II.- Ecuaciones de segundo grado completas

En esta página tenemos ejercicios de ecuaciones de segundo grado completas, aquellas que tienen todos sus términos presentes: ax2+bx+c=0. Tendréis que aplicar distintos métodos de resolución (como la fórmula general o factorización) y resolver varios ejemplos paso a paso.

1.1641.165
\large x^{2}-7x+12=0\large x^{2}-9x+18=0
1.1661.167
\large x^{2}-9x+14=0\large x^{2}-6x+8=0
1.1681.169
\large x^{2}-6x+9=0\large x^{2}+6x-27=0
1.1701.171
\large x^{2}+6x=-9\large x^{2}+9=10x
1.1721.173
\large 2x^{2}+10x-48=0\large x^{2}-x=20
1.1741.175
\large x^{2}-5x+6=0\large x^{2}+8x+15=0
1.1761.177
\large x^{2}+10x+25=0\large 3x^{2}-39x+108=0
1.1781.179
\large 6x^{2}-37x+57=0\large 3x^{2}+2x=8
1.1801.181
\large x^{2}=5x+6\large 2x^{2}-5x+3=0
1.1821.183
\large x^{2}-7x+3\, \frac{1}{4}=0\large x^{2}-5\, \frac{3}{4}\, x=18
1.1841.185
\large 4x^{2}+12x=-9\large 5x^{2}+1=6x
1.1861.187
\large 12x^{2}-17x+6=0\large 10x^{2}+x=3
1.1881.189
\large 6x^{2}-12=x\large x^{2}-512x+6000=0
1.1901.191
\large 2x^{2}+7x+6=0\large 4x^{2}+3=8x
1.1921.193
\large x^{2}+x+\frac{1}{4}=0\large x(3x-2)=65
1.1941.195
\large x+\frac{1}{x-3}=5\large x^{2}-\frac{x}{2}=\frac{1}{3}-\frac{2x}{3}
1.1961.197
\large \frac{\, \, \, x^{2}}{4}+2=\frac{3x}{2}\large x+\frac{1}{x}=\frac{65}{8}
1.1981.199
\large x-2=\frac{4x-9}{x}\large \frac{x}{2}+\frac{3}{x}=\frac{x+13}{3x}
1.2001.201
\large x+1=\frac{8x}{x-1}-4\large \frac{x+2}{x+1}+\frac{x+1}{x+2}=\frac{13}{6}
1.2021.203
\large \frac{16x^{3}-12}{2x^{2}-4}=6+8x\large \frac{24}{x}+1=\frac{24}{x-2}
1.2041.205
\large \frac{2x}{x-2}-\frac{x+2}{2}=1\large x(x+1)+\frac{1}{4}(x+\frac{3}{2})=0
1.206
\large 8x+11-\frac{7}{x}=\frac{21+63x}{7}
1.207
\large 1+\frac{x+4}{3}=\frac{4x+7}{9}+\frac{7-x}{x-3}
1.208
\large \frac{\, \, x^{2}}{2}-\frac{3x}{5}=\frac{1}{5}-\frac{x}{6}
1.209
\large 2x^{2}+\frac{6}{5}=x(x+\frac{31}{5})
1.210
\large \frac{5x^{2}}{8}-\frac{3x}{5}=\frac{\, \, x^{2}}{8}-\frac{x}{6}+\frac{1}{5}
1.211
\large \frac{5x^{2}}{6}-\frac{x}{2}+\frac{3}{4}=8-\frac{2x}{3}-x^{2}+\frac{273}{12}
1.212
\large \frac{2x+1}{x-1}=\frac{5(x-1)}{x+1}

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