1) \(\left(x+\dfrac{1}{3}\right)^3=x^3+3.x^2.\dfrac{1}{3}+3.x.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3\)

\(=x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}\)

2) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.y^2+3.2x.\left(y^2\right)^2+\left(y^2\right)^3\)

\(=8x^3+12x^2y^2+6xy^4+y^6\)

3) \(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y\right)^3=\left(\dfrac{1}{2}x^2\right)^3+3.\left(\dfrac{1}{2}x^2\right)^2.\dfrac{1}{3}y+3.\dfrac{1}{2}x^2.\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)

\(=\dfrac{1}{8}x^6+\dfrac{1}{4}x^4y+\dfrac{1}{6}x^2y^2+\dfrac{1}{27}y^3\)

4) \(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.3x^2.\left(2y\right)^2-\left(2y\right)^3\)

\(=27x^6-54x^4y+36x^2y^2-8y^3\)

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

\(=\dfrac{8}{27}x^6-\dfrac{1}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)

6) \(\left(2x+\dfrac{1}{2}\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{2}+3.2x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)

\(=8x^3+6x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)

7) \(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3\)

\(=x^3-9x^2+27x-27\)

8) \(\left(x+1\right)\left(x^2-x+1\right)\)

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

\(=x^3+1^3\)

\(=x+1\)

9) \(\left(x-3\right)\left(x^2+3x+9\right)\)

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

\(=x^3-3^3\)

\(=x^3-27\)

10) \(\left(x-2\right)\left(x^2+2x+4\right)\)

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

\(=x^3-2^3\)

\(=x^3-8\)

11) \(\left(x+4\right)\left(x^2-4x+16\right)\)

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

\(=x^3+4^3\)

\(=x^3+64\)

12) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)

\(=x^3-\left(3y\right)^3\)

\(=x^3-27y^3\)

13) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)

\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+x^2.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]\)

\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)

\(=x^6-\dfrac{1}{27}\)

14) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)

\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\dfrac{1}{27}x^3+8y^3\)

Bài nào vậy bn bn phải nói rõ chứ

a) 108 . 28 = (10.2)8 = 208

b) 108 : 28 = (10:2)8 = 58

c) 254 . 28 = 58 . 28 = 108

d) 158 . 94 = 158 . 38 = 458

e) 272 : 253 = (33)2 : (52)3 = 33.2 : 52.3 = 36 : 56 = 

https://m.loigiaihay.com/bai-36-trang-22-sgk-toan-7-tap-1-c42a3232.html                       bạn vào đây tham khảo

\(\dfrac{6}{x}+\dfrac{1}{2}=2\)

\(\dfrac{6}{x}=2-\dfrac{1}{2}\)

\(\dfrac{6}{x}=\dfrac{3}{2}\)

\(3x=12\)

\(x=4\)

`6/x + 1/2 = 2` (` x ne 0`)

`=> 6/x = 2- 1/2`

`=> 6/x=3/2`

`=> x= 6 : 3/2`

`=> x= 6 . 2/3`

`=> x=4` (thỏa mãn ĐK)

Vậy `x =4`

\(\dfrac{6}{x}\) + \(\dfrac{1}{2}\)= 2

\(\dfrac{6}{x}\) = 2- \(\dfrac{1}{2}\)

\(\dfrac{6}{x}\) =  3/2

\(\dfrac{3x2}{2x2}\)= 6/x = 4

\(\dfrac{6}{x}\)

BÀI ĐÂY NHA MỌI NGƯỜI 

Bạn có thể kết bạn với mình ko?

C

B

C

C

B

A

Bạn cần hỗ trợ bài nào thì đăng bài đó lên để mọi người trợ giúp nhé.

Câu 3: 

a: Xét ΔABC có AB<BC

nên \(\widehat{ACB}< \widehat{BAC}\)

b: Xét ΔABM có 

AH là đường cao

AH là đường trung tuyến

Do đó: ΔABM cân tại A

mà \(\widehat{B}=60^0\)

nên ΔABM đều

 Mảnh trăng non đầu tháng lơ lửng giữa trời như một cánh diều

quả bóng=)))

quả bóng =)))