Ta có:

Theo tính chất dãy tỉ số bằng nhau ta có:

Ta có:

Mà  nên a, b và c cùng dấu.

Vậy ta tìm được các số a1 = 4; b1 = 6; c1 = 8 hoặc a2 = -4; b2 = -6 và c2 = -8

a) \(\dfrac{a}{5}=\dfrac{b}{4}\Rightarrow\dfrac{a^2}{25}=\dfrac{b^2}{16}\)

Áp dụng tính chất DTSBN :

\(\dfrac{a^2}{25}=\dfrac{b^2}{16}=\dfrac{a^2-b^2}{25-16}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{1}{9}\cdot25=\dfrac{25}{9}\\b^2=\dfrac{1}{9}\cdot16=\dfrac{16}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{5}{3};b=\dfrac{4}{3}\\a=\dfrac{-5}{3};b=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(\left(a;b\right)\in\left\{\left(\dfrac{5}{3};\dfrac{4}{3}\right);\left(-\dfrac{5}{3};-\dfrac{4}{3}\right)\right\}\)

b) \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)

Áp dụng tính chất DTSBN :

\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=4.4=16\\b^2=4.9=36\\c^2=4,16=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=4;=6;c=8\\a=-4;b=-6;c=-8\end{matrix}\right.\)

Vậy (a;b;c) \(\in\left\{\left(4;6;8\right);\left(-4;-6;-8\right)\right\}\)

Sửa \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

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

\(a^2-b^2+2c^2=108\\ \Rightarrow4k^2-9k^2+32k^2=108\\ \Rightarrow27k^2=108\Rightarrow k^2=4\\ \Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4;y=6;z=8\\x=-4;y=-6;z=-8\end{matrix}\right.\)

Ta có:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a^2}{2^2}=\dfrac{b^2}{3^2}=\dfrac{2c^2}{2.4^2}=\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}\)

Áp dụng tcdtsbn , ta có:

\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)

\(\Rightarrow\left\{{}\begin{matrix}a=8\\b=12\\c=16\end{matrix}\right.\)

Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)

$P\le \sqrt{a^2+2\sqrt{a}ab+2b^2}+\sqrt{b^2+2\sqrt{2}bc+2c^2}+\sqrt{c^2+2\sqrt{2}ca+2a^2}$

$P\le \sqrt{{\left(a+\sqrt{2}b\right)}^2}+\sqrt{{\left(b+\sqrt{2}c\right)}^2}+\sqrt{{\left(c+\sqrt{2}a\right)}^2}$

$P\le \left(1+\sqrt{2}\right)\left(a+b+c\right)=1+\sqrt{2}$

Dấu "=" xảy ra khi $\left(a;b;c\right)=\left(0;0;1\right)$ và các hoán vị

ké ý (b) ạ!!!

VP `=(a+b)(a^2-ab+b^2)`

`=a^3-a^2b+ab^2+a^2b-ab^2+b^3`

`=a^3+(a^2b-a^2b)+(ab^2-ab^2)+b^3`

`=a^3+b^3`

.

VP `=(a-b)(a^2+ab+b^2)`

`=a^3+a^2b+ab^2-a^2b-ab^2-b^3`

`=a^3+(a^2b-a^2b)+(ab^2-ab^2)-b^3`

`=a^3-b^3`

đúng rồi mà

Ta có: \(a^3+b^3\)

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

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

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

Ta có: \(a^3-b^3\)

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

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

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

a, \(\widehat{B}_1=\widehat{B_3}\) đối đỉnh

\(\widehat{A}_1=\widehat{B}_1\) theo bài đầu 

Do đó \(\widehat{A_1}=\widehat{B_3}\)

Mặt khác,ta có \(\widehat{A_1}+\widehat{A_4}=180^0\) hai góc kề bù

=> \(\widehat{A_4}=180^0-\widehat{A_1}\)                                  \((1)\)

Và \(\widehat{B_2}+\widehat{B_3}=180^0\) hai góc kề bù

=> \(\widehat{B_2}=180^0-\widehat{B_3}\)                                 \((2)\)

\(\widehat{A_1}=\widehat{B_3}\)                                                      \((3)\)

Từ 1,2,3 ta có : \(\widehat{A_4}=\widehat{B_2}\)

b, \(\widehat{A_2}=\widehat{A_4}\) đối đỉnh

\(\widehat{A_4}=\widehat{B_2}\) theo câu a

Do đó : \(\widehat{A_2}=\widehat{B_2};\widehat{A_1}=\widehat{A_3}\) đối đỉnh

\(\widehat{A_1}=\widehat{B_3}\) câu a

Do đó \(\widehat{A_3}=\widehat{B_3}\). Mặt khác \(\widehat{B_2}=\widehat{B_4}\) hai góc đối đỉnh

\(\widehat{A_4}=\widehat{B_2}\) câu a . Do đó \(\widehat{A_4}=\widehat{B_4}\)

c, \(\widehat{B_1}+\widehat{B_2}=180^0\) hai góc kề bù

\(\widehat{A_1}=\widehat{B_1}\) theo đầu bài

Do đó \(\widehat{A_1}+\widehat{B_2}=180^0\)

Mặt khác \(\widehat{B_2}+\widehat{B_3}=180^0\) kề bù

\(\widehat{A_4}=\widehat{B_2}\) theo câu a . Do đó \(\widehat{A_4}+\widehat{B_3}=180^0\)

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