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

x=0

ban

a, (2x-3)⁴=(2x-3)⁶

=> (2x-3)⁶ : (2x-3)⁴=1

=> (2x-3)³=

=> 2x-3=1

=> 2x=4

=> x=2

b, (3x+5)³=(3x+5)²⁰¹⁶

=> (3x+5)²⁰¹⁶ : (3x+5)³=1

=> (3x+5)²⁰¹³=1

=> 3x+5=1

=> 3x=-4

=> x=-4/3

c, (2x+1)²⁰¹⁵=(2x+1)²⁰¹⁷

=> (2x+1)²⁰¹⁷ : (2x+1)²⁰¹⁵=1

=> (2x+1)²=1

=> 2x+1=1

=> 2x=0

=> x=0

a) \(36x^2-12x-36x^2+27x=30\)

                                                \(15x=30\)

                                                     \(x=2\)

b) \(5x-2x^2+2x^2-2x=15\)

                                          \(3x=15\)

                                             \(x=5\)

4(18-5x)-12(3x-7)=15(2x-16)-6(x+14)
<=>72-20x-36x+84=30x-240x-6x-84

<=>160x=-86

<=>x=-0.0375

a)|3x-2|=|3x+5|

x<-5/3 or x>=2/3

3x-2=3x+5=> loai

-5/3<=x<2/3

3x-2=-3x-5

6x=-3;x=-1/2(n)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(x^3y^5+3x^3y^5+...+\left(2k-1\right)x^3y^5=3249x^3y^5\)

\(\Leftrightarrow x^3y^5\left[1+2+3+...+\left(2k-1\right)\right]=3249x^3y^5\)

\(\Leftrightarrow1+3+5+...+\left(2k-1\right)=3249\)

\(\Leftrightarrow\frac{\left[\left(2k-1\right)+1\right].\left(\frac{\left(2k-1\right)-1}{2}+1\right)}{2}=3249\)

\(\Leftrightarrow\frac{2k.\left(k-1+1\right)}{2}=3249\)

\(\Leftrightarrow\frac{2k^2}{2}=3249\)

\(\Leftrightarrow k^2=3249=57^2\) ( ko xét k = - 57 vì theo quy luật thi k luôn dương )

\(\Rightarrow k=57\)

kết quả k = 57

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

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

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

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

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

Vậy \(x=0\) hoặc \(x=1\)

Chúc bạn học tốt ~ 

a)(x+3)³-x(3x+1)²+(2x+1)(4x²-2x+1-3x²)=54

\(\Rightarrow\)x³+9x²+27x+27-x(9x²+6x+1)+(2x+1)(x²-2x+1)=54

\(\Rightarrow\)x³+9x²+27x+27-9x³-6x²-x+2x³-4x²+2x+x²-2x+1=54

\(\Rightarrow\)-6x³+26x+28=54

\(\Rightarrow\)-6x³+26x=54-28

\(\Rightarrow\)-6x³+26x=26

\(\Rightarrow\)-6x³+26x-26=0

\(\Rightarrow\)-2(3x³+13x+14)