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14 tháng 10 2019
1,\(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)
\(=2x^2-10x-x^2+4x-4-x^2+9\)
\(=\left(2x^2-x^2-x^2\right)+\left(-10x+4x\right)+\left(-4+9\right)\)
\(=-6x+5\)
2,\(\left(x+1\right)^2-3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)
\(=x^2+2x+1-3\left(x^2-25\right)-\left(4x^2-4x+1\right)\)
\(=x^2+2x+1-3x^2+75-4x^2+4x-1\)
\(=-6x^2+6x+75\)
3,\(\left(x-1\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-1\right)^3-\left(x^3-27\right)\)
\(=x^3-3x^2+3x-1-x^3+27\)
\(=-3x^2+3x+26\)
4,\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+2\right)^3\)
\(=\left(x^3+125\right)-\left(x^3+6x^2+12x+8\right)\)
\(=x^3+125-x^3-6x^2-12x-8\)
\(=-6x^2-12x+117\)
5,\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)^2+\left(x+1\right)^2\)
\(=2x^2-14x-\left(x+3\right)\left(x^2-4x+4\right)+x^2+2x+1\)
=\(2x^2-14x-x^3+4x^2-4x-3x^2+12x-12+x^2+2x+1\)
\(=-x^3+4x^2-4x+1\)
6,\(\left(2x+5\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)
\(=2x^2-6x+5x-15-x^2+x-5x+5-x^2+8x-16\)
\(=3x-26\)
7,\(\left(x+5\right)\left(x-5\right)\left(x+2\right)-\left(x+2\right)^3\)
=\(\left(x^2-25\right)\left(x+2\right)-x^3-6x^2-12x-8\)
\(=x^3+2x^2-25x-50-x^3-6x^2-12x-8\)
\(=-4x^2-27x-58\)
Nếu đúng thì tick cho mk nha ^_^
BL
3 tháng 3 2020
Bài 1:
1. \(x-8=3-2\left(x+4\right)\)
\(x-8=3-2x-8\)
\(3x=3\Rightarrow x=1\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(2x+6-3x+3=2\)
\(-x+9=2\Rightarrow x=7\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(4x-20-3x+1=x-19\)
\(0x=0\Rightarrow x=0\)
4. \(7-\left(x-2\right)=5\left(2x-3\right)\)
\(7-x+2=10x-15\)
\(-11x=-24\Rightarrow x=\frac{24}{11}\)
5. \(32-4\left(0,5y-5\right)=3y+2\)
\(32-2y+20=3y+2\)
\(-5y=-50\Rightarrow y=10\)
6. \(3\left(x-1\right)-x=2x-3\)
\(3x-3-x=2x-3\)
\(0x=0\Rightarrow x=0\)
Bài 2:
1. \(\frac{2-x}{3}=\frac{3-2x}{5}\)
\(\frac{\left(2-x\right)5}{15}-\frac{\left(3-2x\right)3}{15}=0\)
\(\frac{10-5x-9+6x}{15}=0\)
\(x+1=0\Rightarrow x=-1\)
2. \(\frac{3-4x}{4}=\frac{x+2}{5}\)
\(\frac{5\left(3-4x\right)}{20}-\frac{4\left(x+2\right)}{20}=0\)
\(\frac{15-20x-4x-8}{20}=0\)
\(7-24x=0\)
\(24x=7\Rightarrow x=\frac{7}{24}\)
NN
2
10 tháng 1 2017
1/ 3(x - 4) - (8 - x) = 12 => 3x - 12 - 8 + x = 12 => 4x = 32 => x = 8
2/ 4(x - 5) - 3(x - 7) = -9 => 4x - 20 - 3x + 21 = -9 => x = -10
3/ 7(x - 3) -5(3 - x) = 11x - 5 => 7x - 21 - 15 + 5x = 11x - 5 => x = 31
4/ -5(2 - x) + 4(x - 3) = 10x => -10 + 5x + 4x - 12 = 10x => x = -22
D
10 tháng 1 2017
1) 3.(x-4)-(8-x) = 12
=> 3x-12-8+x = 12
=> 4x - 20 = 12
=> 4x = 12 +20 = 32
=> x = 32: 4 = 8
2) 4.(x-5)-3(x-7) = -9
=> 4x-20-3x+21 = -9
=>x +1 = -9
=>x = -9-1= -10
3) 7.(x-3)-5(3-x) = 11x-5
=> 7x-21-15+5x = 11x -5
=> 12x - 36 = 11x -5
=> 12x-11x = -5+36
hay x = 31
4) -5.(2-x)+4.(x-3) = 10x
=> -10+5x+4x-12 = 10x
=> -22 + 9x = 10x
=> -22 = 10x -9x
hay x = -22
XONG 
3 tháng 7 2022
