\(\left(3x-48\right).6=27.9+8.9-12\\ \left(3x-48\right).6=243+72-12\\ \left(3x-48\right).6=303\\ 3x-48=\dfrac{101}{2}\\ 3x=\dfrac{101}{2}+48\\ 3x=\dfrac{101+48.2}{2}\\ 3x=\dfrac{197}{2}\\ x=\dfrac{197}{6}\)

= 23 m^2 69 dm^2 1cm^2

a, Xét Δ ABM và Δ ACM có

MB = MC ( M là trung điểm BC)

góc B = góc C ( Δ ABC cân )

AB = AC ( Δ ABC cân )

=> 2 Δ bằng nhau ( c-g-c)

b, => gó BAM = gó CAM 

=> AM là tia phân giác góc A

Ta có Δ ABC cân  mà AM là tia phân giác A

=> AM là đg cao hay AM vg BC   

Gọi O nhấn liền thành gội

3x7 là nhân hay x 

2x(x+1) là nhân hay x

\(a,x\left(x+2\right)=0\\ x=0hoacx+2=0\\ x=0;x=-2\)

\(b,x\left(x^2+1\right)=0\\ x=0\)

\(x^2+1=0\left(voli\right)\)

\(=\dfrac{11}{13}.\dfrac{3}{8}+\dfrac{11}{13}.\dfrac{1}{2}+\dfrac{11}{18}.\dfrac{11}{13}=\dfrac{11}{13}\left(\dfrac{3}{8}+\dfrac{1}{2}+\dfrac{11}{18}\right)=\dfrac{11}{13}.\left(\dfrac{3.6+9+11}{18}\right)=\dfrac{11}{13}.\dfrac{107}{72}=\dfrac{1177}{936}\)

\(a,=\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)\\ b,=\left[4\left(x-y\right)-2y\right]\left[4\left(x-y\right)+2y\right]=\left(4x-4y-2y\right)\left(4x-4y+2y\right)=\left(4x-6y\right)\left(4x-2y\right)\)

\(=\dfrac{\sqrt{7776}}{\sqrt{8.243}}=\dfrac{\sqrt{7776}}{\sqrt{1944}}=\dfrac{36\sqrt{6}}{18\sqrt{6}}=\dfrac{36}{18}=2\)

\(a,=\left(xy^2\right)^2-\left(u^4v^9\right)^2=\left(xy^2-u^4v^9\right)\left(xy^2+u^4v^9\right)\\ b,=\left(3x-5y\right)\left(3x+5y\right)\)