Tìm x biết: (2x-1)^2012=(2x-1)^2010
Những câu hỏi liên quan
Bài 1 tìm x
a) 3x(2x-7)-(6x+1)(x-15)-2010=0
b) 2x(x-2012)-x+2012=0
c) (x+2y)(x^2-2xy+4y^2-8y^3+27=0
d) x^3+x^2-2x-8=0
tìm x:
a) 3x(2x-7)-(6x+1)(x-15)-2010=0
b) 2x(x-2012)-x+2012=0
c) (x+2y)(x^2-2xy+4y^2)-8y^3+27=0
d)x^3+x^2-2x-8=0
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 }
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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 = \(\left\{2012;\frac{1}{2}\right\}\)
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tìm x
a) 3x(2x - 7) - (6x + 1)(x-15)-2010 =0
b) 2x(x-2012)-x + 2012 = 0
b) 2x(x - 2012) - x + 2012 = 0
<=> 2x(x - 2012) - (x - 2012)= 0
<=> (x - 2012)(2x - 1) = 0
<=> \(\begin{bmatrix} x - 2012= 0 & & \\ 2x - 1 = 0 & & \end{bmatrix} \)
<=> \(\begin{bmatrix} x = 2012 & & \\ x = \frac{1}{2} & & \end{bmatrix} \)
Vậy x = 2012 và x= \(\frac{1}{2}\)
pn bỏ dấu ngoặc vuông bên phải nhé
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tính min B= /x-2014/ + /x-2013/ + /x-2012/ + /x-2011/ + /x-2010/ + 2014TÌM X : /x-2/ + /3-2x/ = 2x-1
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LÀM GẤP GIÚP MK HA, MK TICK CHO
Tìm x biết
a, |2x+3| -3 |4-x|= -5
b, |x-2010| + |x-2012| + |x-2014| = 2
chứng minh
a) đa thức x2011-2x2010+3x2009-2 chia hết cho đa thức x-1
b)đa thức (x+1)^2012-x^2012-2x-1 chia hết cho đa thức x(x+1)(2x+1)
tìm x thuộc z (2x-3)2010 = (2x-3)2012
Tim x,biet
\(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)
\(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)
\(\Leftrightarrow\left(2x-1\right)^{2012}-\left(2x-1\right)^{2010}=0\)
\(\Leftrightarrow[\left(2x-1\right)^{2010}.\left(2x-1\right)^2]-\left(2x-1\right)^{2010}=0\)\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-1\right)^2-1]=0\)
\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-1-1\right)\left(2x-1+1\right)]=0\)
\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-2\right)2x]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^{2010}\\2x\left(2x-2\right)=0\end{matrix}\right.=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x=0\\2x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\\x=1\end{matrix}\right.\)
Vậy x \(\in\left\{\dfrac{1}{2};0;1\right\}\)
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\((2x-1)^{2012} = (2x-1)^{2010} \)
\(\)\(\Leftrightarrow\)\((2x-1)^{2012} - (2x-1)^{2010} = 0\)
\(\Leftrightarrow\)\((2x-1)^{2010} . [(2x-1)^{2} - 1] = 0\)
\(\Leftrightarrow\)\((2x-1)^{2010} . (2x-2).2x = 0\)
\(\Leftrightarrow\)\(4 . (2x-1)^{2010} . (x-1) . x = 0\)
\(\Rightarrow\)\(\left[{}\begin{matrix}\left(2x-1\right)^{2010}=0\\x-1=0\\x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)
\(Vậy \) \(x= \)\(\dfrac{1}{2}\); \(x=1\) \(hay\) \(x=0\)
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tìm x biết: |2x|^2011 +|2x-1|^2010 =1