Подбираем такие числа, из которых можно извлечь корень и оцениваем:
√4 < √5 < √9
2 < √5 < 3;
аналогично:
2 < √6 < <span>3
</span>5 < √27 < 6
6 < √37 < 7
Значит, только √27 ∈ [5; 6].
<span>x(x-5)/0.3=0
х</span>²-5х=0*0,3
х²-5х=0
х(х-5)=0
х=0 или х-5=0
х=5
Ответ: х1=0, х2=5
3x-2/1 x=2/3
3x/4-1/2 x=2/3
<span> -2(1-2x)/x x=1/2</span>
![\sqrt{x^2-3x+2}>x+3](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-3x%2B2%7D%3Ex%2B3)
ОДЗ:
![x^2-3x+2\geq0\\x^2-3x+2=0\\x_1+x_2=3\\x_1*x_2=2\\x_1=2\ x_2=1](https://tex.z-dn.net/?f=x%5E2-3x%2B2%5Cgeq0%5C%5Cx%5E2-3x%2B2%3D0%5C%5Cx_1%2Bx_2%3D3%5C%5Cx_1%2Ax_2%3D2%5C%5Cx_1%3D2%5C+x_2%3D1)
///////+//////[1]....-.....[2]////+/////>x
x=0
![x\in (-\infty;1]\cup[2;+\infty)](https://tex.z-dn.net/?f=x%5Cin+%28-%5Cinfty%3B1%5D%5Ccup%5B2%3B%2B%5Cinfty%29)
![\sqrt{x^2-3x+2}>x+3\\x^2-3x+2=x^2+6x+9\\9x=-7\\x=-\frac{7}{9}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-3x%2B2%7D%3Ex%2B3%5C%5Cx%5E2-3x%2B2%3Dx%5E2%2B6x%2B9%5C%5C9x%3D-7%5C%5Cx%3D-%5Cfrac%7B7%7D%7B9%7D)
///+//(-7/9)...............-.............>x
x=0
<span>///////+//////[1]....-.....[2]////+/////>x
Решение неравенства:</span><span>
![x\in(\infty;-\frac{7}{9})](https://tex.z-dn.net/?f=x%5Cin%28%5Cinfty%3B-%5Cfrac%7B7%7D%7B9%7D%29+)
</span>