Ta có : a+b/b+c = c+d/d+a 
=> (a+b)/(c+d)= (b+c)/(d+a) 
=> (a+b)/(c+d)+1=(b+c)/(d+a)+1 
hay: (a+b+c+d)/(c+d)=(b+c+d+a)/(d+a) 
- Nếu a+b+c+d khác 0 thì : c+d=d+a => c=a (1)
- Nếu a+b+c+d = 0 (2) 

Từ (1) và (2) 

\(\RightarrowĐPCM\)

Ta có : \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

\(\Rightarrow\)\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)

\(\Rightarrow\)\(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)

Hoặc \(\frac{a+b+c+d}{c+d}=\frac{b+c+d+a}{d+a}\)

Nếu a + b + c + d khác 0 thì c + d = d + a => c = a ( hoặc a = c )

Nếu a + b + c + d = 0 ( đpcm )

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b) Đặt \(\hept{\begin{cases}\frac{a}{b}=k\Rightarrow a=kb\\\frac{c}{d}=k\Rightarrow c=kd\end{cases}}\)

VT : \(\frac{5a+3b}{5a-3b}\Rightarrow\frac{5kb+3b}{5ka-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\) (1)

VP : \(\frac{5c+3d}{5c-3d}=\frac{5kd+3d}{5kd-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\) (2)

Từ (1) và (2) => đpcm

Ta có 5a = 2b = 3c và a+b-c = -33

=> \(\frac{a}{\frac{1}{5}}=\frac{b}{\frac{1}{2}}=\frac{c}{\frac{1}{.3}}\)và  \(a+b-c=-33\)

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\frac{a}{\frac{1}{5}}=\frac{b}{\frac{1}{2}}=\frac{c}{\frac{1}{3}}=\frac{a+b-c}{\frac{1}{5}+\frac{1}{2}-\frac{1}{3}}=-\frac{33}{\frac{11}{30}}=-33:\frac{11}{30}=-90\)

* \(\frac{a}{\frac{1}{5}}=-90\Rightarrow a=-90\cdot\frac{1}{5}=-18\)

*\(\frac{b}{\frac{1}{2}}=-90\Rightarrow b=-90\cdot\frac{1}{2}=-45\)

* \(\frac{c}{\frac{1}{3}}=-90\Rightarrow c=-90\cdot\frac{1}{3}=-30\)

Vậy a=-18 ; b = -45 ;  c= -30 

Có cách dễ hiểu hơn đó là

Ta có: 5a =2b = 3c => 5a/30 = 2b/30 = 3c/30 ( Tìm BCNN của 5,2,3 lấy làm mẫu số ) => a/6 = b/15 = c/10 ( Rút gọn tử vs mẫu )

=> a + b - c/ 6 + 15 -10 = -33/11 = -3

Do đó: a/6 = -3 => a = -18

           b/15 = -3 => b = -45

           c/10 = -3 => c = -30

Ta có : \(\left(-32\right)^9=-32^9\)

\(\left(-18\right)^{13}=-18^{13}\)

Ta có : \(-32^9=-\left(2^5\right)^9=-2^{45}>-2^{52}=-\left(2^4\right)^{13}=-16^{13}>-18^{13}\)

Mà \(32^{27}< 18^{39}\)

Nên \(\left(-32\right)^9>\left(-18\right)^{19}\)  ( đpcm )

\(\frac{2^5.3^6+6^5.11}{12^5-6^5.2^2}\)

\(=\frac{2^5.3^6+2^5.3^5.11}{2^2.3-2^5.3^5.2^2}\)

\(=\frac{2^5.3^5.\left(3+11\right)}{2^2.3\left(1-2^5.3^4\right)}\)

\(=\frac{2^3.3^4.14}{1-32.81}\)

\(=\frac{2^4.3^4.7}{-2591}\)

\(\frac{2^5\cdot3^6+6^5\cdot11}{12^5-6^5.2^2}\)

\(=\frac{6^5\cdot3+6^5\cdot11}{6^5\cdot2^5-6^5\cdot2^2}\)

\(=\frac{6^5\cdot\left(3+11\right)}{6^5\cdot\left(2^5-2^2\right)}\)

\(=\frac{14}{32-4}\)

\(=\frac{14}{28}=\frac{1}{2}\)

Ta có : a = 2b ; b = -3c

=> a = -6c

\(M=\frac{7a-3b+2c}{2a-5b+c}\)

Thay a = -6c ; b = -3c

\(M=\frac{7\cdot\left(-6c\right)-3\cdot\left(-3c\right)+2c}{2\cdot\left(-6c\right)-5\cdot\left(-3c\right)+c}\)

\(M=\frac{-42c+9c+2c}{-12c+15c+c}\)

\(M=\frac{\left(-42+9+2\right)c}{\left(-12+15+1\right)c}\)

\(M=-\frac{31}{4}\)

-31/4 nha.

\(\frac{2^5\cdot6^{14}\cdot10^7}{2\cdot8^8\cdot3^5\cdot15^7}\)

\(=\frac{2^5\cdot2^{14}\cdot3^{14}\cdot2^7\cdot5^7}{2\cdot2^8\cdot2^8\cdot2^8\cdot3^5\cdot3^7\cdot5^7}\)

\(=\frac{2^{26}\cdot3^{14}\cdot5^7}{2^{25}\cdot3^{12}\cdot5^7}\)

\(=2\cdot3^2\)

= 2 x 9

= 18

\(\frac{2^5\cdot2^{14}\cdot3^{14}\cdot2^7\cdot5^7}{2\cdot\left(2^3\right)^8\cdot3^5\cdot3^7\cdot5^7}=\frac{2^{26}\cdot3^{14}\cdot5^7}{2^{25}\cdot3^{12}\cdot5^7}=2\cdot3^2=18\)