Áp dụng t/c dtsbn:

\(\dfrac{\widehat{A}}{3}=\dfrac{\widehat{B}}{5}=\dfrac{\widehat{C}}{7}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{3+5+7}=\dfrac{180^0}{15}=12^0\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=12^0.3=36^0\\\widehat{B}=12^0.5=60^0\\\widehat{C}=12^0.7=84^0\end{matrix}\right.\)

\(x^2+10x+25=\left(x+5\right)^2\)

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

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

Thiếu r nha bn, ước của 5 còn -1;-5 nữa nhé

a) \(=\left(2x-1\right)^2\)

b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)

\(m-4=2m-3\Leftrightarrow2m-m=-4+3\Leftrightarrow m=-1\)

Giả sử 2 đường thẳng (d), (d') cắt nhau tại \(M\left(x_0;y_0\right)\) trên trục tung

\(\Rightarrow x_0=0\)

Thay tọa độ của M và 2 đường thẳng ta có:

\(\left(d\right):y=m-4\) và \(\left(d'\right):y=2m-3\)

PT hoành độ gia điểm: \(m-4=2m-3\Leftrightarrow m=-1\)

Vậy...

Bài 2:

a) \(=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{x\left(x-2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)

b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)