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b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
⇔\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)
⇔\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)
⇔\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)
⇔\(9x-10=0\)
hay 9x=10
⇔\(x=\frac{10}{9}\)
Vậy: \(x=\frac{10}{9}\)
c) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
⇔\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)
⇔\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)
⇔\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)
⇔\(6x-3-5x+10-3x-21=0\)
⇔\(-2x-14=0\)
⇔\(-2x=14\)
hay x=-7
Vậy: x=-7
d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
⇔\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
⇔\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
⇔\(6x-18+7x-35-13x-4=0\)
⇔\(-21\ne0\)
Vậy: x∈∅
e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)
⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)
⇔\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)
⇔\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)
⇔\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)
⇔\(-12x+96=0\)
⇔\(-12x=-96\)
hay x=8
Vậy: x=8
a, Ta có : \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)
=> \(3\left(2x-1\right)-5\left(x-2\right)=x+7\)
=> \(6x-3-5x+10-x-7=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
b, Ta có : \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
=> \(\frac{3\left(x+3\right)}{6}-\frac{2\left(x-1\right)}{6}=\frac{x+5}{6}+\frac{6}{6}\)
=> \(3\left(x+3\right)-2\left(x-1\right)=x+5+6\)
=> \(3x+9-2x+2-x-5-6=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
=> \(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
=> \(4\left(x+5\right)+3\left(x+12\right)-5\left(x-2\right)=2x+66\)
=> \(4x+20+3x+36-5x+10-2x-66=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
a, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\frac{8x-3}{4}-\frac{6x-4}{4}=\frac{4x-2}{4}+\frac{x+3}{4}\)
\(8x-3-6x-4=4x-2+x+2\)
\(2x-7=5x\Leftrightarrow2x-5x+7=0\Leftrightarrow-3x=-7\Leftrightarrow x=\frac{7}{3}\)
b, \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
\(\frac{3x+9}{6}-\frac{2x-2}{6}=\frac{x+5}{6}+\frac{6}{6}\)
\(3x+9-2x-2=x+5+6\)
\(x+7=x+11\Leftrightarrow-4\ne0\)
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
c,|x-2|=2
=>\(\orbr{\begin{cases}x-2=2\\x-2=-2\end{cases}}\)
<=>\(\orbr{\begin{cases}x=4\\x=0\end{cases}}\)