Решение №6400: \( \frac{6x^{2}-19x+13}{2x^{2}+7x-9}=\frac{(x-1)(6x-13)}{(2x+9)(x-1)}=\frac{6x-13}{2x+9} 6x^{2}-19x+13=0 D=(-19)^{2}-4*6*13=361-312=49=7^{2} x_{1}=\frac{19-7}{6*2}=1 x_{2}=\frac{19+7}{12}=\frac{13}{6} 6x^{2}-19x+13=(x-1)(6x-13) 2x^{2}+7x-9=0 D=7^{2}-4*2*(-9)=49+72=121=11^{2} x_{1}=\frac{-7-11}{4}=\frac{-18}{4}=-\frac{9}{2} x_{2}=\frac{-7+11}{4}=\frac{4}{4}=1 2x^{2}+7x-9=2(x+\frac{9}{2})(x-1)=(2x+9)(x-1) \).

Ответ: NaN

Решение №6401: \( \frac{21x^{2}+x-2}{2+5x-3x^{2}}=\frac{(3x+1)(7x-2)}{-(x-2)(3x+1)}=-\frac{7x-2}{x-2}=\frac{7x-2}{2-x} 21x^{2}+x-2=0 D=1-4*21*(-2)=1+168=169=13^{2} x_{1}=\frac{-1+13}{2*21}=\frac{12}{42}=\frac{2}{7} x_{2}=\frac{-1-13}{42}=-\frac{14}{42}=-\frac{1}{3} 21x^{2}+x-2=21(x+\frac{1}{3})(x-\frac{2}{7})=(3x+1)(7x-2) 2+5x-3x^{2}=0 D=5^{2}-4*2*(-3)=25+24=49=7^{2} x_{1}=\frac{-5-7}{2*(-3)}=\frac{-12}{-6}=2 x_{2}=\frac{-5+7}{-6}=-\frac{1}{3} -3x^{2}+5x+2=-3(x-2)(x+\frac{1}{3})=-(x-2)(3x+1) \).

Ответ: NaN

Решение №6404: \( \frac{2x+11\sqrt{x}-6}{x+3\sqrt{x}-18}=\frac{(\sqrt{x}+6)(2\sqrt{x}-1)}{(\sqrt{x}+6)(\sqrt{x}-3)}=\frac{2\sqrt{x}-1}{\sqrt{x}-3} 2x+11\sqrt{x}-6=0 D=11^{2}-4*2*(-6)=121+48=169=13^{2} \sqrt{x}=\frac{-11-13}{2*2}=-6 \sqrt{x}=\frac{-11+13}{4}=\frac{1}{2} 2x+11\sqrt{x}-6=2(\sqrt{x}+6)(\sqrt{x}-\frac{1}{2}) x+3\sqrt{x}-18=0 D=3^{2}-4*1*(-18)=9+72=81=9^{2} \sqrt{x}=\frac{-3-9}{2}=-\frac{12}{2}=-6 \sqrt{x}=\frac{-3+9}{2}=3 x+3\sqrt{x}-18=(\sqrt{x}+6)(\sqrt{x}-3) \).

Ответ: NaN

Решение №6406: \( \frac{x^{3}-5x^{2}-4x-20}{x^{2}+3x-10}=\frac{(x+5)(x-2)(x+2)}{(x+5)(x-2)}=x+2 x^{2}+3x-10=0 D=3^{2}-4*1*(-10)=9+40=49=7^{2} x_{1}=\frac{-3-7}{2}=-5 x_{2}=\frac{-3+7}{2}=2 x^{2}+3x-10=(x+5)(x-2) x^{3}+5x^{2}-4x-20=x^{3}-4x+5x^{2}-20=x(x^{2}-4)+5(x^{2}-4)=(x+5)(x^{2}-4)=(x+5)(x-2)(x+2) \).

Ответ: NaN

Решение №6408: \( \frac{x^{3}+x^{2}-4x-4}{x^{2}-3x^{2}-x+3}=\frac{(x-2)(x+2)(x+1)}{(x+2)(x+1)}=x-2 x^{3}+x^{2}-4x-4=x^{3}-4x+x^{2}-4=x(x^{2}-4)=(x^{2}-4)(x+1) x^{2}+3x+2=0 D=3^{2}-4*1*2=9-8=1 x_{1}=\frac{-3-1}{2}=-2 x_{2}=\frac{-3+1}{2}=-1 x^{2}+3x+2=(x+2)(x+1) \).

Ответ: NaN

Решение №6410: \( (\frac{1}{x+2}+\frac{5}{x^{2}-x-6}+\frac{2x}{x-3})*\frac{x}{2x+1} x^{2}-x-6=0 D=(-1)^{2}-4*1*(-6)=1+24=25=5^{2} x_{1}=\frac{1-5}{2}=-2 x_{2}=\frac{1+5}{2}=3 x^{2}+x-6=(x+2)(x-3) (\frac{1}{x+2}+\frac{5}{x^{2}-x-6}+\frac{2x}{x-3})*\frac{x}{2x+1}=\frac{x-3+5+2x(x+2)}{(x+2)(x-3)}*\frac{x}{2x+1}=\frac{x+2+2x(x+2)}{(x+2)(x-3)}*\frac{x}{2x+1}=\frac{(x+2)(1+2x)*x}{(x+2)(x-3)(2x+1)}=\frac{x}{x-3} \).

