28.24
a)
\\
(-2+2)^2(x-5)^3=(x-5)(-2+2)^4\\
0\cdot(x-5)^3=(x-5)\cdot0\\
0=0\\
2) x=5==>\\
(x+2)^2(5-5)^3=(5-5)(-x+2)^4\\
(x+2)^2\cdot0=0\cdot(x+2)^4\\
0=0\\;
3)x\in(-\infty;-2)\bigcup(-2;5)\bigcup(5;+\infty);\\
\left \{ {{x\neq-2} \atop {x\neq5}} \right. \\
(x-5)^2=(x+2)^2;\\
x^2-10x+25=x^2+4x+4;\\
14x=21;\\
x= \frac{3}{2}=1,5;\\
x=-2;\ 1,5;\ 5;\\
" alt="(x+2)^2(x-5)^3=(x-5)(x+2)^4;\\
1) x=-2==>\\
(-2+2)^2(x-5)^3=(x-5)(-2+2)^4\\
0\cdot(x-5)^3=(x-5)\cdot0\\
0=0\\
2) x=5==>\\
(x+2)^2(5-5)^3=(5-5)(-x+2)^4\\
(x+2)^2\cdot0=0\cdot(x+2)^4\\
0=0\\;
3)x\in(-\infty;-2)\bigcup(-2;5)\bigcup(5;+\infty);\\
\left \{ {{x\neq-2} \atop {x\neq5}} \right. \\
(x-5)^2=(x+2)^2;\\
x^2-10x+25=x^2+4x+4;\\
14x=21;\\
x= \frac{3}{2}=1,5;\\
x=-2;\ 1,5;\ 5;\\
" align="absmiddle" class="latex-formula">
b)
x=- \frac{1}{2}=-0,5;\\
2) 2x-3=0==> x= \frac{3}{2}=1,5;\\
3) \left \{ {{2x+1\neq0} \atop {2x-5\neq0}} \right.\\
(2x+3)^2=(2x-5)^2;\\
4x^2+6x+9=4x^2-10x+25;\\
16x=16;
x=1;\\
x=-0,5;\ 1;\ 1,5" alt="(2x+1)^3(2x-3)^5=(2x+1)^5(2x-3)^3;\\
1)2x+1=0;==>x=- \frac{1}{2}=-0,5;\\
2) 2x-3=0==> x= \frac{3}{2}=1,5;\\
3) \left \{ {{2x+1\neq0} \atop {2x-5\neq0}} \right.\\
(2x+3)^2=(2x-5)^2;\\
4x^2+6x+9=4x^2-10x+25;\\
16x=16;
x=1;\\
x=-0,5;\ 1;\ 1,5" align="absmiddle" class="latex-formula">