 - 4^2 + 2x 4 - 4 

= 4 - 16 + 4 - 4 

= -12 + 4 - 4 

= - 12 

đây là tính giá trị mà cậu ơi