Tọa độ giao điểm `2` đường thẳng `5x+11y=8` và `10x-7y=74` là:

`{(5x+11y=8),(10x-7y=74):}`

`<=>{(x=6),(y=-2):}`

Để `3` đường thẳng đồng quy thì:

`4m.6+(2m-1).(-2)=m+2`

`<=>24m-4m+2-m-2=0`

`<=>19m=0`

`<=>m=0`

Vậy để `3` đường thẳng đồng quy thì `m=0`

Để đường thẳng `ax+by=6` đi qua `A(1;2)` thì:

`a.1+b.2=6`

`<=>a+2b=6`

Để đường thẳng `ax+by=6` đi qua `N(-2;3)` thì:

`a.(-2)+b.3=6`

`<=>-2a+3b=6`

Từ `(1),(2)` ta có hệ phương trình:

\(\left\{{}\begin{matrix}a+2b=6\\-2a+3b=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{6}{7}\\b=\dfrac{18}{7}\end{matrix}\right.\)

Vậy `a= 6/7 ; b= 18/7`

`(3x-1)^3`

`=(3x)^3 - 3.(3x)^2 .1 +3.3x.1^2 -1^3`

`=27x^3 -3.9x^2 .1 +3.3x.1-1`

`=27x^3 -27x^2 +9x-1`

`(3x-1)^3= 27x^3 +ax^2 +9x-1`

`=>27x^3 -27x^2 +9x-1= 27x^3 +ax^2 +9x-1`

`=>-27x^2 =ax^2`

`=>a=-27`

\(a,VT=\dfrac{-x^2+3x-2}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{-\left(x^2-3x+2\right)}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{-\left(x^2-x-2x+2\right)}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{-\left[x\left(x-1\right)-2\left(x-1\right)\right]}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{-\left(x-2\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{-\left(x-2\right)}{x+2}\\ =\dfrac{-\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{-\left(x-2\right)^2}{x^2-4}\\ =\dfrac{x^2-4x+4}{4-x^2}=VP\left(đpcm\right)\)

\(b,VT=\dfrac{x^3+27}{x^2-3x+9}\\ =\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}\\ =x+3=VP\left(đpcm\right)\)

`x^3 +9x^2 y +27xy^2 +27y^3`

`=x^3 +3.x^2 .3y  + 3.x.(3y)^2 + (3y)^3`

`=(x+3y)^3`

`=-(-x-3y)^3`

`=-(-5)^3`

`=-125`

\(5-\left(\dfrac{-5}{7}\right)^0+\left(\dfrac{1}{3}\right)^2:3\\ =5-1+\dfrac{1}{9}.\dfrac{1}{3}\\ =4+\dfrac{1}{27}\\ =\dfrac{109}{27}\)

\(x^2-4x+3=0\\ \Leftrightarrow x^2-x-3x+3=0\\ \Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(-2x\left(-3x+3\right)-6\left(x+3\right)^2\\ =\left(6x^2-6x\right)-6\left(x^2+6x+9\right)\\ =6x^2-6x-\left(6x^2+36x+54\right)\\ =6x^2-6x-6x^2-36x-54\\ =-42x-54\)

\(\left(2x-3\right)^2-2\left(4x^2-9\right)+\left(2x+3\right)^2\\ =\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)+\left(2x+3\right)^2\\ =\left[\left(2x-3\right)-\left(2x+3\right)\right]^2\\ =\left(2x-3-2x-3\right)^2\\ =\left(-6\right)^2\\ =36\)

\(\dfrac{4}{3-\sqrt{5}}-\dfrac{4}{3+\sqrt{5}}\\ =\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\dfrac{4\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\\ =\dfrac{4\left(3+\sqrt{5}\right)}{3^2-\left(\sqrt{5}\right)^2}-\dfrac{4\left(3-\sqrt{5}\right)}{3^2-\left(\sqrt{5}\right)^2}\\ =\dfrac{4\left(3+\sqrt{5}\right)}{9-5}-\dfrac{4\left(3-\sqrt{5}\right)}{9-5}\\ =\dfrac{4\left(3+\sqrt{5}\right)}{4}-\dfrac{4\left(3-\sqrt{5}\right)}{4}\\ =3+\sqrt{5}-3+\sqrt{5}\\ =2\sqrt{5}\)