Tim x
1/ 2.x2012 = x2014
2/ 2x-1 .8 + 2x-1 . 3 = 44
3/ 22 .3x = 108
K
Khách
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9 tháng 11 2017
Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0
=> 6x2 - 21x - (6x2 + x - 90x - 15) - 2010 = 0
=> 6x2 - 21x - 6x2 + 89x + 15 - 2010 = 0
=> 68x - 1995 = 0
?
b) 2x(x - 2012) - x + 2012 = 0
=> 2x(x - 2012) - (x - 2012) = 0
=> (x - 2012) (2x - 1) = 0
⇔[
| x−2012=0 |
| 2x−1=0 |
⇔[
| x=2012 |
| 2x=1 |
⇔[
| x=2012 |
| x=12 |
Vậy x = {2012;12 }
5 tháng 11 2017
Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0
=> 6x2 - 21x - (6x2 + x - 90x - 15) - 2010 = 0
=> 6x2 - 21x - 6x2 + 89x + 15 - 2010 = 0
=> 68x - 1995 = 0
?
b) 2x(x - 2012) - x + 2012 = 0
=> 2x(x - 2012) - (x - 2012) = 0
=> (x - 2012) (2x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2012=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2012\\2x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2012\\x=\frac{1}{2}\end{cases}}\)
Vậy x = \(\left\{2012;\frac{1}{2}\right\}\)
14 tháng 7 2023
5: =>4x^2-1/9=0
=>(2x-1/3)(2x+1/3)=0
=>x=1/6 hoặc x=-1/6
6: =>x-1=2
=>x=3
7:=>(2x-1)^3=-27
=>2x-1=-3
=>2x=-2
=>x=-1
8: =>1/8(x-1)^3=-125
=>(x-1)^3=-1000
=>x-1=-10
=>x=-9
3: =>(5x-5)^2-4=0
=>(5x-7)(5x-3)=0
=>x=3/5 hoặc x=7/5
4: =>(5x-1)^2=0
=>5x-1=0
=>x=1/5
1: =>(3x-1)(2x-1)=0
=>x=1/3 hoặc x=1/2
2: =>x^2(2x-3)-4(2x-3)=0
=>(2x-3)(x^2-4)=0
=>(2x-3)(x-2)(x+2)=0
=>x=3/2;x=2;x=-2
NN
14 tháng 7 2023
`@` `\text {Answer}`
`\downarrow`
`1,`
\(2x\left(3x-1\right)+1-3x=0\)
`<=> 2x(3x - 1) - 3x + 1 = 0`
`<=> 2x(3x - 1) - (3x - 1) = 0`
`<=> (2x - 1)(3x-1) = 0`
`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy, `S = {1/2; 1/3}`
`2,`
\(x^2\left(2x-3\right)+12-8x=0\)
`<=> x^2(2x - 3) - 8x + 12 =0`
`<=> x^2(2x - 3) - (8x - 12) = 0`
`<=> x^2(2x - 3) - 4(2x - 3) = 0`
`<=> (x^2 - 4)(2x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `S = {+-2; 3/2}`
`3,`
\(25\left(x-1\right)^2-4=0\)
`<=> 25(x-1)(x-1) - 4 = 0`
`<=> 25(x^2 - 2x + 1) - 4 = 0`
`<=> 25x^2 - 50x + 25 - 4 = 0`
`<=> 25x^2 - 15x - 35x + 21 = 0`
`<=> (25x^2 - 15x) - (35x - 21) = 0`
`<=> 5x(5x - 3) - 7(5x - 3) = 0`
`<=> (5x - 7)(5x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy, `S = {7/5; 3/5}`
`4,`
\(25x^2-10x+1=0\)
`<=> 25x^2 - 5x - 5x + 1 = 0`
`<=> (25x^2 - 5x) - (5x - 1) = 0`
`<=> 5x(5x - 1) - (5x - 1) = 0`
`<=> (5x - 1)(5x-1)=0`
`<=> (5x-1)^2 = 0`
`<=> 5x - 1 = 0`
`<=> 5x = 1`
`<=> x = 1/5`
Vậy,` S = {1/5}.`
13 tháng 12 2021
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
27 tháng 7 2018
I don't now
sorry
...................
nha
KT
27 tháng 7 2018
b) \(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)
\(\Leftrightarrow\)\(\left(3x-2\right)\left(3x+3\right)^2\left(3x+8\right)+144=0\)
Đặt: \(3x+3=a\)pt trở thành:
\(\left(a-5\right)a^2\left(a+5\right)+144=0\)
\(\Leftrightarrow\)\(a^4-25a^2+144=0\)
\(\Leftrightarrow\)\(\left(a-4\right)\left(a-3\right)\left(a+3\right)\left(a+4\right)=0\)
đến đây bạn tìm a rồi tính x
c) \(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)
\(\Leftrightarrow\)\(\left(4x-5\right)\left(4x-6\right)\left(4x-4\right)-72=0\)
Đặt \(4x-5=a\)pt trở thành:
\(a\left(a-1\right)\left(a+1\right)-72=0\)
\(\Leftrightarrow\)\(a^3-a-72=0\)
p/s: ktra lại đề
d) \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
\(\Leftrightarrow\)\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)=0\)
\(\Leftrightarrow\)\(\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)
\(\Leftrightarrow\)\(\left(11x+2011\right)^2=0\)
đến đây làm nốt
16 tháng 12 2022
1: =>(x+3)(x-5)=0
=>x=5 hoặc x=-3
2: =>(x-1)(5x-1)=0
=>x=1/5 hoặc x=1
5: =>(x-4)*x=0
=>x=0 hoặc x=4
10: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
9: =>(x-2)(x-4)=0
=>x=2 hoặc x=4
7: =>(x-6)(2x-1)=0
=>x=1/2 hoặc x=6
8: =>(2x-1)(3x-12)=0
=>x=4 hoặc x=1/2
25 tháng 10 2023
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
1/ 2.x2012 = x2014
<=> 2.x2012 - x2014 = 0
<=> x2012 . ( 2 - 1) = 0
<=> x2012 . 1 = 0
<=> x2012 = 0
<=> x = 0
Vậy x = 0
2/ 2x-1 .8 + 2x - 1 .3 = 44
<=> 2x-1 . ( 8 + 3 ) = 44
<=> 2x-1 . 11 = 44
<=> 2x-1 = 44 : 11
<=> 2x-1 = 4
<=> 2x-1 = 22
<=> x - 1 = 2
<=> x = 3
Vậy x = 3
3/ 22 . 3x = 108
<=> 4 . 3x = 108
<=> 3x = 108 : 4
<=> 3x = 27
<=> 3x = 33
<=> x = 3
Vậy x = 3
~ Hok tốt ~