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20 tháng 12 2020
1, \(45+x^3-5x^2-9x=9\left(5-x\right)+x^2\left(x-5\right)\)
\(=\left(9-x^2\right)\left(x-5\right)=\left(3-x\right)\left(x+3\right)\left(x-5\right)\)
3, \(x^4-5x^2+4\)
Đặt \(x^2=t\left(t\ge0\right)\)ta có :
\(t^2-5t+4=t^2-t-4t+4=t\left(t-1\right)-4\left(t-1\right)\)
\(=\left(t-4\right)\left(t-1\right)=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
YN
29 tháng 3 2022
`Answer:`
1. `45+x^3-5x^2-9x`
`=x^3+3x^2-8x^2-24x+15x+45x`
`=x^2 .(x+3)-8x.(x+3)+15.(x+3)`
`=(x+3).(x^2-8x+15)`
`=(x+3).(x^2-5x-3x+15)`
`=(x-3).(x-5).(x-3)`
2. `x^4-2x^3-2x^2-2x-3`
`=x^4+x^3-3x^3+x^2+x-3x-3`
`=x^3 .(x+1)-3x^2 .(x+1)+x.(x+1)-3.(x+1)`
`=(x+1).(x^3-3x^2+x-3)`
`=(x+1).[x^3 .(x-3).(x-3)]`
`=(x+1).(x-3).(x^2+1)`
3. `x^4-5x^2+4`
`=x^4-x^2-4x^2+4`
`=x^2 .(x^2-1)-4.(x^2-1)`
`=(x^2-1).(x^2-4)`
`=(x-1).(x+1).(x-2).(x+2)`
4. `x^4+64`
`=x^4+16x^2+64-16x^2`
`=(x^2+8)^2-16x^2`
`=(x^2+8-4x).(x^2+8+4x)`
5. `x^5+x^4+1`
`=x^5+x^4+x^3-x^3+1`
`=x^3 .(x^2+x+1)-(x^3-1)`
`=x^3 .(x^2+x+1)-(x-1).(x^2+x+1)`
`=(x^2+x+1).(x^3-x+1)`
6. `(x^2+2x).(x^2+2x+4)+3`
`=(x^2+2x)^2+4.(x^2+2x)+3`
`=(x^2+2x)^2+x^2+2x+3.(x^2+2x)+3`
`=(x^2+2x+1).(x^2+2x)+3.(x^2+2x+1)`
`=(x^2+2x+1).(x^2+2x+3)`
`=(x+1)^2 .(x^2+2x+3)`
7. `(x^3+4x+8)^2+3x.(x^2+4x+8)+2x^2`
`=x^6+8x^4+16x^3+16x^2+64x+64+3x^3+12x^2+24x+2x^2`
`=x^6+8x^4+19x^3+30x^2+88x+64`
8. `x^3 .(x^2-7)^2-36x`
`=x[x^2.(x^2-7)^2-36]`
`=x[(x^3-7x)^2-6^2]`
`=x.(x^3-7x-6).(x^3-7x+6)`
`=x.(x^3-6x-x-6).(x^3-x-6x+6)`
`=x.[x.(x^2-1)-6.(x+1)].[x.(x^2-1)-6.(x-1)]`
`=x.(x+1).[x.(x-1)-6].(x-1).[x.(x+1)-6]`
`=x.(x+1).(x-1).(x^2-3x+2x-6).(x^2+3x-2x-6)`
`=x.(x+1).(x-1).[x.(x-3)+2.(x-3)].[x.(x+3)-2.(x+3)]`
`=x.(x+1)(x-1).(x-2).(x+2).(x-3).(x+3)`
9. `x^5+x+1`
`=x^5-x^2+x^2+x+1`
`=x^2 .(x^3-1)+(x^2+x+1)`
`=x^2 .(x-1).(x^2+x+1)+(x^2+x+1)`
`=(x^2+x+1).(x^3-x^2+1)`
10. `x^8+x^4+1`
`=[(x^4)^2+2x^4+1]-x^4`
`=(x^4+1)^2-(x^2)^2`
`=(x^4-x^2+1).(x^4+x^2+1)`
`=[(x^4+2x^2+1)-x^2].(x^4-x^2+1)`
`=[(x^2+1)^2-x^2].(x^4-x^2+1)`
`=(x^2-x+1).(x^2+x+1).(x^4-x^2+1)
11. ` x^5-x^4-x^3-x^2-x-2`
`=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2`
`=x^4 .(x-2)+x^3 ,(x-2)+x^2 .(x-2)+x.(x-2)+(x-2)`
`=(x-2).(x^4+x^3+x^2+x+1)`
12. `x^9-x^7-x^6-x^5+x^4+x^3+x^2-1`
`=(x^9-x^7)-(x^6-x^4)-(x^5-x^3)+(x^2-1)`
