(3x+2)*(x+4)-(3x-1)*(x-5)=0
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30 tháng 1 2022
\(\dfrac{x+1}{x-1}+\dfrac{x-2}{x+2}+\dfrac{x-3}{x+3}+\dfrac{x+4}{x-4}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x-4\right)+\left(x-2\right)\left(x-1\right)\left(x+3\right)\left(x-4\right)+\left(x-3\right)\left(x-1\right)\left(x+2\right)\left(x-4\right)+\left(x+4\right)\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow4x^4+20x-96=0\)
\(\Leftrightarrow4\left(x^4+5x-24\right)=0\)
\(\Leftrightarrow x^4+5x-24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2,45...\\x=1,94...\end{matrix}\right.\)
Vậy: \(S=\left\{-2,45...;1,94...\right\}\)
KT
3
2 tháng 11 2021
\(\left(x-5\right)\left(x-7\right)=0\)
=>\(\left\{{}\begin{matrix}x-5=0\\x-7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)
Vậy S ={5,7}
JP
1
Gửi bạn
8 tháng 6 2019
Tìm x:
1. 3x (2x + 3) - (2x + 5).(3x - 2) = 8
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=0 \)
\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)
Vậy x = 5
2. 4x (x -1) - 3(x2 - 5) -x2 = (x - 3) - (x + 4)
\(\Leftrightarrow4x^2-4x-3x^2+15-x^2=x-3-x-4\)
\(\Leftrightarrow-4x+15=-7\)
\(\Leftrightarrow-4x=-22\Leftrightarrow x=\frac{11}{2}\)
Vậy x = \(\frac{11}{2}\)
3. 2 (3x -1) (2x +5) - 6 (2x - 1) (x + 2) = -6
\(\Leftrightarrow2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)
\(\Leftrightarrow12x^2+30x-4x-10-12x^2-24x+6x+12=-6\)
\(\Leftrightarrow8x=-8\Leftrightarrow x=-1\)
Vậy x = -1
4. 3 ( 2x - 1) (3x - 1) - (2x - 3) (9x - 1) - 3 = -3
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-18x^2+2x+27x-3-3=-3\)
\(\Leftrightarrow18x^2-6x-9x+3-18x^2+2x+27x-6=-3\)
\(\Leftrightarrow14x=0\Leftrightarrow x=0\)
Vậy x = 0
5. (3x - 1) (2x + 7) - ( x + 1) (6x - 5) = (x + 2) - (x - 5)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=7\)
\(\Leftrightarrow18x=9\Leftrightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
6. 3xy (x + y) - (x + y) (x2 + y2 + 2xy) + y3 = 27
\(\Leftrightarrow3x^2y+3xy^2-\left(x+y\right)^3+y^3=27\)
\(\Leftrightarrow3x^2y+3xy^2-x^3-y^3-3x^2y-3xy^2+y^3=27\)
\(\Leftrightarrow-x^3=27\)
\(\Leftrightarrow x=-3\)
Vậy x = -3
7. 3x (8x - 4) - 6x (4x - 3) = 30
\(\Leftrightarrow24x^2-12x-24x^2+12x=30\)
\(\Leftrightarrow0=30\) ( vô lý)
Vậy pt vô nghiệm
8. 3x (5 - 2x) + 2x (3x - 5) = 20
\(\Leftrightarrow15x-6x^2+6x^2-10x=20\)
\(\Leftrightarrow5x=20\Leftrightarrow x=4\)
Vậy x = 4
3 tháng 2 2017
a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)
\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)
\(\Leftrightarrow-297-99x=0\)
\(\Leftrightarrow x=3\)
Vậy \(n_0\) của PT là: x=3
b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)
\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)
\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)
\(\Leftrightarrow-64-36x=250-30x\)
\(\Leftrightarrow-6x=314\)
\(\Leftrightarrow x=-\frac{157}{3}\)
Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)
c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)
\(\Leftrightarrow-3x=4x-\frac{23}{5}\)
\(\Leftrightarrow7x=\frac{23}{5}\)
\(\Leftrightarrow x=\frac{23}{35}\)
Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)
d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow x=-\frac{5}{12}\)
Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)
14 tháng 3 2020
1) \(\frac{4x-8}{2x^2+1}=0\)
<=> \(\frac{4\left(x-2\right)}{2x^2+1}=0\)
<=> 4(x - 2) = 0
<=> x - 2 = 0
<=> x = 2
2) \(\frac{x^2-x-6}{x-3}=0\)
<=> \(\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
<=> x + 2 = 0
<=> x = -2
3) xem ở đây Câu hỏi của Vương Thanh Thanh
4) \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
<=> \(\frac{12}{\left(1+3x\right)\left(1-3x\right)}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
<=> 12 = (1 - 3x)2 - (1 + 3x2)
<=> 12 = 1 - 6x + 9x2 - 1 - 6x - 9x2
<=> 12 = -12x
<=> x = -1
5) ĐKXĐ: \(x\ne1,x\ne3\)
\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
<=> \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 5)(x - 3) = (x + 1)(x - 1) - 8
<=> x2 - 3x + 5x - 15 = x2 - x + x - 1 - 8
<=> x2 + 2x - 15 = x2 - 9
<=> x2 + 2x - 15 - x2 = -9
<=> 2x - 15 = -9
<=> 2x = -9 + 15
<=> 2x = 6
<=> x = 3 (ktm)
=> pt vô nghiệm
6) ĐKXĐ: \(x\ne\pm2\)
\(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\)
<=> \(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{\left(x-2\right)\left(x+2\right)}+1\)
<=> (x + 1)(x + 2) - 5(x - 2) = 12 + (x - 2)(x + 2)
<=> x2 + 2x + x + 2 - 5x + 10 = 12 + x2 + 2x - 2x - 4
<=> x2 - 2x + 12 = x2 + 8
<=> x2 - 2x + 12 - x2 = 8
<=> -2x + 12 = 8
<=> -2x = 8 - 12
<=> -2x = -4
<=> x = 2 (ktm)
=> pt vô nghiệm
9 tháng 7 2017
Ai chẳng biết chuyển vế đổi dấu :v
a) \(x-7=4x+10\)
\(x-4x=10+7\)
\(-3x=17\)
\(x=\dfrac{17}{-3}\)
Vậy \(x=\dfrac{17}{-3}\)
b) \(2x+5=-3x+7\)
\(2x+3x=7-5\)
\(5x=2\)
\(x=\dfrac{2}{5}\)
Vậy \(x=\dfrac{2}{5}\)
c) \(x-\left(3x+7\right)=6x-1\)
\(x-3x-7=6x-1\)
\(-2x-7=6x+1\)
\(-7-1=6x+2x\)
\(-8=8x\)
\(x=\dfrac{-8}{8}=-1\)
Vậy \(x=-1\)
d) \(x+\left(5x-1\right)=15\)
\(x+5x-1=15\)
\(6x=15+1\)
\(6x=16\)
\(x=\dfrac{16}{6}=\dfrac{8}{3}\)
Vậy \(x=\dfrac{8}{3}\)
HT
9 tháng 7 2017
1 , x - 7 = 4x + 10
x - 4x = 10 + 7
- 3x = 17
x = 17 : ( - 3 )
x = \(\dfrac{-17}{3}\)
2 , 2x + 5 = -3x + 7
2x + 3x = 7 -5
5x = 2
x = 2 : 5
x =\(\dfrac{2}{5}\)
3 , x - ( 3x + 7 ) = 6x - 1
x - 3x - 7 = 6x - 1
x - 3x -6x = -1 +7
-8x = 6
x = 6 : ( -8 )
x = \(\dfrac{-3}{4}\)
4 , x + ( 5x -1 ) = 15
x + 5x - 1 = 15
x + 5x = 15 + 1
6x = 16
x = 16 : 6
x = \(\dfrac{8}{3}\)
5 , / x + 1 / = / 2x - 5 /
TH 1 : x + 1 = 2x - 5
x - 2x = -5 -1
- x = -4
= > x = 4
TH 2 : -x -1 = -2x + 5
-x + 2x = 5 + 1
x = 6
6 , / 3x + 8 / - / x -10 / = 0
3x + 8 - x + 10 = 0
3x - x = 0 - 10 - 8
2 x = -18
x = -18 : 2
x = - 9
16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
Bạn tham khảo cách làm!
`(3x+2)*(x+4)-(3x-1)*(x-5)=0`
\(\Leftrightarrow3x\left(x+4\right)+2\left(x+4\right)-3x\left(x-5\right)+1\left(x-5\right)=0\)
\(\Leftrightarrow3x^2+3x\cdot4+2x+2\cdot4-3x^2+3x\cdot5+x-5=0\)
\(\Leftrightarrow3x^2+12x+2x+8-3x^2+15x+x-5=0\)
\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(12x+2x+15x+x\right)+\left(8-5\right)=0\)
\(\Leftrightarrow30x+3=0\)
\(\Leftrightarrow30x=0-3\)
`=> 30x=-3`
`-> x=-3 \div 30`
`-> x=-1/10 `