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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)
13 tháng 6 2018
1/ (x+1)(-3)+5(x-4)=-3
\(\Leftrightarrow\)-3x - 3 + 5x - 20= -3
\(\Leftrightarrow\)2x - 23=-3
\(\Leftrightarrow\)x=10
2/3(5x-1) -x (x+1)+x2=14
\(\Leftrightarrow\)15x - 3 - x2 -x + x2=14
\(\Leftrightarrow\)14x=17
\(\Leftrightarrow\)x=17/14
3/2(x-1)-x(3-x)=x2
\(\Leftrightarrow\)2x - 2 - 3x + x2=x2
\(\Leftrightarrow\)2x-3x+x2-x2=2
\(\Leftrightarrow\)x= -2
4/ 3x(x+5)-2(x+5)=3x2
\(\Leftrightarrow\)3x2 + 15x - 2x - 10=3x2
\(\Leftrightarrow\)13x = 10
\(\Leftrightarrow\)x=10/13
5/ 4x(x+2)+x(4-x)=3x2+12
\(\Leftrightarrow\)4x2 + 8x + 4x - x2 = 3x2 + 12
\(\Leftrightarrow\)12x=12
\(\Leftrightarrow\)x=1
3 tháng 5 2021
=) vào ngay quả bảng phá dấu GTTĐ, cay thế :<
a, \(3x+\frac{2x}{3}-3=\frac{5}{2}x-2\Leftrightarrow\frac{18x+4x-18}{6}=\frac{15x-12}{6}\)
\(\Rightarrow22x-18=15x-12\Leftrightarrow7x=6\Leftrightarrow x=\frac{6}{7}\)
Vậy pt có nghiệm x = 6/7
b, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
\(\Leftrightarrow\frac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\frac{x+7}{12}\)
\(\Rightarrow18x+9-10x-6+4x+4=x+7\)
\(\Leftrightarrow12x+7=x+7\Leftrightarrow11x=0\Leftrightarrow x=0\)
Vậy pt có nghiệm là x = 0
c, \(\frac{3x}{x-3}-\frac{x-3}{x+3}=2\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{3x\left(x+3\right)-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow3x^2+9x-x^2+6x-9=2\left(x^2-9\right)\)
\(\Leftrightarrow2x^2+15x-9=2x^2-18\Leftrightarrow15x+9=0\Leftrightarrow x=-\frac{9}{15}=-\frac{3}{5}\)
Vậy pt có nghiệm là x = -3/5
d, Sửa đề : \(\frac{x+10}{2003}+\frac{x+6}{2007}+\frac{x+2}{2011}+3=0\)
\(\Leftrightarrow\frac{x+10}{2003}+1+\frac{x+6}{2007}+1+\frac{x+2}{2011}+1=0\)
\(\Leftrightarrow\frac{x+2013}{2003}+\frac{x+2013}{2007}+\frac{x+2013}{2011}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2003}+\frac{1}{2007}+\frac{1}{2011}\ne0\right)=0\Leftrightarrow x=-2013\)
Vậy pt có nghiệm là x = -2013
3 tháng 5 2021
e, \(4\left(x+5\right)-3\left|2x-1\right|=10\)
\(\Leftrightarrow4x+20-3\left|2x-1\right|=10\Leftrightarrow-3\left|2x-1\right|=-10-4x\)
\(\Leftrightarrow\left|2x-1\right|=\frac{10+4x}{3}\)
ĐK : \(\frac{10+4x}{3}\ge0\Leftrightarrow10+4x\ge0\Leftrightarrow x\ge-\frac{10}{4}=-\frac{5}{2}\)
TH1 : \(2x-1=\frac{10+4x}{3}\Rightarrow6x-3=10+4x\Leftrightarrow2x=13\Leftrightarrow x=\frac{13}{2}\)( tm )
TH2 : \(2x-1=\frac{-10-4x}{3}\Rightarrow6x-3=-10-4x\Leftrightarrow10x=-7\Leftrightarrow x=-\frac{7}{10}\)( tm )
f, để mình xem lại đã, quên cách phá GTTĐ rồi :v :>
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
\(3/\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)
\(=\left(x^2+x+1\right)^2+x^2+x+1-12\)
\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-3\left(x^2+x+1\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-3\left(x^2+x+1+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
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})`
NH
11 tháng 11 2020
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
MC
Tìm x
a) (12x-5)(3x-1)-(18x-1)(2x+3)=5
b) (x+2)(x-3)-(x-2)(x+5)=2(x+3)
c) (2x+3)(2x-1)-(2x+5)-(2x-3)=12
0
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)
20 tháng 3 2020
\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\left(x\ne1;x\ne3\right)\)
\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-4x+3}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\frac{x^2-1}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x-15-x^2+1+8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow2x-4=0\)
<=> 2x=4
<=> x=2 (tmđk)
Vậy x=2
20 tháng 3 2020
b) \(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{x+1}{x-2}-\frac{5}{x+2}-\frac{12}{\left(x-2\right)\left(x+2\right)}-1=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-4}{x^2-4}=0\)
\(\Leftrightarrow\frac{x^2+3x+2-5x+10-12-x^2+4}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-2x+2}{\left(x-2\right)\left(x+2\right)}=0\)
=> -2x+2=0
<=> -2x=-2
<=> x=1 (tmđk)
Vậy x=1
(x + 1) (x + 3) - x (x + 5) = 12
(x + 1) (x + 3) - x (x + 5) - 12 = 0
x2 + 3x + x + 3 - x2 - 5x - 12 = 0
-x - 9 = 0
9 - x = 0
=> 9 - x = 0
* 9 - x = 0 => x = 9
(sai thì thôi)
#Học tốt!!!
ko cần k sai nữa đâu, mik biết nó sai rồi