\(1,\left(2x+1\right)\left(x-1\right)-x\left(2x-3\right)+3=0\)

\(\Rightarrow2x^2-2x+x-1-\left(2x^2-3x\right)+3=0\)

\(\Rightarrow2x^2-2x+x-1-2x^2+3x+3=0\)

\(\Rightarrow2x=-2\Rightarrow x=-1\)

\(2,\left(x^2+x-2\right)\left(x^2-x-2\right)-x^2\left(x^2-2\right)+8=0\)

\(\Rightarrow[\left(x^2\right)^2-\left(x-2\right)^2]-x^2\left(x^2-2\right)+8=0\)

\(\Rightarrow x^4-\left(x^2-4x+4\right)-x^4+2x^2+8=0\)

\(\Rightarrow x^4-x^2+4x-4-x^4+2x^2+8=0\)

\(\Rightarrow x^2+4x+4=0\)

\(\Rightarrow\left(x+2\right)^2=0\Rightarrow x=-2\)

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

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)

1. \(\left(5x-1\right)\left(x-1\right)+6x-44=0\)

=> \(5x^2-5x-x+1+6x-44=0\)

=> \(5x^2-43=0\)

=> \(5x^2=43\)

=> \(x^2=43:5\)

=> \(x^2=\frac{43}{5}\)

=> \(\orbr{\begin{cases}x=\sqrt{\frac{43}{5}}\\x=-\sqrt{\frac{43}{5}}\end{cases}}\)

2. \(\left(2x+1\right)\left(4x^2-2x+1\right)+9=0\)

=> \(8x^3+1+9=0\)

=> \(8x^3+10=0\)

=> \(8x^3=-10\)

=> \(x^3=-10:8\)

=> \(x^3=-\frac{5}{4}\)

=> ko bt

2,\(\left(2x+1\right)\left(4x^2-2x+1\right)+9=0\)

\(\Rightarrow8x^3-4x^2+4x^2-2x+1+9=0\)

\(\Rightarrow8x^3+10=0\)

\(\Rightarrow x^3=\frac{-5}{4}\) 

\(\Rightarrow x=\sqrt[3]{\frac{-5}{4}}\) 

Vậy....

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

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

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

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

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

Vậy \(x=0\) hoặc \(x=1\)

Chúc bạn học tốt ~ 

ta có : \(5^x.5^{\left(x+1\right)}.5^{\left(x+2\right)}\le\dfrac{100...0}{2^{18}}\) \(\Leftrightarrow5^{3x+3}\le\dfrac{10^{18}}{2^{18}}=5^{18}\)

\(\)\(\Leftrightarrow3x+3\le18\Leftrightarrow x\le5\) vậy \(x\le5\)