Упр.728 ГДЗ Колягин Ткачёва 8 класс (Алгебра)

1) {(5x-4?x-3
-2x+11>x+1
12-3x>4-5x)+
2) {(3x?5-6x
-3x+1?4x-1
7-2x>2x+9)+
3) {(3x-2>2(x-3)+5x
2x^2+(5+x)^2>3(x-5)(x+5) )+
4) {(8x(x+2)(x-2)<(2x-3)(4x^2+6x+9)-5x (1/4 x+2)(2-1/4 x)-(3-1/4 x)(1/4 x+2)>-3)+
5) {(2(x-1/2)(x+3)>2x(x+3)
(x+3)/3>(3x+4)/2)+
6) {((3x+1/2)(2-x)+1/2 (x+1)>3(3-x)(3+x)-1
2-(2x+3)^2+(3+2x)(2x-3)<-2 1/3 (9+x)+1/3)+ Упростить выражение: 1) (x/(y-x)-x/(y+x))•(x+y)^2/(2x^2 ); 2) (1/(a-1)-1-1/(a+1))•(a^2-1); 3) (a/b-b/a)•ab/(a-b); 4) (a+b)(1/a-1/b) :(a^2-b^2)/(a^2 b^2 ).

1. Решим каждое неравенство и найдём пересечение решений:

   $$\begin{aligned} 5x-4&\ge x-3,\\ -2x+11&>x+1,\\ 12-3x&>4-5x. \end{aligned}$$

   $$\begin{aligned} 4x&\ge 1,\\ -3x&>-10,\\ 2x&>-8. \end{aligned}$$

   $$x\ge \frac14,\quad x<\frac{10}{3},\quad x>-4.$$

   Пересечение:

   $$\frac14\le x<\frac{10}{3}.$$

2. $$\begin{aligned} 3x&\le 5-6x,\\ -3x+1&\le 4x-1,\\ 7-2x&>2x+9. \end{aligned}$$

   $$\begin{aligned} 9x&\le 5,\\ -7x&\le -2,\\ -4x&>2. \end{aligned}$$

   $$x\le \frac59,\quad x\ge \frac27,\quad x<-\frac12.$$

   Общих решений нет.

3. $$\begin{aligned} 3x-2&>2(x-3)+5x,\\ 2x^2+(5+x)^2&>3(x-5)(x+5). \end{aligned}$$

   $$\begin{aligned} 3x-2&>2x-6+5x,\\ 2x^2+25+10x+x^2&>3(x^2-25). \end{aligned}$$

   $$\begin{aligned} -4x&>-4,\\ 10x&>-100. \end{aligned}$$

   $$x<1,\quad x>-10.$$

   Пересечение:

   $$(-10;1).$$

4. $$\begin{aligned} 8x(x+2)(x-2)&<(2x-3)(4x^2+6x+9)-5x,\\ \left(\frac14x+2\right)\left(2-\frac14x\right)-\left(3-\frac14x\right)\left(\frac14x+2\right)&>-3. \end{aligned}$$

   $$\begin{aligned} 8x(x^2-4)&<8x^3-27-5x,\\ 4-\frac1{16}x^2-\frac34x-6+\frac1{16}x^2+\frac24x&>-3. \end{aligned}$$

   $$\begin{aligned} -32x&<-27-5x,\\ -\frac14x&>-1. \end{aligned}$$

   $$x>1,\quad x<4.$$

   Пересечение:

   $$\left(1;4\right).$$

5. $$\begin{aligned} 2\left(x-\frac12\right)(x+3)&>2x(x+3),\\ \frac{x+3}{3}&>\frac{3x+4}{2}. \end{aligned}$$

   $$\begin{aligned} (2x-1)(x+3)&>2x(x+3),\\ 2(x+3)&>3(3x+4). \end{aligned}$$

   $$\begin{aligned} 2x^2+6x-x-3-2x^2-6x&>0,\\ 2x+6&>9x+12. \end{aligned}$$

   $$x<-3,\quad 7x<-6.$$

   Пересечение:

   $$x<-3.$$

6. $$\begin{aligned} \left(3x+\frac12\right)(2-x)+\frac12(x+1)&>3(3-x)(3+x)-1,\\ 2-(2x+3)^2+(3+2x)(2x-3)&<-\frac73(9+x)+\frac13. \end{aligned}$$

   $$\begin{aligned} 6x-3x^2+1-\frac12x+\frac12x+\frac12&>27-3x^2-1,\\ 2-4x^2-12x-9+4x^2-9&<-\frac73(9+x)+\frac13. \end{aligned}$$

   $$\begin{aligned} 6x&>24.5,\\ -12x-18&<-21-\frac73x+\frac13. \end{aligned}$$

   $$x>\frac{49}{12},\quad x>\frac{29}{9}.$$

   Пересечение:

   $$\left(\frac{49}{12};+\infty\right).$$

Ответ

1. $$\left[\frac14;\frac{10}{3}\right)$$
2. решений нет
3. $$(-10;1)$$
4. $$\left(1;4\right)$$
5. $$x<-3$$
6. $$\left(\frac{49}{12};+\infty\right)$$

Ход решения

1. $$\left(\frac{x}{y-x}-\frac{x}{y+x}\right)\cdot \frac{(x+y)^2}{2x^2}$$

   $$\begin{aligned} \frac{x(y+x)-x(y-x)}{(y-x)(y+x)}\cdot \frac{(x+y)^2}{2x^2} &=\frac{x(y+x-y+x)}{(y-x)(y+x)}\cdot \frac{(x+y)^2}{2x^2}\\ &=\frac{2x^2}{(y-x)(y+x)}\cdot \frac{(x+y)^2}{2x^2}\\ &=\frac{y+x}{y-x}. \end{aligned}$$

2. $$\left(\frac1{a-1}-1-\frac1{a+1}\right)(a^2-1)$$

   $$\begin{aligned} \frac{a+1-(a-1)(a+1)-(a-1)}{(a-1)(a+1)}\cdot (a^2-1) &=\frac{a+1-a^2+1-a+1}{a^2-1}\cdot (a^2-1)\\ &=3-a^2. \end{aligned}$$

3. $$\left(\frac{a}{b}-\frac{b}{a}\right)\cdot \frac{ab}{a-b}$$

   $$\begin{aligned} \frac{a^2-b^2}{ab}\cdot \frac{ab}{a-b} &=\frac{(a-b)(a+b)\cdot ab}{ab\cdot (a-b)}\\ &=a+b. \end{aligned}$$

4. $$ (a+b)\left(\frac1a-\frac1b\right):\frac{a^2-b^2}{a^2b^2} $$

   $$\begin{aligned} (a+b)\cdot \frac{b-a}{ab}:\frac{(a-b)(a+b)}{a^2b^2} &=(a+b)\cdot \frac{b-a}{ab}\cdot \frac{a^2b^2}{(a-b)(a+b)}\\ &=-ab. \end{aligned}$$

Ответ

1. $$\frac{y+x}{y-x}$$
2. $$3-a^2$$
3. $$a+b$$
4. $$-ab$$