GIÚP VS tui đang cần gấp
giúp tui làm từ câu V->câu VI vs ạ tui đang cần gấp
theng kiuu bbi trc nhoaaaa=333
<333333
I am not keen on listening to loud music
Jenny is not fond of doing housework
I am really interested in reading book
She doesn't like watching films on TV
No, they didn't
Until the programs finished
Yes. they did
Life is much more comfortable
T - T
TL:
Ta có:
10 có mũ bao nhiêu đi nữa, thì + 109 hay 108 +..v.v
Thì số cuối của dãy số vẫn băng 0 nên
=> 109+108+107 chia hết cho 2
Học Tốt👍
chỉ tui bài 2 vs bài 3 với, tui đang cần gấp
`# \text {Ryo}`
`2,`
`a)`
`x^3 -9x^2 + 27x - 27`
`= (x)^3 - 3*x^2 * 3 + 3*x*3^2 - (3)^3`
`= (x - 3)^3`
`b)`
`- (x^3)/8 + 3/4x^2 - 3/2x + 1`
`= - ( (x^3)/8 - 3/4x^2 + 3/2x - 1)`
`= - [ (x/2)^3 - 3*(x/2)^2 * 1 + 3*x/2*1^2 - 1^3]`
`= - (x/2 - 1)^3`
`c)`
Phiền bạn ghi lại đề giúp mình với ạ! Số mũ của biến 3 số sau mình kh đọc được.
`3,`
`a)`
`A = x^3 - 6x^2 + 12x - 8`
`= (x)^3 - 3*x^2*2 + 3*x*2^2 - (2)^3`
`= (x - 2)^3`
`b)`
`B = 1 - (3x)/2 + (3x^2)/4 - (x^3)/8` phải k c? (Mình thấy biến phần cuối hơi mờ).
`= 1^3 - 3*1^2*x/2 + 3* 1 * (x/2)^2 - (x/2)^3`
`= (1 - x/2)^3`
__
Cả bài 2 và 3, bạn sử dụng CT:
`A^3 - 3A^2B + 3AB^2 - B^3 = (A - B)^3`
Giúp tui vs tui cần gấp !!!
Giúp tui vs tui đang cần hấp 😟😭😭😭❤
33 arranging flowers interesting
34 about 2 hours each day doing my homework
35 collecting glass bottles is interesting
36 as dangerous as mountain climbing
37 To stay healthy you eat more vegetables and eat less high-fat food
38 Linh doesn't like playing board games
39 Lan work less than Nam
40 Are you interested in listening to music
giúp tui với, tui đang cần gấp
giúp tui vs ak tui đag cần gấp
giúp tui với tui đang cần gấp ạ xong tui tick
Bài 1:
\(a,x^2-y^2-2x+2y=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)
\(b,2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
\(c,3a^2-6ab+3b^2-12c^2=3\left(a-b\right)^2-12c^2=3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
\(d,x^2-25+y^2+2xy=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\)
Bài 1:
\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b-c\right)\left(a+b\right)\)
\(f,x^2-2x-4y^2-4y=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-2y-2\right)\left(x+2y\right)\)
\(g,x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)
\(h,x^2\left(x-1\right)+16\left(1-x\right)=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
Bài 2:
1) \(4x^2-25+\left(2x+7\right)\left(5-2x\right)=\left(2x-5\right)\left(2x+5\right)-\left(2x+7\right)\left(2x-5\right)=\left(2x-5\right)\left(-2\right)\)
2) \(3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\)
3) \(5x^2-5y^2-10x+10y=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=5\left(x-y\right)\left(x+y-2\right)\)
4) \(x^2-xy+x-y=\left(x-y\right)\left(x+1\right)\)
Tui đang cần gấp ai giúp tui với xogn tui tick
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) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\dfrac{1}{27}x^3+8y^3\)