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Упростите выражение: \(\frac{2x^{2}+1}{x^{3}-1}-\frac{x}{x^{2}+x+1}\)

Решение №12073: \(\frac{2x^{2}+1}{x^{3}-1}-\frac{x}{x^{2}+x+1}=\frac{2x^{2}+1}{(x-1)(x^{2}+x+1)}-\frac{x}{x^{2}+x+1}=\frac{2x^{2}+1-x(x-1)}{(x-1)(x^{2}+x+1)}=\frac{2x^{2}+1-x^{2}+x}{(x-1)(x^{2}+x+1)}=\frac{x^{2}+x+1}{(x-1)(x^{2}+x+1)}=\frac{1}{x-1}\)

Ответ: \(\frac{1}{x-1}\)

Упростите выражение: \(\frac{6c^{2}+48}{c^{3}+64}-\frac{3c}{c^{2}-4c+16}\)

Решение №12075: \(\frac{6c^{2}+48}{c^{3}+64}-\frac{3c}{c^{2}-4c+16}=\frac{6c^{2}+48}{(c+4)(c^{2}-4c+16)}-\frac{3c}{c^{2}-4c+16}=\frac{6c^{2}+48-3c(c+4)}{(c+4)(c^{2}-4c+16)}=\frac{6c^{2}+48-3c^{2}-12c}{(c+4)(c^{2}-4c+16)}=\frac{3c^{2}-12c+48}{(c+4)(c^{2}-4c+16)}=\frac{3(c^{2}-4c+16)}{(c+4)(c^{2}-4c+16)}=\frac{3}{c+4}\)

Ответ: \(\frac{3}{c+4}\)

Упростите выражение: \(c^{2}-cd+d^{2}-\frac{c^{3}-d^{3}}{c+d}\)

Решение №12077: \(c^{2}-cd+d^{2}-\frac{c^{3}-d^{3}}{c+d}=c^{2}-cd+d^{2}-\frac{(c-d)(c^{2}+cd+d^{2})}{(c+d)}=\frac{(c+d)(c^{2}-cd+d^{2})-c^{3}+d^{3}}{(c+d)}=\frac{c^{3}+d^{3}-c^{3}+d^{3}}{c+d}=\frac{2d^{3}}{c+d}\)

Ответ: \(\frac{2d^{3}}{c+d}\)

Упростите выражение: \(\frac{c}{c^{2}+3c+9}-\frac{1}{c-3}+\frac{27}{c^{3}-27}\)

Решение №12084: \(\frac{c}{c^{2}+3c+9}-\frac{1}{c-3}+\frac{27}{c^{3}-27}=\frac{c}{c^{2}+3c+9}-\frac{1}{c-3}+\frac{27}{(c-3)(c^{2}+3c+9)}=\frac{c(c-3)-(c^{2}+3c+9)+27}{(c-3)(c^{2}+3c+9)}=\frac{c^{2}-3c-c^{2}-3c-9+27}{c^{3}-27}=\frac{18}{c^{3}-27}\)

Ответ: \(\frac{18}{c^{3}-27}\)

Упростите выражение: \(1-\frac{2d-1}{4d^{2}-2d+1}-\frac{2d}{2d+1}\)

Решение №12085: \(1-\frac{2d-1}{4d^{2}-2d+1}-\frac{2d}{2d+1}=\frac{(2d+1)(4d^{2}-2d+1)-(2d-1)(2d+1)}{(2d+1)(4d^{2}-2d+1)}-\frac{2d(4d^{2}-2d+1)}{(2d+1)(4d^{2}-2d+1)}=\frac{8d^{3+1-(4d^{2}-1)-(8d^{3}-4d^{2}+2d)}{8d^{3}+1}=\frac{8d^{3}+1-4d^{2}+1-8d^{3}+4d^{2}-2d}{8d^{3}+1}=\frac{2-2d}{8d^{3}+1}\)

Ответ: \(\frac{2-2d}{8d^{3}+1}\)

Упростите выражение: \(\frac{1}{(2m-5n)^{2}}-\frac{2}{25n^{2}-4m^{2}}+\frac{1}{(5n+2m)^{2}}\)

Решение №12092: \(\frac{1}{(2m-5n)^{2}}-\frac{2}{25n^{2}-4m^{2}}+\frac{1}{(5n+2m)^{2}}=\frac{1}{(2m-5n)^{2}-\frac{2}{(5n-2m)(5n+2m)}+\frac{1}{(5n+2m)^{2}}=\frac{1}{(5n-2m)^{2}}-\frac{2}{(5n-2m)(5n+2m)}+\frac{1}{(5n+2m)^{2}}=\frac{(5n+2m)^{2}}{(5n-2m)^{2}(5n+2m)^{2}}-\frac{2(5n-2m)(5n+2m}{(5n-2m)^{2}(5n+2m)^{2}}+\frac{(5n-2m)^{2}}{(5n+2m)^{2}(5n-2m)^{2}}=\frac{25n^{2}+20mn+4m^{2}}{(5n-2m)^{2}(5n+2m)^{2}}-\frac{2(25n^{2}-4m^{2})}{(5n-2m)^{2}(5n+2m)^{2}}+\frac{25n^{2}-20mn+4m^{2}}{(5n-2m)^{2}(5n+2m)^{2}}=\frac{25n^{2}+20mn+4m^{2}-50n^{2}+8m^{2}+25n^{2}-20mn+4m}{(5n-2m)^{2}(5n+2m)^{2}}=\frac{16m^{2}}{(5n-2m)^{2}(5m+2m)^{2}}=\frac{16m^{2}}{(25m^{2}-4m^{2})^{2}}=\frac{16m^{2}}{(4m^{2}-25m^{2})^{2}}\)

