i: =20*(-29)+(-20)*111

=20(-29-111)

=-20*140=-2800

k: \(=-152\cdot125+152\cdot333-125\cdot333+125\cdot152\)

=27*333

=8991

m: \(=63\left(83+17\right)=6300\)

n: \(=126-7\cdot\left(-16\right)=126+112=238\)

p: \(=35\left(71\cdot2-7-13\right)=35\cdot122=4270\)

q: \(=18\cdot23-18\cdot17-13\cdot23-13\cdot18=5\cdot23-18\cdot30=115-540=-425\)

Bài `4`

`1, 2y(x+2)-3x-6`

`=2y(x+2) -(3x+6)`

`=2y(x+2) -3(x+2)`

`=(x+2)(2y-3)`

`2, 3(x+4) -x^2-4x`

`=3(x+4)-(x^2+4x)`

`=3(x+4) -x(x+4)`

`=(x+3)(3-x)`

`3, 2(x+5) -x^2-5x`

`=2(x+5)-(x^2+5x)`

`=2(x+5)-x(x+5)`

`=(x+5)(2-x)`

`4, x^2 +6x-3(x+6)`

`= (x^2+6x) -3(x+6)`

`=x(x+6)-3(x+6)`

`=(x+6)(x-3)`

`5, x(x+y) -5x-5y`

`=x(x+y) -(5x+5y)`

`=x(x+y)-5(x+y)`

`=(x+y)(x-5)`

`6,x(x-y)+2x-2y`

`=x(x-y)+2(x-y)`

`=(x-y)(x+2`

Bạn tách ra từng bài đi ạ. Làm all trong 1 câu nhiều lắm.

Em ko muốn bài giải vậy Thầy xóa đây. Em tự xem lại cách học của mình nhé!

`a,-2022.(x+8)=0`

`=>x+8=0:(-2022)`

`=>x+8=0`

`=>x=0-8`

`=>x=-8`

`b,(7-x)(x+3)=0`

`@ TH1`

`7-x=0`

`=>x=7-0`

`=>x=7`

`@ TH2`

`x+3=0`

`=>x=0-3`

`=>x=-3`

`c,2023x .(14-x)=0`

`@ TH1`

`2023x=0`

`=>x=0:2023`

`=>x=0`

`@ TH2`

`14-x=0`

`=>x=14-0`

`=>x=14`

`d,x^2-x=0`

`=>x(x-1)=0`

`@ TH1`

`x=0`

`@ TH2`

`x-1=0`

`=>x=0+1`

`=>x=1`

\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(\left(-2020\right).\left(x+8\right)=0\)

\(=>x+8=0=>x=-8\)

\(\left(7-x\right).\left(x+3\right)=0\)

\(=>\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(2023x.\left(14-x\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\14-x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\)

\(x^2-x=0\)

\(x\left(x-1\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(36\cdot11\cdot25\cdot9\\=6\cdot6\cdot11\cdot5\cdot5\cdot9\\=(6\cdot5)\cdot(6\cdot5)\cdot(11\cdot9)\\=30\cdot30\cdot99\\=900\cdot99\\=89100\)

36.11.25.9

= 396.25.9

=9900.9= 89100

31.11.25.9

= 6.6.11.5.5.9 

={6.5} . {6.5} . {9.11}

=30.30.99 

=900.99

=89100 

Ta có \(x\inƯ\left(30\right)\)\(\left(ĐKXĐ:x\le8\right)\)

\(< =>x\in\left\{1;2;3;5;6;10;15;30\right\}\)

Do \(x\le8\)suy ra ta có bộ số x thỏa mãn sau :

\(x\in\left\{1;2;3;5;6\right\}\)

Trả lời :

Theo bài ta có :

\(30⋮x\Rightarrow x\inƯ\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\)

Mà \(x< 8\Rightarrow x\in\left\{1;2;3;5;6\right\}\)

Cảm ơn các bạn nhé

Câu 3:

a: \(25x-30x^2=5x\left(5-6x\right)\)

b: \(8x\left(x+4y\right)+16\left(x+4y\right)=8\left(x+4y\right)\left(x+2\right)\)

c: \(24x^8y^7-15x^6y^5+18x^6y^3\)

\(=3x^6y^3\left(8x^2y^4-5y^2+6\right)\)

câu 1:

a) 4x( 5x + 9 )

 = 4x5x + 4x9

 = 20x²+36x

b) (3x - 2) ( x+7 )

 = 3xx + 3x7 - 2x - 2 . 7

 = 3x²+ 21x - 2x - 14

 = 3x² +19x - 14

c) (2x- 5 )(x²-7x + 4 )

 = 2xx² - 2x7x + 2x4 - 5x² +5 . 7x - 5.4

 = 2x³- 14x²+8x - 5x ²+ 35x - 20

 = 2x³+ ( - 14x²- 5x² )+(8x+35x) - 20

 = 2x³ - 9x² + 43x - 20

Câu 2 :

a) (5x - 3x)²

  = (5x)² - 2.5x3y + (3y)²

  = 25x² - 30xy +9y

b) (2x - 5)(2x + 5)

 = (2x)² - 5²

  = 4x² - 25

c) (3x - 7y)³

 = ( 3x)³ - 3(3x)²7y + 3.3x(7y)² - (7y)²

 = 27x³ - 189x²y + 441xy²- 343y³

d) (x+5)(x²-5x+25)

 = (x+5)(x²-5x+5²)

 = x³+5³

 = x³+125

Câu 3 : 

a) 25x-30x²

 = 5.5x-5.6xx

 = 5x(5-6x)

b) 8x(x+4y)+16(x+4y)

 = (x+4y)(8x+16)

c) 24x⁸y⁷ -15x⁶y⁵+18x⁶y³

 = 3.8x⁶x²y³y⁴-3.5x⁶y³y²+3.6x⁶y³

 = 3x⁶y³(8x²y⁴-5y²+6)

Câu 3: 

a: \(25x-30x^2=5x\left(5-6x\right)\)

b: \(8x\left(x+4y\right)+16\left(x+4y\right)=8\left(x+4y\right)\left(x+2\right)\)