Решите пожалуйста5x²-3x-2=02x²-5x+3=0x²+3x+1=02x²-9x+4=03x+8x-3=02x-7x+3=0. Created by Neумник. matematika-ru

Решите пожалуйста5x²-3x-2=02x²-5x+3=0x²+3x+1=02x²-9x+4=03x+8x-3=02x-7x+3=0...

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5x^2-3x-2= 0
D=9+40= 49
x(1,2) = 3+-7/10= 1;-0,4

2x^2 - 5x+3=0
D= 25-24=1
x(1,2) = 5+-1/4= 1; 3/2

x^2+3x+1=0
D= 9-4= 5;
x(1,2) = -3+-√5/2 = 0,5(-3+√5); 0,5 (-3-√5)

2x^2-9x+4= 0
D= 81-32= 49;
x(1,2) = 9+-7/4= 4;1/2

11x-3= 0
x= 3/11

-5x+3=0
x= 3/5

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1) 5x² - 3x-2= 0
D= b²- 4ac
D= 9+ 40= 49
√D= 7
x₁= $\frac{-b+ \sqrt{D} }{2a} = \frac{3+7}{10}= 1$
x₂= $\frac{-b- \sqrt{D} }{2a} = \frac{3-7}{10} = -\frac{4}{10}= - \frac{2}{5}$
Ответ: x₁= 1; x₂= $\frac{2}{5}$
2) 2x²- 5x+ 3= 0
D= b² -4ac
D= 25- 24 = 1
√D= 1
x₁= $\frac{-b+ \sqrt{D} }{2a} = \frac{5+1}{4} = 1,5$
x₂= $\frac{-b- \sqrt{D} }{2a}= \frac{5-1}{4} = 1$
Ответ: x₁= 1,5 ; x₂= 1
3) x²+ 3x+ 1= 0
D= b² -4ac
D= 9- 4= 5
√D= √5
x₁= $\frac{-b+ \sqrt{D} }{2a} = \frac{-3+ \sqrt{5} }{2}$
x₂= $\frac{-b- \sqrt{D} }{2a} = \frac{3- \sqrt{5} }{2}$
4) 2x² -9x+ 4 = 0
D= b² -4ac
D= 81 - 32= 49
√D= 7
x₁= $\frac{-b+ \sqrt{D} }{2a} = \frac{9+ 7}{4} = 4$
x₂= $\frac{-b- \sqrt{D} }{2a} = \frac{9-7}{4} = \frac{1}{2}$
5) 3x² + 8x- 3=0
D= b² -4ac
D= 64+ 36= 100
√D= 10
x₁= $\frac{-b+ \sqrt{D} }{2a} = \frac{-8+10}{6} = \frac{2}{6} = \frac{1}{3}$
x₂= $\frac{-b- \sqrt{D} }{2a} = \frac{-8-10}{6} = -3$
6) 2x² -7x+ 3= 0
D= b² -4ac
D= 49- 24 = 25
√D= 5
x₁= $\frac{-b+ \sqrt{D} }{2a} = \frac{7+5}{4} = 3$
x₂= $\frac{-b- \sqrt{D} }{2a} = \frac{7-5}{4} = \frac{1}{2}$

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