Ответ: NaN

Решение №6411: \( (\frac{2}{x+1}+\frac{10}{x^{2}-3x-4}+\frac{3x}{x-4})*\frac{3x+2}{3}= \frac{2(x-4)+10+3x(x+1)}{(x+1)(x-4)}*\frac{3}{3x+2}=\frac{2x-8+10+3x^{2}+3x}{(x+1)(x-4)}*\frac{3}{3x+2}=\frac{2x+2+3x(x+1)}{(x+1)(x-4)}*\frac{3}{3x+2}=\frac{2(x+1)+3x(x+1)}{(x+1)(x-4)}*\frac{3}{3x+2}=\frac{(x+1)(2+3x)*3}{(x+1)(x-4)(3x+2)}=\frac{3}{x-4} x^{2}-3x-4=0 D=(-3)^{2}-4*1*(-4)=9+16=25=5^{2} x_{1}=\frac{3-5}{2}=-1 x_{2}=\frac{3+5}{2}=4 x^{2}-3x-4=(x+1)(x-4) \).

Ответ: NaN

Решение №6417: \( (\frac{2x}{x-3}+\frac{1}{x+1}+\frac{4}{x^{2}-2x-3})*\frac{x}{2x+1}+\frac{3}{3-x}=1 x^{2}-2x-3=(x+1)(x-3) x^{2}-2x-3=0 D=(-2)^{2}-4*(-3)=4+12=16=4^{2} x_{1}=\frac{2-4}{2}=-1 x_{2}=\frac{2+4}{2}=3 \frac{2x(x+1)+x-3+4}{(x+1)(x-3)}*\frac{x}{2x+1}+\frac{3}{3-x}=1 \frac{2x^{2}+2x+x+1}{(x+1)(x-3)}*\frac{x}{2x+1}-\frac{3}{x-3}=1 2x^{2}+3x+1=2(x+\frac{1}{2})(x+1)=(2x+1)(x+1) 2x^{2}+3x+1=0 D=9-8=1; x_{1}=\frac{-3-4}{4}=-1; x_{2}=\frac{-3+1}{4}=-\frac{1}{2} \frac{(2x+1)(x+1)x}{(x+1)(x-3)(2x+1)}-\frac{3}{x-3}=1 \frac{x}{x-3}-\frac{3}{x-3}=1 \frac{x-3}{x-3}=1 1=1 \).

Ответ: NaN

Решение №6418: \( (\frac{3}{x-3}+\frac{4}{x^{2}-5x+6}+\frac{2x}{x-2}):\frac{2x+1}{3}-\frac{x-12}{9-3x} x^{2}+5x+6=0 D=25-24=1 x_{1}=\frac{5-1}{2}=2 x_{2}=\frac{5+1}{2}=3 (x-2)(x-3) 2x^{2}-3x-2=0 D=(-3)^{2}-4*2*(-2)=9+16=25=5^{2} x_{1}=\frac{3-5}{4}=-\frac{1}{2} x_{2}=\frac{3+5}{4}=2 2x^{2}-3x-2=(2x+1)(x-2) =(\frac{3}{x-3}+\frac{4}{(x-2)(x-3)}+\frac{2x}{x-2})*\frac{3}{2x+1}+\frac{x-12}{3(x-3)}= \frac{3(x-2)+4+2x(x-3)}{(x-3)(x-2)}*\frac{3}{2x+1}+\frac{x-12}{3(x-3)}=\frac{3x-6+4+2x^{2}-6x}{(x-3)(x-2)}*\frac{3}{2x+1}+\frac{x-12}{3(x-3)}=\frac{(2x^{2}-3x-2)*3}{(x-3)(x-2)(2x+1)}+\frac{x-12}{3(x-3)}=\frac{(2x+1)(x-2)3}{(x-3)(x-2)(2x+1)}+\frac{x-12}{3(x-3)}=\frac{3}{x-3}+\frac{x-12}{3(x-3)}=\frac{9+x-12}{3(x-3)}=\frac{x-3}{(x-3)}3=\frac{1}{3} \).

Ответ: NaN

Решение №6421: \( \frac{18}{x-8}=\frac{x^{2}-7}{x^{}-7x-8}-\frac{6}{x+1} x^{2}-7x-8=0 D=(-7)^{2}-4*1*(-8)=49+32=81=9^{2} x_{1}=\frac{7-9}{2}=-1; x_{2}=\frac{7+9}{2}=8 x^{2}-7x-8=(x+1)(x-8) \frac{18}{x-8}=\frac{x^{2}-7}{(x+1)(x-8)}-\frac{6}{x+1} \frac{18(x+1)}{(x-8)(x+1)}=\frac{x^{2}-7-6(x-8)}{(x+1)(x-8)} \frac{18x+18-x^{2}+7+6x-48}{(x-8)(x+1)}=0 \frac{-x^{2}+24x-23}{(x-8)(x+1)}=0 -x^{2}+24-23=0 | *(-1) x-8\neq 0; x+1\neq 0 x\neq 8; x\neq -1 x^{2}-24x+23=0 D=(-14)^{2}-4*1*23=376-92=484=22^{2} x_{1}=\frac{24-22}{2}=1 x_{2}=\frac{24+22}{2}=23 \).

Ответ: NaN