Вправа 634 алгебра Бевз гдз 7 клас
№ 6341) $\frac{1}{8} x-2 x^3=\frac{1}{8} x\left(1-16 x^2\right)=\frac{1}{8} x(1-4 x)(1+4 x)$;
б) $\frac{2}{3} \mathrm{a}^4-\frac{3}{8} \mathrm{a}^2 \mathrm{x}^4=\frac{16}{24} \mathrm{a}^4-\frac{9}{24} \mathrm{a}^2 \mathrm{x}^4=\mathrm{a}^2\left(\frac{16}{24} \mathrm{a}^2-\frac{9}{24} \mathrm{x}^4\right)=$ $=\frac{a^2}{24}\left(16 a^2-9 x^4\right)=\frac{a^2}{24}\left(4 a-3 x^2\right)\left(4 a+3 x^2\right)$;
в) $\frac{1}{4} \mathrm{ac}^2-\mathrm{ac}+\mathrm{a}=\mathrm{a}\left(\frac{1}{4} \mathrm{c}^2-\mathrm{c}+1\right)=\mathrm{a}\left(\frac{1}{2}-1\right)^2$;
г) $2 \frac{1}{4} x^6-\frac{4}{9} x^2 y^4=\frac{9}{4} x^6-\frac{4}{9} x^2 y^4=x^2\left(\frac{9}{4} x^4-\frac{4}{9} y^4\right)=$ $=x^2\left(\frac{3}{2} x^2-\frac{2}{3} y^2\right)\left(\frac{3}{2} x^2+\frac{2}{3} y^2\right)$.
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