\(\dfrac{a}{5}+\dfrac{1}{10}=\dfrac{1}{b}\\ =>\dfrac{2a+1}{10}=\dfrac{1}{b}\\ =>b\left(2a+1\right)=10\)

\(Vì\ :\ a,b\ là\ các\ số\ nguyên\ dương=>2a+1,b\in N*;2a+1>=3\)

\(Ta\ có\ bảng\ giá\ trị :\)

\(Vậy (a;b)=(2;2)\)
 

\(8-27x^3=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

\(Để\ phương\ trình\ có\ nghiệm\ x=3\ thì :\)

\(\left(2m-3\right).3-2=\left(m+1\right)-3\\ < =>6m-9-2=m-2\\ < =>6m-11=m-2\\ < =>5m=9\\ < =>m=\dfrac{9}{5}\)

\(b=a-1=>a-b=1\)

\(VT=\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)....\left(a^{64}+b^{64}\right)\\ =1.\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)....\left(a^{64}+b^{64}\right)\\ =\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)....\left(a^{64}+b^{64}\right)\\ =\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)....\left(a^{64}+b^{64}\right)\\ =\left(a^4-b^4\right)\left(a^4+b^4\right)....\left(a^{64}+b^{64}\right)\\ =\left(a^8-b^8\right)....\left(a^{64}+b^{64}\right)\\ ...\\ =\left(a^{64}-b^{64}\right)\left(a^{64}+b^{64}\right)\\ =a^{128}-b^{128}=VP\\ =>DPCM\)

\(A=\left(\dfrac{2x}{x^2-x-6}+\dfrac{x}{x-3}\right):\dfrac{x}{x-3}\left(ĐK:x\ne\left\{3;-2;0\right\}\right)\\ =\left(\dfrac{2x}{\left(x-3\right)\left(x+2\right)}+\dfrac{x}{x-3}\right).\dfrac{x-3}{x}\\ =\dfrac{2x+x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}.\dfrac{x-3}{x}\\ =\dfrac{2x+x^2+2x}{x\left(x+2\right)}\\ =\dfrac{x\left(x+4\right)}{x\left(x+2\right)}=\dfrac{x+4}{x+2}\)

\(B=\dfrac{x-1}{x-3}+\dfrac{x}{1-x}-\dfrac{2}{x^2-4x+3}\left(ĐK:x\ne\left\{3;1\right\}\right)\\ =\dfrac{x-1}{x-3}-\dfrac{x}{x-1}-\dfrac{2}{\left(x-1\right)\left(x-3\right)}\\ =\dfrac{\left(x-1\right)^2-x\left(x-3\right)-2}{\left(x-1\right)\left(x-3\right)}\\ =\dfrac{x^2-2x+1-x^2+3x-2}{\left(x-1\right)\left(x-3\right)}\\ =\dfrac{x-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{1}{x-3}\)

\(A=\sqrt{\left(1-2\sqrt{5}\right)^2}-\sqrt{20}\\ =\left|1-2\sqrt{5}\right|-2\sqrt{5}\left(1\right)\)

\(Vì : 1<2\sqrt 5=>1-2\sqrt 5<0=>|1-2\sqrt5|=-(1-2\sqrt5)=2\sqrt5 -1\)

\(\left(1\right)=>2\sqrt{5}-1-2\sqrt{5}=-1\)

\(Vậy A=-1\)

\(C=\dfrac{1}{3-\sqrt{x}}+\dfrac{\sqrt{x}}{3+\sqrt{x}}-\dfrac{x+9}{x-9}\left(x>=0;x\ne9\right)\\ =-\dfrac{1}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{-\left(\sqrt{x}+3\right)+\sqrt{x}\left(\sqrt{x}-3\right)-\left(x+9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{-4\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{-4\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=-\dfrac{4}{\sqrt{x}-3}=\dfrac{4}{3-\sqrt{x}}\)

\(GTNN\ là:\ -1\ khi\ x=-4+\sqrt 5\ hoặc\ x=-4-\sqrt5\)

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\\ =\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\\ =\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\left(1\right)\)

\(Đặt : x^2+8x+11=t\)

\(\left(1\right)=>\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1>=-1\)

\(Dấu ''=''\ xảy\ ra\ khi :\)

\(t^2=0< =>t=0\\ =>x^2+8x+11=0\\ < =>\left(x+4\right)^2-5=0\\ =>\left[{}\begin{matrix}x=-4+\sqrt{5}\\x=-4-\sqrt{5}\end{matrix}\right.\)

\(Vậy\ GTLN\ của\ biểu\ thức\ là:-1\ khi\ x=-4+\sqrt5 hoặc x=-4-\sqrt5\)

\(Để\ hai\ đường\ thẳng\ trùng\ nhau\ thì:\)

\(\left\{{}\begin{matrix}k^2=3k-2\\m+3=5-m\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}k^2-3k+2=0\\m+m=5-3\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}\left(k-2\right)\left(k-1\right)=0\\2m=2\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}\left[{}\begin{matrix}k-2=0\\k-1=0\end{matrix}\right.\\m=1\end{matrix}\right.\)

\(Vậy\ các\ giá\ trị\ của\ k\ và\ m\ thỏa\ mãn\ là:\ (k;m)=(2;1);(1;1)\)