Tìm x
( 2x – 1 )6 = ( 2x – 1 )8
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17 tháng 5 2019
a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84
<=> -80x = -480
<=> x = 6
b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1
<=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1
<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0
<=> -4x = 0
<=> x = 0
c) 2(5x-8)-3(4x-5)=4(3x-4)+11
= 10x - 16 - 12x + 15 = 12x - 16 + 11
= -14x = -4
= x =\(\frac{2}{7}\)
d) 5x-3{4x-2[4x-3(5x-2)]}=182
= 5x - 3 . [4x - 2(4x - 15x + 6)]
= 5x - 3 . (4x - 8x + 30x - 12)
= 5x - 12x + 24x - 90x + 36
= -73x + 36 = 182
=> -73x = 182 - 36 = 146
=> x = 146 : (-73) = -2
~Hok tốt~
14 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\) phải k bạn nhỉ? `11/8` k có bậc lũy thừa nào `=5` á.
`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)
`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{27}{8}\)
`=>`\(\left(\dfrac{x}{2}-1\right)^3=\left(-\dfrac{3}{2}\right)^3\)
`=>`\(\dfrac{x}{2}-1=-\dfrac{3}{2}\)
`=>`\(\dfrac{x}{2}=-\dfrac{3}{2}+1\)
`=>`\(\dfrac{x}{2}=-\dfrac{1}{2}\)
`=> x=1`
Vậy, `x=1`
`b)`
\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\0,75-1\dfrac{1}{2}x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\-\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-3\\-3x\cdot100=2\cdot75\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x\cdot100=150\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x=1,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x={-3/2; -1/2}.`
NT
1
6 tháng 8 2018
\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
6 tháng 8 2018
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
19 tháng 8 2017
1.\(\left(2x-1\right)\left(x-1\right)=0\)
2. \(\left(x+1\right)\left(2x-5\right)=0\)
3. \(\left(2x-1\right)\left(x+4\right)=0\)
4. Vô nghiệm vì VT > 0 \(\forall\)x
TN
1
20 tháng 8 2021
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
8 tháng 7 2019
a) 2x - 3 = x + 1/2
<=> 2x - 3 = 1/2x + 1/2
<=> 2x - 3 - 1/2x = 1/2
<=> 3/2x - 3 = 1/2
<=> 3/2x = 1/2 + 3
<=> 3/2x = 7/2
<=> x = 7/2 : 3/2
<=> x = 7/3
=> x = 7/3
8 tháng 7 2019
\(a,2x-3=x+\frac{1}{2}\)
\(2x-3=\frac{1}{2}x+\frac{1}{2}\)
\(2x-3-\frac{1}{2}x=\frac{1}{2}\)
\(\frac{3}{2}x-3=\frac{1}{2}\)
\(\frac{3}{2}x=\frac{1}{2}+3\)
\(\frac{3}{2}x=\frac{7}{2}\)
\(x=\frac{7}{2}:\frac{3}{2}\)
\(x=\frac{14}{6}=\frac{7}{3}\)
\(\)B làm tương tự
PP
1
MN
27 tháng 1 2020
Ta có :
\(\left(2x^2-3x+1\right)-\left(2x^2-3x+4\right)=0\)
\(\Leftrightarrow2x^2-3x+1-2x^2+3x-4=0\)
\(\Leftrightarrow-3=0\left(ktm\right)\)
\(\Leftrightarrow x\in\varnothing\)
(2x - 1)6 = (2x - 1)8
=> (2x - 1)8 - (2x - 1)6 = 0
=> (2x - 1)6 . [(2x - 1)2 - 1] = 0
\(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\\left(2x-1\right)^2=1\end{cases}\Rightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}}\)
• Nếu 2x - 1 = 0 \(\Rightarrow x=\frac{1}{2}\)
• Nếu (2x - 1)2 = 1 \(\Rightarrow\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=2\\2x=0\end{cases}\Rightarrow}}\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy, \(x\in\left\{0;\frac{1}{2};1\right\}\)
\(1=\left(2x-1\right)^2\)
\(1=4x^2-4x+1\)
\(4x\left(x-1\right)=0\)
\(\orbr{\begin{cases}x-1=0\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
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