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Ejercicios – BCH1-CCSS

23/11/2021

$\displaystyle \log 3x=\log 6+2 \log x$

$\displaystyle 5 \log x=3 \log x+2 \log 6$

$\displaystyle \log (2x+3)-\log (x-2)=\log 36$

$\displaystyle \log (4-5x)+\log (2x-2)=\log \left(2x-x^{2}\right)+1$

$\displaystyle 2 \log (2x-2)-\log (x-1)=1$

$\displaystyle \log \left(65-x^{2}\right)=3 \log (5-x)$

$\displaystyle \log x=\log 6+2 \log \frac{x}{3}$

$\displaystyle \log 10^{\sqrt{20 x+320}}=10 \sqrt{x}$

$\displaystyle 3 \log _{x} 2+\log _{x} 4=-5$

$\displaystyle \log \sqrt{7x+51}-1=\log 9-\log \sqrt{2x-67}$

22/11/2021

$\displaystyle \sqrt{x+2}-x+4=0$

$\displaystyle \sqrt{x+1}-\sqrt{4 x-3}=-5$

$\displaystyle \sqrt{x+4}+\sqrt{x-1}=5$

$\displaystyle x+\sqrt{10+x^{2}}=5$

$\displaystyle \sqrt{x+7}+\sqrt{2x}=\sqrt{x+23}$

$\displaystyle x^{2}-\sqrt{3x^{2}-2}=4$

$\displaystyle \frac{x-1}{\sqrt{x}}=x-1$

$\displaystyle \sqrt{3 \sqrt{16-x}}=\sqrt{2x-5}$

$\displaystyle \sqrt{2+\sqrt{x-4}}=\sqrt{12-x}$

$\displaystyle \frac{\sqrt{2x-1}}{4}=\frac{3}{\sqrt{2x-1}}$

$\displaystyle \frac{1}{\sqrt{1-x}+\sqrt{x}}=\frac{1}{\sqrt{x}}$

$\displaystyle \sqrt{x+\sqrt{4x+40}}=5$

21/11/2021

$\displaystyle x+\frac{2}{x}=-3$

$\displaystyle \frac{11x+11}{9}=2x-\frac{12}{2-x}-7$

$\displaystyle \frac{4}{x+2}+\frac{4}{x}=3$

$\displaystyle \frac{6x+7}{x+3}=\frac{x}{x-1}$

$\displaystyle \frac{2x}{3}+\frac{2x+3}{x-1}=\frac{7}{3x-3}$

$\displaystyle \frac{2x}{x-2}+\frac{3x}{x+2}=\frac{6x^{2}}{x^{2}-4}$

$\displaystyle \frac{2}{x^{2}-1}+\frac{3x}{x-1}=\frac{x}{x+1}$

$\displaystyle \frac{x^{3}-8}{x-1}=\frac{24x+16}{x+2}$

$\displaystyle \frac{1}{x}+\frac{2}{x}+\frac{3}{x^{2}}=\frac{9}{4}$

$\displaystyle \frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}=-1$

$\displaystyle \frac{1}{2x}+\frac{2}{3x}+\frac{3}{4x}=\frac{13}{36}$

$\displaystyle 2x-\frac{12}{2-x}=7+\frac{11x+11}{9}$

$\displaystyle \frac{x+9}{x}-\frac{5+x}{x+2}=\frac{12x+12}{x^{2}+2x}$

$\displaystyle \frac{x+2}{x+1}-\frac{x+1}{x+2}=\frac{9}{20}$

$\displaystyle \frac{x+1}{2x}=\frac{x^{2}-1}{x-1}$

$\displaystyle \frac{4}{x}+\frac{4}{x+2}=3$

$\displaystyle \frac{x+1}{x-1}=\frac{4x+12}{3x+3}$

$\displaystyle \frac{4}{x-2}-\frac{6}{x+3}=\frac{1}{3}$

$\displaystyle \frac{x+1}{3x-2}+\frac{2x+1}{x+5}=\frac{3}{2}$

$\displaystyle \frac{4x+2}{x^{2}+2x+1}+\frac{3}{2}=\frac{x+5}{x+1}$

$\displaystyle \frac{3}{x+1}-\frac{6}{x+4}=\frac{-2}{4x-8}$

$\displaystyle \frac{4x+5}{3}=\frac{1}{2 x+3}$

$\displaystyle \frac{x^{2}-3}{2}=\frac{-3}{2 x^{2}+1}$

$\displaystyle \frac{3}{x-2}+\frac{8}{x+5}=\frac{x+1}{x^{2}+3 x-10}$

Resuelve la siguiente ecuación aplicando el cambio de variable
$z=x^{2}-3 x \smallskip$