`=x^7 .(x^2-1)-x^4 .(x^2-1)-x^3 .(x^2-1)+(x^2-1)`
`=(x^2-1).(x^7-x^4-x^3+1)`
`=(x-1)(x+1)(x^3-1)(x^4-1)`
`=(x-1)(x+1)(x^2+x+1)(x-1)(x^2-1)(x^2+1)`
`=(x-1)^2 .(x+1)(x^2+x+1)(x-1)(x+1)(x^2+1)`
`=(x-1)^3 .(x+1)^2 .(x^2+x+1)(x^2+1)`
13. `(x^2-x)^2-12(x^2-x)+24`
`=[ (x^2-x)^2-2.6(x^2-x)+6^2]-12`
`=(x^2-x+6)^2-12`
`=(x^2-x+6-\sqrt{12})(x^2-x+6+\sqrt{12})`
28 tháng 8 2019
\(a,\left(2x-3\right)^3-\left(x-1\right)^3-2\left(x-2\right)\left(x+2\right)\)
\(=8x^3+36x^2+27x+27-\left(x^3-3x^2+3x-1\right)-2\left(x^2-4\right)\)
\(=8x^3+36x^2+27x+27-x^3+3x^2-3x+1-2x^2+8\)
\(=7x^3+37x^2+24x+36\)
\(b,\left(x-3\right)^2-2\left(x+2\right)^3-4\left(x+3\right)\left(x-3\right)\)
\(=x^2-6x+9-2\left(x^3+6x^2+12x+8\right)-4\left(x^2-9\right)\)
\(=x^2-6x+9-2x^3-12x^2-24x-16-4x^2+36\)
\(=-15x^2-30x-2x^3+45\)
\(c,\left(2x-5\right)^3-4\left(x-2\right)\left(x+2\right)-2\left(x+1\right)^3\)
\(=8x^3-10x^2+50x-25-4\left(x^2-4\right)-2\left(x^3+3x^2+3x+1\right)\)
\(=8x^3-10x^2+50x-25-4x^2+16-2x^3-6x^2-6x-2\)
\(=6x^3-20x^2+44x-11\)
30 tháng 9 2020
Bài 1.
1) ( x - 1 )3 - x( x - 3 )2 + 1
= x3 - 3x2 + 3x - 1 - x( x2 - 6x + 9 ) + 1
= x3 - 3x2 + 3x - x3 + 6x2 - 9x
= 3x2 - 6x = 3x( x - 2 )
2) ( x + 2 )2 - x2( x + 6 )
= x2 + 4x + 4 - x3 - 6x2
= -x3 - 5x2 + 4x + 4
3) ( x + 2 )3 - ( x - 2 )3
= x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 )
= x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8
= 12x2 + 16 ( có phụ thuộc vào biến )
Bài 2.
1) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
2) ( x - 2 )3 - x( x + 1 )( x - 1 ) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5
<=> x3 + 12x - 8 - x3 + x = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
30 tháng 9 2020
Bài 1:
a) \(\left(x-1\right)^3-x\left(x-3\right)^2+1\)
\(=x^3-3x^2+3x-1-x^3+6x^2-9x+1\)
\(=3x^2-6x\)
b) \(\left(x+2\right)^2-x^2\left(x+6\right)\)
\(=x^2+4x+4-x^3-6x^2\)
\(=-x^3-5x^2+4x+4\)
c) \(\left(x+2\right)^3-\left(x-2\right)^3\)
\(=x^3+6x^2+12x+8-x^3+6x^2-12x+8\)
\(=12x^2+16\)
=> BT phụ thuộc vào biến
16 tháng 3 2020
\(x\left(x-1\right)=x\left(x+3\right)\)
\(x^2-x=x^2+3x\)
\(x^2+x-x^2-3x=0\)
\(-2x=0\)
\(x=0\)
\(\left(x-1\right)\left(x+3\right)=x^2-4\)
\(x^2+3x-x-3=x^2-4\)
\(x^2+2x-3=x^2-4\)
\(x^2+2x-3-x^2+4=0\)
\(2x+1=0\)
\(2x=1\)
\(x=\frac{1}{2}\)
cj lm nốt nha , cj lm ms ý nghĩa , cố lên !
= x ^2 - 5x + 6
học tốt