Ta có \(5a+6b+10c=7a-2a+7b-b+7c+3c\)

\(=7a+7b+7c-\left(2a+b-3c\right)=7\left(a+b+c\right)-\left(2a+b-3c\right)\)

Mà \(\left\{{}\begin{matrix}7\left(a+b+c\right)⋮7\\2a+b-3c⋮7\end{matrix}\right.\) \(\Rightarrow7\left(a+b+c\right)-\left(2a+b-3c\right)⋮7\)

Hay \(5a+6b+10c⋮7\) (đpcm)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-5\right)^2+12=0\)

Khi \(a=1;b=2;c=5\)

Good luck :3

\(5a=6b\Leftrightarrow\frac{a}{6}=\frac{b}{5}\)và a + b = 22

theo tính chất dãy tỉ số bằng nhau

\(\frac{a}{6}=\frac{b}{5}=\frac{a+b}{6+5}=\frac{22}{11}=2\)

\(\hept{\begin{cases}\frac{a}{6}=2\Leftrightarrow a=6.2=12\\\frac{b}{5}=2\Leftrightarrow b=5.2=10\end{cases}}\)

Ta có : \(5a=6b\Rightarrow a=\frac{6}{5}b\)

\(a+b=22\Rightarrow\frac{6}{5}b+b=22\Rightarrow\frac{11}{5}b=22\Rightarrow b=10\)

\(a+b=22\Rightarrow a+10=22\Rightarrow a=12\)

Vậy a = 12 ; b = 10 

\(\)

b=10

a=12

a² +b² +c² +42 = 2a+8b+610c

a² -2a+1 + b²-8b+16 +c² -10c + 24 =0

(a-1)² +(b-4)²+(c-5)²=0

suy ra a= 1 ;b= 4; c= 5

vậy a+b+c = 10

Đặt a/1=b/2=k

=>a=k; b=2k

\(Q=\dfrac{-5a+6b}{5a+6b}=\dfrac{-5k+12k}{5k+12k}=\dfrac{7}{17}\)