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10 tháng 9 2018
1/(x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)
=(x+2)(x-2+7)(x+3)(x-3+7)
=[(x+2)(x-2)+7x+14][(x+3)(x-3)+7x+21]
=(x2-4+7x+14)(x2-9+7x+21)
=(x2+10+7x)(x2+12+7x)
2/(x2+x)2+4(x2+x)-12
=(x2+x)2+4(x2+x)+22-16
=(x2+x+2)2-42
=(x2+x+2+4)(x2+x+2-4)
=(x2+x+6)(x2+x-2)
3/(x2+x+1)(x2+x+2)-12
=(x2+x+1)(x2+x+-1+3)-12
=(x2+x+1)(x2+x+-1)+3(x2+x+1)-12
=(x2+x)-1+3(x2+x)+3-12
=(x2+x)(x2+x+3)-10
làm đến đây thì mk bí, bạn giúp suy nghĩ nốt nha
4/nó là nhân tử sẵn rồi mà
9 tháng 8 2021
\(3/\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)
\(=\left(x^2+x+1\right)^2+x^2+x+1-12\)
\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-3\left(x^2+x+1\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-3\left(x^2+x+1+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
D
11 tháng 9 2018
\(\left(1+2\right),y^2-13y+12=y^2-12y-y-12=y\left(y-12\right)+\left(y-12\right)=\left(y+1\right)\left(y-12\right)\)
\(3,x^2-x-30=x^2-6x+5x-30=x\left(x-6\right)+5\left(x-6\right)=\left(x+5\right)\left(x-6\right)\)
\(4,y^2+y-42=y^2-6y+7y-42=y\left(y-6\right)+7\left(y-6\right)=\left(y+7\right)\left(y-6\right)\)
\(5,x^2+3x-10=x^2-2x+5x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)
\(6,x^2-8x+15=x^2-5x-3x+15=x\left(x-5\right)-3\left(x-5\right)=\left(x-3\right)\left(x-5\right)\)
PT
10 tháng 9 2018
1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=a\), ta có:
\(=\left(a+1\right)\left(a-1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+6x+x+6\right)\left(x^2+7x+16\right)\)
\(=\left[x\left(x+6\right)+\left(x+6\right)\right]\left(x^2+7x+16\right)\)
\(=\left(x+6\right)\left(x+1\right)\left(x^2+7x+16\right)\)
2) \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right).2+4-4-12\)
\(=\left(x^2+x+2\right)^2-16\)
\(=\left(x^2+x+2\right)^2-4^2\)
\(=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
3) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=a\), ta được
\(=a\left(a+1\right)-12\)
\(=a^2+a-12\)
\(=a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-12\)
\(=\left(a+\dfrac{1}{2}\right)^2-\dfrac{49}{4}\)
\(=\left(a+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2\)
\(=\left(a+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(a+\dfrac{1}{2}+\dfrac{7}{2}\right)\)
\(=\left(a-3\right)\left(a+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
4) \(\left(a^2-4\right)\left(a^2+6a+5\right)\)
\(=\left(a-2\right)\left(a+2\right)\left(a^2+5a+a+5\right)\)
\(=\left(a-2\right)\left(a+2\right)\left[a\left(a+5\right)+\left(a+5\right)\right]\)
\(=\left(a-2\right)\left(a+2\right)\left(a+5\right)\left(a+1\right)\)
NM
1
30 tháng 6 2017
Ta có : x4 + x3 + 6x2 + 5x + 5
= (x4 + 5x2) + (x3 + 5x) + (x2 + 5)
= x2(x2 + 5) + x(x2 + 5) + (x2 + 5)
= (x2 + 5)(x2 + x + 1)
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})`
6 tháng 7 2017
1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)
Câu 2: Y^2+4y+4=16=>(y+2)^2=16 =>y=2;hoặc y=6
pt mới 1: X^2+x+1/4=2+1/4=9/4=>x=2 hoặc x=-5/2
pt mới 2: X^+x+1/4=1/4-5/2=>vô nghiệm
@ngonhuminh Xin lỗi nhưng tui không hiểu. Làm gì có Y, là X mà