Bài 1:
a)
\(\left(x+1\right)\cdot\left(2x-3\right)=\left(2x-1\right)\cdot\left(x+5\right)\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow2x^2-x-3-2x^3-9x+5=0\\ \Leftrightarrow2-10x=0\\ \Rightarrow x=\frac{2}{10}=\frac{1}{5}\)
b)
\(\left(x-2\right)^3+\left(3x-1\right)\cdot\left(3x+1\right)=\left(x+1\right)^3\\ \Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\\ \Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\\ \Leftrightarrow9x-9=0\\ \Rightarrow x=1\)
c)
\(2x\cdot\left(x+2\right)^2-8x^2=2\cdot\left(x-2\right)\cdot\left(x^2+2x+4\right)\\ \Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\\ \Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\\ \Leftrightarrow8x+16=0\\ \Rightarrow x=-\frac{16}{8}=-2\)
d)
\(\left(x+1\right)\cdot\left(x^2-x+1\right)-2x=x\cdot\left(x+1\right)\cdot\left(x-1\right)\\ \Leftrightarrow x^3+1-2x=x^3-x\\ \Leftrightarrow x^3+1-2x-x^3+x=0\\ \Leftrightarrow1-x=0\\ \Rightarrow x=1\)
Bài 2:
a)
\(x^2+\left(x+2\right)\cdot\left(11x-7\right)=4\\ \Leftrightarrow x^2+11x^2-7x+22x-14-4=0\\ \Leftrightarrow12x^2+15x-18=0\\ \Leftrightarrow3\cdot\left(4x^2+5x-6\right)=0\\ \Leftrightarrow3\cdot\left(4x^2+8x-3x-6\right)=0\\ \Leftrightarrow3\cdot\left[4x\cdot\left(x+2\right)-3\cdot\left(x+2\right)\right]=0\\ \Leftrightarrow3\cdot\left(4x-3\right)\cdot\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{4}\\x=-2\end{matrix}\right.\)
b)
\(2x^3+5x^2-3x=0\\ \Leftrightarrow x\cdot\left(2x^2+5x-3\right)=0\\ \Leftrightarrow x\cdot\left(2x^2+6x-x-3\right)=0\\ \Leftrightarrow x\cdot\left[2x\cdot\left(x+3\right)-\left(x+3\right)\right]=0\\ \Leftrightarrow x\cdot\left(2x-1\right)\cdot\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=-3\end{matrix}\right.\)
Bài 3:
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\\ \Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0\\ \Leftrightarrow\frac{4x-10x-15x-3x+60}{12}=0\\ \Leftrightarrow\frac{60-24x}{12}=0\\ \Rightarrow60-24x=0\\ \Rightarrow x=\frac{60}{24}=\frac{5}{2}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\\ \Leftrightarrow\frac{8x-3}{4}-\frac{6x-4}{4}-\frac{4x-2}{4}-\frac{x+3}{4}=0\\ \Leftrightarrow8x-3-6x+4-4x+2-x-3=0\\ \Leftrightarrow-3x=0\\ \Rightarrow x=0\)
c)
\(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\\ \Leftrightarrow\frac{15x-15}{30}-\frac{2x+2}{30}-\frac{10x-65}{30}=0\\ \Leftrightarrow15x-15-2x-2-10x+65=0\\ \Leftrightarrow3x+48=0\\ \Rightarrow x=-\frac{48}{3}=-16\)
d)
\(\frac{3\cdot\left(3-x\right)}{8}+\frac{2\cdot\left(5-x\right)}{3}=\frac{1-x}{2}-2\\ \Leftrightarrow\frac{27-9x}{24}+\frac{80-16x}{24}-\frac{12-12x}{24}+\frac{48}{24}=0\\ \Leftrightarrow27-9x+80-16x-12+12x+48=0\\ \Leftrightarrow143-13x=0\\ \Rightarrow x=\frac{143}{13}=11\)
e)
\(\frac{3\cdot\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\cdot\left(x-7\right)\\ \Leftrightarrow\frac{45x-18}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60x-420}{12}=0\\ \Leftrightarrow45x-18-24-28x+60x-420=0\\ \Leftrightarrow77x-462=0\\ \Rightarrow x=\frac{462}{77}=6\)
f)
\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\\ \Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1-\frac{x+5}{31}+1-\frac{x+7}{29}+1=0\\ \Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\\ \Leftrightarrow\left(x+36\right)\cdot\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\\ \Rightarrow x+36=0\\ \Rightarrow x=-36\)
Bài 3 câu g:
\(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{20}\)
\(\Leftrightarrow\left(\frac{x-10}{1994}-1\right)+\left(\frac{x-8}{1996}-1\right)+\left(\frac{x-6}{1998}-1\right)+\left(\frac{x-4}{2000}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2000}{4}-1\right)+\left(\frac{x-1998}{6}-1\right)+\left(\frac{x-1996}{8}-1\right)+\left(\frac{x-1994}{10}-1\right)\)
\(\Leftrightarrow\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
\(\Leftrightarrow\left(x-2004\right)\cdot\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}\right)=\left(x-2004\right)\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{10}\right)\)
\(\Leftrightarrow\left(x-2004\right)\cdot\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)\)
\(\Rightarrow x-2004=0\\ \Rightarrow x=2004\)
Bài 4:
a)
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\\ \Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}-\frac{7x^2-14x-5}{15}=0\\ \Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\\ \Leftrightarrow36x+3=0\\ \Rightarrow x=-\frac{3}{36}==-\frac{1}{12}\)
b)
\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\cdot\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\\ \Leftrightarrow\frac{8x^2-32x+32}{24}-\frac{12x^2-27}{24}+\frac{4x^2-32x+64}{24}=0\\ \Leftrightarrow8x^2-32x+32-12x^2+27+4x^2-32x+64=0\\ \Leftrightarrow96-64x=0\\ \Rightarrow x=\frac{96}{64}=\frac{3}{2}\)