(x - 3)² = (2x - 1)²

=> |x - 3| = |2x - 1|

=> x - 3 = 2x-1 hoặc x - 3 = -(2x - 1)

x - 2x = -1 + 3 hoặc x - 3 = -2x + 1

-x = 2 hoặc x + 2x = 1 + 3

x = -2 hoặc 3x = 4

x = -2 hoặc x = 4 : 3

x = -2 hoặc x = 1,(3)

a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=-5\)

Vậy phương trình có nghiệm x = - 5 .

a, \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Rightarrow3x^2-3x-\left(3x^2-2x\right)=5\)

\(\Rightarrow3x^2-3x-3x^2+2x=5\)

\(\Rightarrow5x=5\Rightarrow x=1\)

Câu b,c làm tương tự! Cứ tách ra là làm được à!

b ) \(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)

\(\Leftrightarrow16x^2+8x-8x^2+12x=-40\)

\(\Leftrightarrow20x=-40\)

\(\Leftrightarrow x=-2\)

Vậy phương trình có nghiệm x = - 2

\(b)\) \(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)

\(\Leftrightarrow\)\(\left(2x-1\right)^{2010}.\left(2x-1\right)^2=\left(2x-1\right)^{2010}\)

\(\Leftrightarrow\)\(\left(2x-1\right)^2=1\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=2\\2x=0\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{2}\\x=\frac{0}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)

Vậy \(x=0\) hoặc \(x=1\)

Chúc bạn học tốt ~ 

a) \(36x^2-12x-36x^2+27x=30\)

                                                \(15x=30\)

                                                     \(x=2\)

b) \(5x-2x^2+2x^2-2x=15\)

                                          \(3x=15\)

                                             \(x=5\)

a) \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)

b) \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)

c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)

d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)

vậy phương trình vô nghiệm

a) (x-1)² = 0

<=> x-1 = 0

<=> x = 1

b) (x-2)² - 1 = 0

<=> (x-2)² = 1

<=> \(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c) (2x-1)³ = -8

<=> (2x-1)³ = -2³

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = \(-\dfrac{1}{2}\)

d) (x+2)² + 1 = 0

<=> (x+2)² = -1

<=> x+2 = -1

<=> x = -3

a, \(\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2=0^2\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ......

b, \(\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy .....

c, \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy ...

d, \(\left(x+2\right)^2+1=0\)

\(\Leftrightarrow\left(x+2\right)^2=-1\)

\(\Leftrightarrow\) ko tìm dc giá trị của x thỏa mãn (do \(\left(x+2\right)^2\ge0\))

Khi \(f\left(x\right)=-2\)

Ta có : \(f\left(-2\right)=\left(-2\right)^3+2.\left(-2\right)^2+a.\left(-2\right)+1\)

                           \(=-8+8+a.\left(-2\right)+1\)

                              \(=a.\left(-2\right)+1\)

kun ơi, nó là;

x³ + 2x² +ax +1 = -2

bài này bn ấy cho thiếu x=? thì làm sao tính dc a? pk kun

\(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)

\(\Leftrightarrow\left(2x-1\right)^{2012}-\left(2x-1\right)^{2010}=0\)

\(\Leftrightarrow[\left(2x-1\right)^{2010}.\left(2x-1\right)^2]-\left(2x-1\right)^{2010}=0\)\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-1\right)^2-1]=0\)

\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-1-1\right)\left(2x-1+1\right)]=0\)

\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-2\right)2x]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^{2010}\\2x\left(2x-2\right)=0\end{matrix}\right.=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x=0\\2x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\\x=1\end{matrix}\right.\)

Vậy x \(\in\left\{\dfrac{1}{2};0;1\right\}\)

\((2x-1)^{2012} = (2x-1)^{2010} \)

\(​​\)\(\Leftrightarrow\)\((2x-1)^{2012} - (2x-1)^{2010} = 0\)

\(\Leftrightarrow\)\((2x-1)^{2010} . [(2x-1)^{2} - 1] = 0\)

\(\Leftrightarrow\)\((2x-1)^{2010} . (2x-2).2x = 0\)

\(\Leftrightarrow\)\(4 . (2x-1)^{2010} . (x-1) . x = 0\)

\(\Rightarrow\)\(\left[{}\begin{matrix}\left(2x-1\right)^{2010}=0\\x-1=0\\x=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)

\(Vậy \) \(x= \)\(\dfrac{1}{2}\); \(x=1\) \(hay\) \(x=0\)