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a)
![(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\
0cdot(x-5)^3=(x-5)cdot0\
0=0\
2) x=5==>\
(x+2)^2(5-5)^3=(5-5)(-x+2)^4\
(x+2)^2cdot0=0cdot(x+2)^4\
0=0\;
3)xin(-infty;-2)bigcup(-2;5)bigcup(5;+infty);\
left { {{xneq-2} atop {xneq5}} 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;\](https://tex.z-dn.net/?f=%28x%2B2%29%5E2%28x-5%29%5E3%3D%28x-5%29%28x%2B2%29%5E4%3B%5C%0A1%29+x%3D-2%3D%3D%26gt%3B%5C%0A%28-2%2B2%29%5E2%28x-5%29%5E3%3D%28x-5%29%28-2%2B2%29%5E4%5C%0A0cdot%28x-5%29%5E3%3D%28x-5%29cdot0%5C%0A0%3D0%5C%0A2%29+x%3D5%3D%3D%26gt%3B%5C%0A%28x%2B2%29%5E2%285-5%29%5E3%3D%285-5%29%28-x%2B2%29%5E4%5C%0A%28x%2B2%29%5E2cdot0%3D0cdot%28x%2B2%29%5E4%5C%0A0%3D0%5C%3B%0A3%29xin%28-infty%3B-2%29bigcup%28-2%3B5%29bigcup%285%3B%2Binfty%29%3B%5C%0A+left+%7B+%7B%7Bxneq-2%7D+atop+%7Bxneq5%7D%7D+right.+%5C%0A%28x-5%29%5E2%3D%28x%2B2%29%5E2%3B%5C%0Ax%5E2-10x%2B25%3Dx%5E2%2B4x%2B4%3B%5C%0A14x%3D21%3B%5C%0Ax%3D+frac%7B3%7D%7B2%7D%3D1%2C5%3B%5C%0Ax%3D-2%3B+1%2C5%3B+5%3B%5C%0A "(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\
0cdot(x-5)^3=(x-5)cdot0\
0=0\
2) x=5==>\
(x+2)^2(5-5)^3=(5-5)(-x+2)^4\
(x+2)^2cdot0=0cdot(x+2)^4\
0=0\;
3)xin(-infty;-2)bigcup(-2;5)bigcup(5;+infty);\
left { {{xneq-2} atop {xneq5}} 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;\")
b)
![(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+1neq0} atop {2x-5neq0}} 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](https://tex.z-dn.net/?f=%282x%2B1%29%5E3%282x-3%29%5E5%3D%282x%2B1%29%5E5%282x-3%29%5E3%3B%5C%0A1%292x%2B1%3D0%3B%3D%3D%26gt%3Bx%3D-+frac%7B1%7D%7B2%7D%3D-0%2C5%3B%5C%0A2%29+2x-3%3D0%3D%3D%26gt%3B+x%3D+frac%7B3%7D%7B2%7D%3D1%2C5%3B%5C%0A3%29+left+%7B+%7B%7B2x%2B1neq0%7D+atop+%7B2x-5neq0%7D%7D+right.%5C%0A+%282x%2B3%29%5E2%3D%282x-5%29%5E2%3B%5C%0A4x%5E2%2B6x%2B9%3D4x%5E2-10x%2B25%3B%5C%0A16x%3D16%3B%0Ax%3D1%3B%5C%0Ax%3D-0%2C5%3B+1%3B+1%2C5 "(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+1neq0} atop {2x-5neq0}} 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")
a)
b)
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