Ответ: \(\frac{16m^{2}}{(4m^{2}-25m^{2})^{2}}\)

Докажите тождество: \(\frac{1}{2z^{2}+5z}-\frac{2}{25-10z}-\frac{4}{4z^{2}-25}=\frac{1}{5z}\)

Решение №12095: \(\frac{1}{2z^{2}+5z}-\frac{2}{25-10z}-\frac{4}{4z^{2}-25}=\frac{1}{5z}=\frac{1}{z(2z+5)}-\frac{2}{5(5-2z)}-\frac{4}{(2z-5)(2z+5)}=\frac{(5-2z) \cdot 5}{5z(5+2z)(5-2z)}-\frac{2z(5+2z)}{5(5-2z)(5+2z)z}+\frac{4 \cdot 5 \cdot z}{5z(5-2z)(5+2z)}=\frac{25-10z-10z-4z^{2}+20z}{5z(5+2z)(5-2z)}=\frac{25-4z^{2}}{5z(5+2z)(5-2z}=\frac{25-4z^{2}}{5z(25-4z^{2})}=\frac{1}{5z}\)

Ответ: NaN

Найдите значения переменных, при которых не определена дробь: \(\frac{a+3b}{\frac{a}{a+b}+\frac{b}{a-b}}\)

Решение №12099: \(\frac{a+3b}{\frac{a}{a+b}+\frac{b}{a-b}}=\frac{a+3b}{\frac{a(a-b)+b(a+b)}{(a+b)(a-b)}}=\frac{(a+3b)(a^{2}-b^{2})}{a^{2}-ab+ab+b^{2}}=\frac{(a+3b)(a^{2}-b^{2})}{a^{2}+b^{2}}; a+b \neq 0, a \neq -b; a-b \neq 0, a \neq b\)

Ответ: \(a \neq b\)

Упростите выражение: \((\frac{b}{c}+\frac{c}{b})^{2}+(\frac{c}{a}+\frac{a}{c})^{2}+(\frac{a}{b}+\frac{b}{a})^{2}-(\frac{b}{c}+\frac{c}{b})(\frac{c}{a}+\frac{a}{c})(\frac{a}{b}+\frac{b}{a})\)

Решение №12107: \((\frac{b}{c}+\frac{c}{b})^{2}+(\frac{c}{a}+\frac{a}{c})^{2}+(\frac{a}{b}+\frac{b}{a})^{2}-(\frac{b}{c}+\frac{c}{b})(\frac{c}{a}+\frac{a}{c})(\frac{a}{b}+\frac{b}{a})=(\frac{b^{2}+c^{2}}{bc})^{2}+(\frac{c^{2}+a^{2}}{ac})^{2}+(\frac{a^{2}+b^{2}}{ab})^{2}-(\frac{b^{2}+b^{2}}{ab})^{2} \cdot (\frac{c^{2}+a^{2}}{ac})(\frac{a^{2}+b^{2}}{ab})=\frac{(b^{2}+c^{2})^{2}}{b^{2}c^{2}}+\frac{(c^{2}+a^{2})^{2}}{a^{2}c^{2}}+\frac{(a^{2}+b^{2})^{2}}{a^{2}b^{2}}-\frac{(b^{2}+c^{2})(c^{2}+a^{2})(a^{2}+b^{2})}{a^{2}b^{2}c^{2}}=\frac{b^{4}+2b^{2}+c^{4}}{b^{2}c^{2}}+\frac{c^{4}+2c^{2}a^{2}+a^{4}}{a^{2}c^{2}}+\frac{a^{4}+2a^{2}b^{2}+b^{4}}{a^{2}b^{2}}-\frac{(b^{2}c^{2}+a^{2}b^{2}+c^{4}+a^{2}c^{2})(a^{2+b^{2})}{a^{2}b^{2}c^{2}}=\frac{a^{2}(b^{4}+2b^{2}c^{2}+c^{4})+b^{2}(c^{4}+2c^{2}a^{2}+a^{4})+c^{2}(a^{4}+2a^{2}b^{2}+b^{4}}{a^{2}b^{2}c^{2}}-\frac{a^{2}b^{2}c^{2}+a^{4}b^{2}+a^{2}c^{4}+a^{4}c^{2}+b^{4}c^{2}+a^{2}b^{4}+b^{2}c^{4}+a^{2}b^{2}c^{2}}{a^{2}b^{2}c^{2}}=\frac{6a^{2}b^{2}c^{2}-2a^{2}b^{2}c^{2}}{a^{2}b^{2}c^{2}}=\frac{4a^{2}b^{2}c^{2}}{a^{2}b^{2}c^{2}}=4\)

Ответ: \(4\)

Упростите выражение: \(\frac{4}{3y^{3}} \cdot \frac{y^{8}}{18}\)

Решение №12118: \(\frac{4}{3y^{3}} \cdot \frac{y^{8}}{18}=frac{4 \cdot y^{8}}{3y^{3} \cdot 18}=\frac{2 \cdot y^{5}}{3 \cdot 9}=\frac{2y^{5}}{27}\)

Ответ: \(\frac{2y^{5}}{27}\)