$\displaystyle \frac{x^{2}-3 x-1}{\left(x^{2}-3 x\right)^{2}-1}=\frac{x^{2}-3 x-2}{\left(x^{2}-3 x\right)^{2}}$

20/11/2021

$\displaystyle x^{4}-20 x^{2}+64=0$

$\displaystyle x^{4}+6 x^{2}+8=0$

$\displaystyle x^{4}-5 x^{2}+4=0$

$\displaystyle x^{4}+10 x^{2}+9=0$

$\displaystyle \frac{24}{x^{2}}=3x^{2}-6$

$\displaystyle x^{4}-34 x^{2}-72=0$

$\displaystyle 2(x+1)^{4}-8x^{3}-8(x+3)+8=0$

$\displaystyle x^{4}-4x^{2}-12=0$

$\displaystyle 2x^{4}-6x^{2}-20=0$

$\displaystyle x\left((x+2)^{2}-4(x-1)-2\right)=\frac{5}{x}$

$\displaystyle x^{3}-6 x^{2}+9x-4=0$

$\displaystyle x^{4}-5x^{3}-39x^{2}+256x-350=0$

$\displaystyle x^{3}-5x^{2}+6x=0$

$\displaystyle x^{6}+x^{4}=2x^{5}+2x^{3}$

$\displaystyle x^{6}-2x^{3}+1=0$

$\displaystyle 6x^{4}+13x^{3}-8x^{2}-17x+6=0$

$\displaystyle 8x^{4}+10x^{3}-17x^{2}-7x+6=0$

$\displaystyle 2x^{4}-6x^{3}-32x^{2}+96x=0$

$\displaystyle x^{3}-5x-2=0$

$\displaystyle x^{10}-31 x^{5}-32=0$

19/11/2021

$\displaystyle x^{2}-x=0$

$\displaystyle 3x^{2}+18x=0$

$\displaystyle 16x^{2}-25=0$

$\displaystyle -5x^{2}+7x=0$

$\displaystyle 16x^{2}-25=0$

$\displaystyle 3x^{2}-27=0$

$\displaystyle (x+2)^{2}+2(x-1)^{2}=2x(x-4)-10$

$\displaystyle \frac{3x-1}{5}=\frac{13}{4x+5}$

$\displaystyle 6x^{2}-1+\frac{2x(-x+3)}{3}=\frac{5x^{2}-2}{6}-4x^{2}+\frac{59}{6}$

$\displaystyle \frac{3(-x+2)}{5}+4x\left(\frac{-2x+1}{3}\right)=x(-3x+1)-\frac{1}{2}$

18/11/2021

Comprobar las soluciones con las fórmulas de Cardano-Vieta

$\displaystyle x^{2}-6x-7=0$

$\displaystyle -2x(x-1)=x^{2}-5x$

$\displaystyle x^{2}-6x+9=0$

$\displaystyle x^{2}-x(3x+1)=3$

$\displaystyle (x-1)\left(\frac{3-x}{2}\right)=4$

$\displaystyle \frac{x^{2}+1}{2}-\frac{2x-3}{4}+\frac{x^{2}}{6}=\frac{59}{12}$

$\displaystyle \frac{3x^{2}}{2}-\frac{4x-1}{4}=\frac{2x(x-3)}{6}+\frac{17}{12}$

$\displaystyle \frac{2x-1}{x+7}-\frac{2x+1}{x-7}=\frac{5}{8}$

$\displaystyle \frac{1}{x}+\frac{1}{1+\frac{1}{x}}=\frac{7}{6}$

$\displaystyle \frac{3x-1}{5}=\frac{13}{4x+5}$

17/11/2021

$\displaystyle -5(2x-1)+3 x-2=-(6x-4)+7$

$\displaystyle 3(2x-5)+8 x-6=\frac{x}{2}-(5x+3) \smallskip$

$\displaystyle \frac{3(x+3)}{2}-2(2-3x)=8x-1-2(x+3)$

$\displaystyle \frac{-3x}{4}+\frac{1}{2}=-5x+26$

$\displaystyle \frac{x-4}{5}-4(-2x+1)-\frac{(-4x+2)}{10}=2(x-3)+\frac{5x+6}{2}$

$\displaystyle \frac{2x}{3}-\frac{x-2}{12}+\frac{x+3}{2}=2x-\frac{1}{6}$

$\displaystyle \frac{3(x-2)}{5}+2(-3x+1)-\frac{2}{5}=\frac{-4x+3}{15}+\frac{16}{3}$

$\displaystyle \frac{x+3}{6}+\frac{2x-1}{3}+\frac{1}{4}=\frac{x-5}{12}-\frac{2}{3}$

$\displaystyle \frac{6-2(x-3)}{7x}=-\frac{8}{4}$

$\displaystyle \frac{4x-3}{5x+1}=\frac{9}{16}$

Matemáticasbch1-ccss

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