làm rõ từng bước ra cho mình nhé. Cảm ơn
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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: \(=18\cdot23-18\cdot17-13\cdot23-13\cdot18=5\cdot23-18\cdot30=115-540=-425\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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`
![](https://rs.olm.vn/images/avt/0.png?1311)
`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`
\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(36\cdot11\cdot25\cdot9\\=6\cdot6\cdot11\cdot5\cdot5\cdot9\\=(6\cdot5)\cdot(6\cdot5)\cdot(11\cdot9)\\=30\cdot30\cdot99\\=900\cdot99\\=89100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 :
\(x\in\left\{1;2;3;5;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(24x^8y^7-15x^6y^5+18x^6y^3\)
\(=3x^6y^3\left(8x^2y^4-5y^2+6\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 3:
a: \(25x-30x^2=5x\left(5-6x\right)\)
b: \(8x\left(x+4y\right)+16\left(x+4y\right)=8\left(x+4y\right)\left(x+2\right)\)