Упр.12.8 ГДЗ Мордкович Семенов 11 класс (Алгебра)

1. а)

   $$\left(\sqrt{50}\right)^x \cdot \left(\sqrt{2}\right)^x = 0{,}1$$
   $$\left(\sqrt{50}\cdot \sqrt{2}\right)^x = 0{,}1$$
   $$\left(\sqrt{100}\right)^x = 10^{-1}$$
   $$10^x = 10^{-1}$$
   $$x=-1$$

2. б)

   $$3\left(\frac15\right)^{3x+7}-10\left(\frac15\right)^{3x+8}=25$$
   $$3\left(\frac15\right)^{3x+7}-10\cdot \frac15 \left(\frac15\right)^{3x+7}=25$$
   $$3\left(\frac15\right)^{3x+7}-2\left(\frac15\right)^{3x+7}=25$$
   $$\left(\frac15\right)^{3x+7}=25$$
   $$\left(\frac15\right)^{3x+7}=\left(\frac15\right)^{-2}$$
   $$3x+7=-2$$
   $$3x=-9$$
   $$x=-3$$

3. в)

   $$\left(\sqrt3\right)^{2x}\cdot \left(2\sqrt3\right)^{2x}=36$$
   $$\left(\sqrt3\cdot 2\sqrt3\right)^{2x}=36$$
   $$6^{2x}=6^2$$
   $$2x=2$$
   $$x=1$$

4. г)

   $$5\left(\frac16\right)^{5x+3}+12\left(\frac16\right)^{5x+4}=42$$
   $$5\left(\frac16\right)^{5x+3}+12\cdot \frac16 \left(\frac16\right)^{5x+3}=42$$
   $$5\left(\frac16\right)^{5x+3}+2\left(\frac16\right)^{5x+3}=42$$
   $$7\left(\frac16\right)^{5x+3}=42$$
   $$\left(\frac16\right)^{5x+3}=6$$
   $$\left(\frac16\right)^{5x+3}=\left(\frac16\right)^{-1}$$
   $$5x+3=-1$$
   $$5x=-4$$
   $$x=-0{,}8$$

5. а)

   $$y=\cos 3x$$
   $$y’=-3\sin 3x$$
   $$y’\left(\frac{\pi}{2}\right)=-3\sin \frac{3\pi}{2}=3$$

6. б)

   $$y=(3-10x)^2$$
   $$y’=-10\cdot 2(3-10x)$$
   $$y'(0{,}1)=-20(3-1)=-40$$

7. в)

   $$y=\sqrt{2x+1}$$
   $$y’=\frac{2}{2\sqrt{2x+1}}=\frac{1}{\sqrt{2x+1}}$$
   $$y'(4)=\frac{1}{\sqrt{9}}=\frac13$$

8. г)

   $$y=\tg 2x$$
   $$y’=\frac{2}{\cos^2 2x}$$
   $$y’\left(\frac{\pi}{8}\right)=\frac{2}{\cos^2 \frac{\pi}{4}}=4$$

9. д)

   $$y=(0{,}5x+3)^3$$
   $$y’=0{,}5\cdot 3(0{,}5x+3)^2$$
   $$y'(-2)=1{,}5( -1+3)^2=6$$

10. е)

    $$y=\sqrt{3-4x}$$
    $$y’=-\frac{4}{2\sqrt{3-4x}}=-\frac{2}{\sqrt{3-4x}}$$
    $$y'(-5{,}5)=-\frac{2}{\sqrt{3+22}}=-\frac{2}{5}=-0{,}4$$

Ответ

а) $$-1$$; б) $$-3$$; в) $$1$$; г) $$-0{,}8$$.
а) $$3$$; б) $$-40$$; в) $$\frac13$$; г) $$4$$; д) $$6$$; е) $$-0{,}4$$.