MN GIÚP MIK VS Ạ! MIK ĐG CẦN GẤP Ạ!!!!!!!
Bài 3:
\(A=\dfrac{m^2+5m+n^2+5n+2mn-6}{m^2+6m+n^2+6n+2mn}\\ A=\dfrac{\left(m+n\right)^2+6\left(m+n\right)-\left(m+n\right)-6}{\left(m+n\right)^2+6\left(m+n\right)}\\ A=\dfrac{\left(m+n\right)\left(m+n+6\right)-\left(m+n\right)-6}{\left(m+n\right)\left(m+n+6\right)}\\ A=1-\dfrac{m+n+6}{\left(m+n\right)\left(m+n+6\right)}\\ A=1-\dfrac{1}{m+n}=\dfrac{m+n-1}{m+n}=\dfrac{2012}{2013}\)
Bài 4:
\(M=\dfrac{x^2+ax^2+a+a^2+a^2x^2+1}{x^2-ax^2-a+a^2+a^2x^2+1}\\ M=\dfrac{x^2\left(a^2+1\right)+a\left(x^2+1\right)+\left(a^2+1\right)}{x^2\left(a^2+1\right)-a\left(x^2+1\right)+\left(a^2+1\right)}\\ M=\dfrac{\left(a^2+1\right)\left(x^2+1\right)+a\left(x^2+1\right)}{\left(a^2+1\right)\left(x^2+1\right)-a\left(x^2+1\right)}\\ M=\dfrac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(x^2+1\right)\left(a^2-a+1\right)}=\dfrac{a^2+a+1}{a^2-a+1}\left(đpcm\right)\)
Câu 5: (câu a bạn sửa vế phải \(=\dfrac{x^2+2x+2}{y-1}\)
\(a,VT=\dfrac{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}{y\left(x^2-2x+2\right)-\left(x^2-2x+2\right)}\\ =\dfrac{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}{\left(x^2-2x+2\right)\left(y-1\right)}=\dfrac{x^2+2x+2}{y-1}=VP\)
\(b,VT=\dfrac{\left(3n-2\right)-m\left(3n-2\right)}{\left(1-m\right)^3}=\dfrac{\left(1-m\right)\left(3n-2\right)}{\left(1-m\right)^3}=\dfrac{3n-2}{\left(1-m\right)^2}=VP\)
