Quang Duy

a, Ta có:

\(\widehat{yOm}+\widehat{mOy'}=180^o\)

mà \(\widehat{yOm}=\widehat{y'On}\) (đối đỉnh)

nên \(\widehat{y'On}+\widehat{mOy'}=180^o\)

hay \(\widehat{mOn}=180^o\)

\(\Rightarrow\) Om và On là hai tia đối nhau.

b, Ta có:

\(\widehat{xOy}+\widehat{xOy'}=180^o\\ \Rightarrow\widehat{xOy'}=180^o-45^o=135^o\)

mà \(\widehat{xOy'}=\widehat{x'Oy}\) (đối đỉnh)

nên \(\widehat{x'Oy}=135^o\)

mặt khác:

\(\widehat{xOy}=\widehat{x'Oy'}\) (đối đỉnh)

nên \(\widehat{x'Oy'}=45^o\)

Vì Om và On lần lượt là phân giác của góc xOy là x'Oy' nên:

\(\widehat{xOm}=\widehat{yOm}=\dfrac{\widehat{xOy}}{2}=\dfrac{45^o}{2}=22,5^o\\ \widehat{x'On}=\widehat{y'On}=\dfrac{\widehat{x'Oy'}}{2}=\dfrac{45^o}{2}=22,5^o\)

Vậy......

\(\left(\dfrac{2}{3}x^2y\right)\left(\dfrac{9}{2}xy\right)=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)\left(x^2.x\right)\left(y.y\right)=3x^3y^2\)

a, \(y^2-2y+1=y^2-y-y+1\\ =y.\left(y-1\right)-\left(y-1\right)=\left(y-1\right)^2\)

b, \(x^2+x+\dfrac{1}{4}=x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}\\ =x.\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}.\left(x+\dfrac{1}{2}\right)=\left(x+\dfrac{1}{2}\right)^2\)

c, \(x^2-6x+9=x^2-3x-3x+9\\ =x.\left(x-3\right)-3.\left(x-3\right)=\left(x-3\right)^2\)

d, \(4y^2+4y+1=4y^2+2y+2y+1\\ =2y.\left(2y+1\right)+\left(2y+1\right)=\left(2y+1\right)^2\)

Mấy câu còn lại làm tương tự!

a, \(0,\left(27\right)+0,\left(72\right)=0,\left(01\right).27+0,\left(01\right).72\)

\(=\dfrac{1}{99}.\left(27+72\right)=\dfrac{1}{99}.99=1\)

b, \(0,\left(22\right).\dfrac{9}{2}=0,\left(01\right).22.\dfrac{9}{2}=\dfrac{1}{99}.22.\dfrac{9}{2}=1\)

c, \(\left[0,\left(11\right).9\right]^{2003}=\left[0,\left(01\right).11.9\right]^{2003}\)

\(=\left[\dfrac{1}{99}.11.9\right]^{2003}=1^{2003}=1\)

\(0,5=\dfrac{1}{2}\)

\(1,\left(45\right)=1+0,\left(45\right)=1+45.0,\left(01\right)\)

\(=1+45.\dfrac{1}{99}=\dfrac{16}{11}\)

\(0,2\left(13\right)=0,2+0,0\left(13\right)=0,2+0,\left(13\right):10\)

\(=0,2+0,\left(01\right).13:10=0,2+\dfrac{1}{99}.13:10=\dfrac{211}{990}\)

\(0,04\left(3\right)=0,04+0,00\left(3\right)\)

\(=0,04+0,\left(3\right):100=0,04+0,\left(1\right).3:100\)

\(=0,04+\dfrac{1}{9}.3:100=\dfrac{13}{300}\)

a, \(x^3+8=x^3+2x^2-2x^2-4x+4x+8\)

\(=x^2.\left(x+2\right)-2x.\left(x+2\right)+4.\left(x+2\right)\)

\(=\left(x+2\right).\left(x^2-2x+4\right)\)

Câu b hình như sai đề bạn xem lại nha!

Bạn xem lại đề đi

a, \(3x^2-3x+2x^3-2x^2=0\)

\(\Rightarrow3x.\left(x-1\right)+2x^2.\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right).\left(3x+2x^2\right)=0\)

\(\Rightarrow\left(x-1\right).x.\left(3+2x\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x=0\\3+2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=0\\x=\dfrac{-3}{2}\end{matrix}\right.\)

Vậy......

Câu b tương tự!!

Bài này tra mạng có đó bạn mà nhớ không nhầm đây là bài trong sách giảo khoa!

Gọi các tập con có hai chữ số là A

Ta có:

\(A=\left\{\left(a;b\right);\left(a;c\right);\left(b;c\right)\right\}\)

Vậy.....

c, \(4x^2+4xy+y^2\)

\(=4x^2+2xy+2xy+y^2\)

\(=2x.\left(2x+y\right)+y.\left(2x+y\right)\)

\(=\left(2x+y\right)^2\)

d, \(x^2+12xy+36y^2\)

\(=x^2+6xy+6xy+36y^2\)

\(=x.\left(x+6y\right)+6y.\left(x+6y\right)\)

\(=\left(x+6y\right)^2\)

e, \(x^2-12xy+36y^2\)

\(=x^2-6xy-6xy+36y^2\)

\(=x.\left(x-6y\right)-6y.\left(x-6y\right)\)

\(=\left(x-6y\right)^2\)

\(\dfrac{151515}{161616}+\dfrac{1}{17}-\dfrac{1500}{1600}-\dfrac{16}{17}\)

\(=\dfrac{10101.15}{10101.16}+\dfrac{1}{17}-\dfrac{15.100}{16.100}-\dfrac{16}{17}\)

\(=\dfrac{15}{16}+\dfrac{1}{17}-\dfrac{15}{16}-\dfrac{16}{17}=\dfrac{-15}{17}\)

Thay \(x=-\dfrac{3}{5}\) vào biểu thức ta có:

\(\dfrac{1}{2}+\dfrac{1}{3}.\dfrac{-3}{5}-\dfrac{1}{6}.\dfrac{-3}{5}\)

\(=\dfrac{1}{2}+\dfrac{-3}{5}.\left(\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{1}{2}+\dfrac{-3}{5}.\dfrac{1}{6}\)

\(=\dfrac{1}{2}+\dfrac{-1}{10}=\dfrac{2}{5}\)

\(A=\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\)

\(A=\sqrt{x^2-3x-3x+9}+\sqrt{x^2+3x+3x+9}\)

\(A=\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+3\right)^2}\)

\(A=x-3+x+3=2x\)