a: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
=>1+3x-6=3-x
=>3x-5=3-x
=>4x=8
hay x=2(loại)
b: \(\Leftrightarrow8-x-8\left(x-7\right)=-26\)
=>8-x-8x+56=-26
=>-9x+64=-26
=>-9x=-90
hay x=10(nhận)
c: \(\dfrac{1}{x-2}+\dfrac{1}{x-3}=\dfrac{2}{x-1}\)
\(\Leftrightarrow\dfrac{x-3+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=2\left(x^2-5x+6\right)\)
\(\Leftrightarrow2x^2-5x-2x+5=2x^2-10x+12\)
=>-7x+10x=12-5
=>3x=7
hay x=7/3(nhận)
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})`
HY
6 tháng 2 2020
\(a.-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-5\right)^2\\ =-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-20x+25\right)\\ =-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+20x-25\\ =-3x^3+x^3-12x^2+3x^2-4x^2-12x-x+20x-3-25\\ =-2x^3-13x^2+7x-28\\ \)\(b.2\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\\ =2\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\\ =2\left(x^3+2x^2-9x-18\right)-x^3+3x+x^2-3-5x^3-40x^2-80x-x^2+10x-25\\ =2x^3+4x^2-18x-36-x^3+3x+x^2-3-5x^3-40x^2-80x-x^2+10x-25\\ =2x^3-x^3-5x^3+4x^2+x^2-40x^2-x^2-18x+3x-80x+10x-36-3-25\\ =-4x^3-36x^2-85x-64\)
EC
6 tháng 9 2016
a)\(\left(x-5\right)\left(x+5\right)=\left(x-2\right)\)
\(x^2-25-x+2=0\)
\(x^2-23-x=0\)
\(x.\left(x-1\right)=23\)
Bài này vô lý quá
b)\(\left(3-2x\right)^2-\left(x-5\right)\left(4x+3\right)=2\left(x+5\right)\)
\(9-12x+4x^2-4x^2-3x+20x+15=2x+10\)
\(5x+24=2x+10\)
\(5x+24-2x-10=0\)
\(3x-14=0\)
\(3x=14\)
\(x=\frac{14}{3}\)
Vậy \(x=\frac{14}{3}\)
EC
6 tháng 9 2016
c)\(\left(7-x\right)\left(2x-5\right)-\left(7-x\right)2x=3\left(-5+x\right)\)
\(\left(7-x\right)\left[\left(2x-5\right)-2x\right]=\left(-15\right)+3x\)
\(5x-35=\left(-15\right)+3x\)
\(5x-35+15+3x=0\)
\(8x-20=0\)
\(8x=20\)
\(x=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}\)
PA
19 tháng 2 2017
a)
\(\frac{1}{x-2}+3=3-\frac{x}{x-2}\)
<=> \(\frac{1}{x-2}=-\frac{x}{x-2}\)
<=> x = - 1
Vậy S = {- 1}
b)
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
<=> \(\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\)
<=> (x + 5)2 - (x - 5)2 = 20
<=> (x + 5 - x + 5)(x + 5 + x - 5) = 20
<=> 10 . 2x = 20
<=> x = 20 : 20
<=> x = 1
Vậy S = {1}
c)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{2\left(x-3\right)\left(x+1\right)}\)
<=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{2x}{2\left(x-3\right)\left(x+1\right)}\)
<=> x(x + 1) + x(x - 3) = 2x
<=> x2 + x + x2 - 3x - 2x = 0
<=> 2x2 - 4x = 0
<=> 2x(x - 2) = 0
<=> \(\left[\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy S = {0; 2}
19 tháng 2 2017
Bạn có sửa đề cũng phải báo chứ:
làm vậy có ai đó vào thấy đúng copy pas đến chỗ khác thành sai=> mất kiểm soát.
Tam sao thất bản mà.
Ngàn Sao thì ....
p/s: xem bài chứng tỏ bạn là đời f(0)
hiihi nói vui nhé xin đừng chém.
(X-5).(x-3)
= x^2 - 5x - 3x + 15
= x^2 - 8x +15
`(x-5)(x-3)`
`=x (x-3)-5(x-3)`
`= x^2 - 3x-5x+15`
`=x^2 + (-3x-5x)+15`
`= x^2 - 8x+15`