\(\widehat{xOy}-\widehat{yOx'}=30^o\\ \Rightarrow\widehat{xOy}=\widehat{yOx'}+30^o\)

Ta có:

\(\widehat{xOy}+\widehat{yOx'}=180^o\)(2 góc kề bù)

\(\Rightarrow\widehat{yOx'}+30^o+\widehat{yOx'}=180^o\\ \Rightarrow2\widehat{yOx'}=150^o\\ \Rightarrow\widehat{yOx'}=75^o\)

\(\widehat{xOy}=\widehat{yOx'}+30^o=75^o+30^o=105^o\)

Ta có:

\(\widehat{yOx'}=\widehat{xOy'}=75^o\) (2 góc đối đỉnh)

\(\widehat{xOy}=\widehat{x'Oy'}=105^o\)(2 góc đối đỉnh)

`(3x-3)^2`

`=(3x)^2 - 2.3x.3+3^2`

`=9x^2 - 18x+9`

\(b,A=1.2.3+2.3.4+...+15.16.17\\ \Rightarrow4A=1.2.3.4+2.3.4.4+...+15.16.17.4\\ \Rightarrow4A=1.2.3.4+2.3.4.\left(5-1\right)+...+15.16.17.\left(18-14\right)\\ \Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+15.16.17.18-14.15.16.17\\ \Rightarrow4A=15.16.17.18\\ \Rightarrow A=\dfrac{15.16.17.18}{4}\\ \Rightarrow A=18360\)

\(A=1.2+2.33.4+...+15.16\\ \Rightarrow3A=1.2.3+2.3.3+3.4.3+...+15.16.3\\ \Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+15.16.\left(17-14\right)\\ \Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+15.16.17-14.15.16\\ \Rightarrow3A=15.16.17\\ \Rightarrow A=\dfrac{15.16.17}{3}\\ \Rightarrow A=1360\)

\(G=4\left(x-2\right)\left(x+2\right)+\left(2+x\right)\left(1-x\right)\\ =4\left(x^2-2^2\right)+\left(2+x-2x-x^2\right)\\ =4\left(x^2-4\right)+\left(2-x-x^2\right)\\ =4x^2-16+2-x-x^2\\ =\left(4x^2-x^2\right)-x-\left(16-2\right)\\ =3x^2-x-14\)

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

Ta có:
`(1) <=> x^2 +2xy + y^2 +2(x+y)=0`

`<=>(x+y)^2 + 2(x+y)=0`

`<=>(x+y)(x+y+2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\x=-y-2\end{matrix}\right.\)

Thay `x=-y` vào `(2)` ta có:
`(-y)^2 + y^2 + 4(-y)-2y+4=0`

`<=>y^2 +y^2 -4y-2y+4=0`

`<=>2y^2 -6y+4=0`

`<=>y^2 -3y+2=0`

\(\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-1\\y=2\Rightarrow x=-2\end{matrix}\right.\)

Thay `x=-y-2` vào `(2)` ta có:

`(-y-2)^2 + y^2 + 4(-y-2)-2y+4=0`

`<=>y^2 +4y+4 + y^2 -4y-8-2y+4=0`

`<=>2y^2-2y=0`

`<=>2y(y-1)=0`

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

Vậy `(x,y) in {(-1,1);(-2,2);(-2,0);(-3,1)}`

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

\(\left\{{}\begin{matrix}x^2-3xy+2y^2=0\left(1\right)\\2x^2-3xy+5=0\left(2\right)\end{matrix}\right.\)

Ta có:

`(1)<=>x^2 -xy-2xy+2y^2 =0`

`<=>x(x-y)-2y(x-y)=0`

`<=>(x-y)(x-2y)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=2y\end{matrix}\right.\)

Thay `x=y` vào `(2)` ta có:
`2x^2 -3.x.x+5=0`

`<=>2x^2 -3x^2 +5=0`

`<=>-x^2 =-5`

`<=>x^2 =5`

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\Rightarrow y=\sqrt{5}\\x=-\sqrt{5}\Rightarrow y=-\sqrt{5}\end{matrix}\right.\)

Thay `x=2y` vào `(2)` ta có:
`2.(2y)^2 - 3.2y.y+5=0`

`<=>2.4y^2 - 6y^2 +5=0`

`<=>8y^2 -6y^2 = -5`

`<=>2y^2 = -5`(Vô lí)

Vậy `(x,y) in {(sqrt5 , sqrt5);(-sqrt5 , -sqrt5)}`

\(\left(\dfrac{2}{4}x+y\right)^2\\ =\left(\dfrac{2}{4}x\right)^2+2.\dfrac{2}{4}x.y+y^2\\ =\dfrac{4}{16}x^2+\dfrac{2}{2}xy+y^2\\ =\dfrac{1}{4}x^2+xy+y^2\)

\(B=\dfrac{7}{1\times4}+\dfrac{7}{4\times7}+...+\dfrac{7}{46\times49}\\ =7\times\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{46\times49}\right)\\ =\dfrac{7}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{46\times49}\right)\\ =\dfrac{7}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{46}-\dfrac{1}{49}\right)\\ =\dfrac{7}{3}\times\left(1-\dfrac{1}{49}\right)\\ =\dfrac{7}{3}\times\dfrac{48}{49}\\ =\dfrac{16}{7}\)