\(A^2+B^2=\left(A+B\right)^2-2AB=5\)

\(A^3+B^3=\left(A+B\right)^3-3AB\left(A+B\right)=9\)

\(A^5+B^5=\left(A^2+B^2\right)\left(A^3+B^3\right)-\left(AB\right)^2\left(A+B\right)=5.9-2^2.3=...\)

B.

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

\(A^6+B^6=\left(A^2\right)^3+\left(B^2\right)^3=\left(A^2+B^2\right)^3-3\left(AB\right)^2\left(A^2+B^2\right)=2^3-3.1^2.2=...\)

Ta có: \(A^2+B^2=\left(A+B\right)^2-2AB=3^2-2.2=5\)

\(A^5+B^5=\left(A^3+B^3\right)\left(A^2+B^2\right)-A^2B^2\left(A+B\right)=\left(A+B\right)\left(A^2-AB+B^2\right)\left(A^2+B^2\right)-A^2B^2\left(A+B\right)=3\left(5-2\right).5-2^2.3=33\)

Xong đống này hơi lâu đấy, thui tui giải cho phần a nhé

Bài giải

Ta có : ab : (a+b) = 5 dư 10 

...(Làm các bước trên như câu đầu thầy giải)

Vì a.5 có giá trị lớn nhất 9.5 = 45

\(\Rightarrow\) ... (Tự làm)

Kết quả bằng:

ab = 65

a: ta có: \(M=\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\)

\(=\dfrac{a\left(\sqrt{ab}-a\right)+b\left(\sqrt{ab}+b\right)}{\left(\sqrt{ab}+b\right)\left(\sqrt{ab}-a\right)}-\dfrac{a+b}{\sqrt{ab}}\)

\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)\cdot\sqrt{a}\cdot\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{a+b}{\sqrt{ab}}\)

\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}-\dfrac{a^2-b^2}{\sqrt{ab}\left(a-b\right)}\)

\(=\dfrac{-\sqrt{ab}}{\sqrt{ab}\left(a-b\right)}\)

\(=-\dfrac{1}{a-b}\)

b: Thay \(a=\sqrt{5}+1\) và \(b=\sqrt{5}-1\) vào M, ta được:

\(M=\dfrac{-1}{\sqrt{5}+1-\sqrt{5}+1}=\dfrac{-1}{2}\)

a: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

=>(a+5)(b-6)=(a-5)(b+6)

=>ab-6a+5b-30=ab+6a-5b-30

=>-6a+5b=6a-5b

=>-12a=-10b

=>6a=5b

=>\(\dfrac{a}{b}=\dfrac{5}{6}\)

b: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)

\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)

Do đó: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)

#)Giải :

\(a^2+b^2\le1+ab\)

\(\Leftrightarrow a^2-ab+b^2\le1\)

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

\(\Leftrightarrow a^3+b^3\le a+b\)

\(\Leftrightarrow\left(a^3+b^3\right)\left(a^3+b^3\right)\le\left(a+b\right)\left(a^5+b^5\right)\left(a^3+b^3=a^5+b^5\right)\)

\(\Leftrightarrow a^6+2a^3b^3+b^6\le a^6+ab^5+a^5b+b^6\)

\(\Leftrightarrow a^5b+ab^5\ge2a^3b^3\)

\(\Leftrightarrow a^5b+ab^5-2a^3b^3\ge0\)

\(\Leftrightarrow ab\left(a^4-2a^2b^2+b^4\right)\ge0\)

\(\Leftrightarrow ab\left(a^2-b^2\right)^2\ge0\)( luôn đúng \(\forall a;b>0\))

Vậy \(a^2+b^2\le1+ab\left(đpcm\right)\)

P/s : Bài này mk tham khảo trên mạng ( tại thấy rảnh nên chép hộ ^^ )

a⁶ + b⁶ = (a² + b²)(a⁴ - a² b² + b⁴) = [(a + b)² - 2ab][(a² + b²)² - 3a² b²] = [(a + b)² - 2ab]{[(a + b)² - 2ab]²- 3a² b²​} 

Thế số vô là ra

\(a^5+b^2+ab+6\ge3a^2b+6\)

\(\Rightarrow P\le\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{\sqrt{a^2b+2}}+\dfrac{1}{\sqrt{b^2c+2}}+\dfrac{1}{\sqrt{c^2a+2}}\right)\le\sqrt{\dfrac{1}{a^2b+2}+\dfrac{1}{b^2c+2}+\dfrac{1}{c^2a+2}}=\sqrt{Q}\)

\(Q=\dfrac{c}{a+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}=\dfrac{1}{2}\left(1-\dfrac{a}{a+2c}+1-\dfrac{b}{b+2a}+1-\dfrac{c}{c+2b}\right)\)

\(Q=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{a^2}{a^2+2ac}+\dfrac{b^2}{b^2+2ab}+\dfrac{c^2}{c^2+2bc}\right)\)

\(Q\le\dfrac{3}{2}-\dfrac{1}{2}\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)

\(\Rightarrow P\le\sqrt{1}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)