`~rai~`
\(a)\text{Đặt }\dfrac{m}{n}=\dfrac{p}{q}=k\\\Rightarrow m=nk;p=qk.\\\text{Xét:}\dfrac{2m+p}{2n+q}=\dfrac{2nk+q}{2n+q}=\dfrac{k(2n+q)}{2n+q}=k.(1)\\\dfrac{2m-p}{2n-q}=\dfrac{2nk-qk}{2n-q}=\dfrac{k(2n-q)}{2n-q}=k.(2)\\\text{Từ (1) và (2)}\Rightarrow \dfrac{2m+p}{2n+q}=\dfrac{2m-p}{2n-q}.(đpcm)\\b)\dfrac{m^2-n^2}{m^2-n^2}=\dfrac{p^2-q^2}{p^2-q^2}\\\text{Ta có:}\dfrac{m^2-n^2}{m^2-n^2}=1.(1)\\\dfrac{p^2-q^2}{p^2-q^2}=1.(2)\\\text{Từ (1) và (2)}\Rightarrow\dfrac{m^2-n^2}{m^2-n^2}=\dfrac{p^2-q^2}{p^2-q^2}.(đpcm)\\\text{Sửa đề:}\dfrac{m^2-n^2}{m^2+n^2}=\dfrac{p^2-q^2}{p^2+q^2}\\\text{Đặt }\dfrac{m}{n}=\dfrac{p}{q}=k\\\Rightarrow m=nk;p=qk.\\\text{Xét:}\\\dfrac{m^2-n^2}{m^2+n^2}=\dfrac{(nk)^2-n^2}{(nk)^2+n^2}=\dfrac{n^2k^2-n^2}{n^2k^2+n^2}=\dfrac{n^2(k^2-1)}{n^2(k^2+1)}=\dfrac{k^2-1}{k^2+1}.(1)\\\dfrac{p^2-q^2}{p^2+q^2}=\dfrac{(qk)^2-q^2}{(qk)^2+q^2}=\dfrac{q^2k^2-q^2}{q^2k^2+q^2}=\dfrac{q^2(k^2-1)}{q^2(k^2+1)}=\dfrac{k^2-1}{k^2+1}.(2)\\\text{Từ (1) và (2)}\Rightarrow\dfrac{m^2-n^2}{m^2+n^2}=\dfrac{p^2-q^2}{p^2+q^2}.(đpcm)\)
