\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

\(a^3+b^3+c^3-3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc\)

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

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

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

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\Rightarrow\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)

\(\left(a-b\right)^2>=0\Rightarrow a^2-2ab+b^2>=0\Rightarrow a^2+b^2>=2ab\)

tương tự \(a^2+c^2>=2ac;b^2+c^2>=2bc\)

\(\Rightarrow a^2+b^2+a^2+c^2+b^2+c^2>=2ab+2ac+2bc\Rightarrow2\left(a^2+b^2+c^2\right)>=2\left(ab+ac+bc\right)\)

\(\Rightarrow a^2+b^2+c^2.=ab+ac+bc\)dấu = xảy ra khi a=b=c

mà nếu \(a^2+b^2+c^2-ab-ac-bc=0\Rightarrow a^2+b^2+c^2=ab+ac+bc\Rightarrow a=b=c\)

th1:a+b+c=0

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

\(M=\frac{ab^2}{a^2+b^2-c^2}+\frac{bc^2}{b^2+c^2-a^2}+\frac{ca^2}{c^2+a^2-b^2}=\frac{ab^2}{a^2+b^2-\left(-c\right)^2}+\frac{bc^2}{b^2+c^2-\left(-a\right)^2}+\frac{ca^2}{c^2+a^2-\left(-b\right)^2}\)

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

\(=\frac{ab^2}{a^2+b^2-a^2-2ab-b^2}+\frac{bc^2}{b^2+c^2-b^2-2bc-c^2}+\frac{ca^2}{c^2+a^2-c^2-2ac-a^2}\)

\(=\frac{ab^2}{-2ab}+\frac{bc^2}{-2bc}+\frac{ca^2}{-2ac}=\frac{b}{-2}+\frac{c}{-2}+\frac{a}{-2}=\frac{a+b+c}{-2}=\frac{0}{-2}=0\)

th2:a=b=c tự lm nhá

K là điểm gì đấy bạn ơi

Kẻ \(BH\perp CD\left(H\in CD\right)\)

Ta có: ABHD là hình chữ nhật => BH=AD=12 và DH=AB=11

Áp dụng định lí Pytago vào tam giác vuông BHC tại H có: \(HC=\sqrt{BC^2-BH^2}=\sqrt{13^2-12^2}=5\)

=> CD=DH+HC=11+5=16

Áp dụng định lí Pytago vào tam giác vuông ADC tại D có: \(AC=\sqrt{AD^2+CD^2}=\sqrt{12^2+16^2}=20\)

Vậy AC=20cm

mk chỉ làm câu a thôi nha câu b mk ko hiểu đề

a) ( x+2) ^2  -  9  =0 

<=> (x+2)^2 = 9 

<=> (x+2)^2  =  3^2 =( -3)^2

TH1 (x+2)^2  =  3^2                                                 TH2 (x+2)^2  =  (-3)^2 

      x+2 = 3 => x =1                                                           x+2 = -3 => x= -5

Vậy x=1 hoặc x= -5    CHÚC BẠN HOK TỐT

mik ko biết

Ta có: a³+b³+c³=3abc

<=> (a+b+c)(a²+b²+c²-ab-bc-ca)=0

<=> (a+b+c)(2a²+2b²+2c²-2ab-2bc-2ca)=0

<=> (a+b+c)[(a-b)²+(b-c)²+(c-a)² ] = 0

<=> \(\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)

Vì a,b,c phân biệt nên a+b+c=0 => \(\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(c+a\right)\\c=-\left(a+b\right)\end{cases}}\)(*)

Lại có: \(M=\frac{ab^2}{a^2+b^2-c^2}+\frac{bc^2}{b^2+c^2-a^2}+\frac{ca^2}{c^2+a^2-b^2}\)

Thay (*) vào M ta được:

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

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

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

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

Mà a+b+c=0

=> M=0

Vậy M=0

\(H=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)-2^{16}=\left(2-1\right).\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)-2^{16}\)

    \(=\left(2^2-1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)-2^{16}=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)-2^{16}\)

     \(=\left(2^8-1\right).\left(2^8+1\right)-2^{16}=\left(2^{16}-1\right)-2^{16}=-1\)

a. Ta có: \(\Delta CED\infty\Delta CAB\left(g-g\right)\Rightarrow\frac{CE}{CD}=\frac{CA}{CB}\Leftrightarrow\frac{CE}{CD}=\frac{CA}{2CE}\Leftrightarrow2CE^2=CA.CD\)

b. Áp dụng định lí Pytago vào tam giác vuông ABC tại A ta có: \(CA=\sqrt{BC^2-AB^2}=\sqrt{40^2-24^2}=32\)

Và \(BE=CE=\frac{CB}{2}=\frac{40}{2}=20\)

Từ phần a ta có: \(\frac{ED}{CE}=\frac{AB}{CA}\Leftrightarrow DE=\frac{CE.AB}{CA}=\frac{20.24}{32}=15\left(cm\right)\)

Theo phần a lại có: \(2CE^2=CA.CD\Leftrightarrow CD=\frac{2CE^2}{CA}=\frac{2.20^2}{32}=25\left(cm\right)\)

\(\Rightarrow DA=AC-CD=32-25=7\left(cm\right)\)