`@` `\text {Ans}`

`\downarrow`

\(x^n\cdot y^{n+2}\cdot\left(xy+x^2y+1\right)\)

`=`\(x^n\cdot y^n\cdot y^2\left(xy+x^2y+1\right)\)

`=`\(\left(xy\right)^n\cdot y^2\cdot xy+y^2\cdot x^2y+y^2\)

`=`\(\left(xy\right)^n\cdot xy^3+x^2y^3+y^2\)

x\(^{n+1}\)y\(^{n+1}\)+2xy+x\(^{ }\)\(^{n+2}\)y\(\)\(^{n+1}\)+2x\(^2\)y+x\(^n\)y\(^n\)+2

\(\left(\dfrac{1}{5}\right)^{25}\cdot\left(\dfrac{1}{5}\right)^{30}\)

\(=\left(\dfrac{1}{5}\right)^{25+30}\)

\(=\left(\dfrac{1}{5}\right)^{55}\)

\(\left(\dfrac{1}{16}\right)^3:\left(\dfrac{1}{8}\right)^2\)

\(=\left[\left(\dfrac{1}{2}\right)^4\right]^3:\left[\left(\dfrac{1}{2}\right)^3\right]^2\)

\(=\left(\dfrac{1}{2}\right)^{12}:\left(\dfrac{1}{2}\right)^6\)

\(=\left(\dfrac{1}{2}\right)^{12-6}=\left(\dfrac{1}{2}\right)^6\)

\(\left(x^3\right)^2:\left(x^2\right)^3\)

\(=x^{3\cdot2-2\cdot3}\)

\(=x^0=1\)

32 < 211 < 128 ( xem lại đề bài em nhé)

4< 2N < 216

\(\dfrac{4}{2}< \) N < \(\dfrac{216}{2}\)

2 < N < 108

vì N \(\in\) Z nên N \(\in\) { 3; 4; 5; 6; 7;...; 107}

`@` `\text {Ans}`

`\downarrow`

`-1/27 = ( (-1)^3/(-3)^3) =(-1/3)^3`

`8/729 = ( (2^3)/(9^3) ) = (2/9)^3`

`16/685 = ( (4^2)/( \sqrt {685}) ) = (4/(\sqrt {685}) )^2`

(2 x X - 1)mũ 3 = (2 x X - 2)mũ 2

\(x+\dfrac{1}{2}=1-x\\ \Rightarrow x+x=1-\dfrac{1}{2}\\ \Rightarrow2x=\dfrac{2}{2}-\dfrac{1}{2}\\ \Rightarrow2x=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{2}:2\\ \Rightarrow x=\dfrac{1}{2}\times\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{4}\)

Góc nhọn: \(\widehat{ABD};\widehat{ADB};\widehat{BEC};\widehat{EDC};\widehat{BDC};\widehat{ECD};\widehat{BCE};\widehat{BCD}\)

Góc vuông: \(\widehat{BAD}\)

Góc tù: \(\widehat{DBC};\widehat{EBC};\widehat{DEC}\)

a, Góc nhọn: \(\widehat{ABD}\);  \(\widehat{BCE}\); \(\widehat{BCD}\); \(\widehat{DCE}\); \(\widehat{BDC}\); \(\widehat{ADB}\); \(\widehat{BEC}\)

b, Góc vuông: \(\widehat{BAD}\); \(\widehat{ADC}\)

c, Góc tù: \(\widehat{ABC}\);  \(\widehat{CBD}\); \(\widehat{CED}\)

a, Các cặp tia đối nhau chung gốc A lần lượt là:

Ax và AO;  Ax và AB;  Ax và AY 

b, Vì OA và OB là hai tia đối nhau nên O nằm giữa A và B

      ⇒ OA + OB = AB 

      ⇒ OB = AB - OA

         Độ dài đoạn OB là: 6 - 3 = 3 (cm)

  c, Vì O nằm giữa A và B mà OA = OB = 3 cm nên O là trrung điểm AB