(3x²-5x) ²-5(3x²-5x)+6=0​

Ответ:

Объяснение:

(3x²-5x) ²-5(3x²-5x)+6=0​;

Обозначим 3x²-5x через t.  Тогда уравнение примет вид

t²-5t+6=0.

По т. Виета

t1+t2=5;

t1*t2=6;

t1=2;

t2=3.

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1)  3x²-5x =2;

3x²-5x -2=0

a=3; b=-5; c=-2.

D=b^2-4ac = (-5)^2 -4*3*(-2) = 25+24 = 49>0 - 2 корня.

x1,2=(-b±√D)/2a = (-(-5)±√49)/2*3 = (5±7)/6;

x1=(5+7)/6 = 12/6 = 2;

x2=(5-7)/6 = -2/6 = -1/3;

2) 3x²-5x =3;

3x^2-5x-3=0;

a=3;  b=-5;  c=-3;

D=b^2-4ac = (-5)^2-4*3*(-3) = 25 + 36 =  61>0 - 2 корня.

x3,4=(-b±√D)/2a = (-(-5)±√61)/2*3 = (5±√61)/6;

x3=(5+√61)/6 ≈2.1;

x4=(5-√61)/6 ≈ - 0.47.

[tex](3 {x}^{2} - 5x) {}^{2} - 5(3 {x}^{2} - 5x) + 6 = 0 \\ 3x {}^{2} - 5x = a \\ {a}^{2} - 5a + 6 = 0[/tex]

По теореме Виета:

[tex] {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c [/tex]

[tex]a_{1} + a_{2} = 5 \\ a_{1}a_{2} =6  \\ a_{1} =  2\\ a_{2} = 3 \\ \\ 1) \: 3 {x}^{2} - 5x = 2 \\ 3 {x}^{2} - 5x - 2 = 0 \\ a =3 \\ b = - 5 \\ c = - 2 \\ D = {b}^{2} - 4ac = ( - 5) {}^{2} - 4 \times 3 \times ( -2 ) = 25 + 24 = 49 \\ x_{1} = \frac{5 - 7}{2 \times 3} = - \frac{2}{6} = - \frac{1}{3} \\ x_{2} = \frac{5 + 7}{2 \times 3} = \frac{12}{6} = 2 \\ \\ 2) \: 3 {x}^{2} - 5x = 3 \\ 3 {x}^{2} - 5x - 3 = 0 \\ a =3 \\ b = - 5\\ c = - 3\\ D = {b}^{2} - 4ac = ( - 5) {}^{2} - 4 \times 3 \times ( - 3) = 25 + 36 = 61 \\ x_{3} = \frac{5 - \sqrt{61} }{2 \times 3} = \frac{5 - \sqrt{61} }{6} \\ x_{4} = \frac{5 + \sqrt{61} }{2 \times 3} = \frac{5 + \sqrt{61} }{6} [/tex]