Question

VA rog rezolvare completa !!
$\frac{29}{5}$ $x^{2$-$\frac{1}{2}$$x^{2}$+ 2,(3)$x^{2}$-$\frac{7}{6}$$x^{2}$+ 2 $\frac{1}{10}$ $x^{2}$=

Answer 1

$\frac{29}{5}x^2- \frac{1}{2}x^2+2,(3)x^2- \frac{7}{6}x^2+2 \frac{1}{10}x^2= \frac{58x^2-5x^2}{10} + \frac{23-2}{9}x^2- \frac{7}{6}x^2+ \\ +\frac{2*10+1}{10}x^2= \frac{53x^2}{10} + \frac{21}{9}x^2- \frac{7}{6}x^2+ \frac{21}{10}x^2= \frac{74x^2}{10}+ \frac{42x^2-21x^2}{18}= \frac{74x^2}{10}+ \\+ \frac{21x^2}{18}= \frac{37x^2}{5}+ \frac{7x^2}{6} = \frac{222x^2+35x^2}{30}= \frac{257x^2}{30}$

Answer 2

=$\frac{29y}{5}$-$\frac{y}{2}$+$\frac{7y}{3}$-$\frac{7y}{6}$+$\frac{21y}{10}$=         unde y=x^2

prima, a doua si a patra fractie prin aducere la n.c. si prin simplificare da
 $\frac{37y}{5}$
a treia fr.si ultima da  $\frac{7y}{6}$

adunand cele doua fr. obtinute si n.c. vom avea   = $\frac{257y}{30}$

deci == 257x^2/30