Question

Решите квадратные уравнения (а,в,д,ж) через дискриминант или по теореме Виета.​​

Answer 1

Объяснение:

а) x(5x-3)=3(2x-1 1/3)

5x²-3x=6x-3×4/3

5x²-3x=6x-4

5x²-3x-6x+4=0

5x²-9x+4=0

a=5 b=-9 c=4

D= 81-4×5×4=1

x1=(9-1)/2×5=8/10=4/5=0.8

x2=(9+1)/2×5=1

в) (3x-2)(3x+2)=4x(x-1)

9x²-4=4x²-4x

9x²-4x²+4x-4=0

5x²+4x-4=0

D=16-4×5×(-4)=16+80=96

$x1 = ( - 4 - 4 \sqrt{6} ) \div 10 \\ x2 = ( - 4 + 4 \sqrt{6} ) \div 10$

д)

4m²+12m+9=m²-1

3m²+12m+10=0

D=144-4×3×10=24

$m1 = ( - 12 - \sqrt{24} ) \div 6 \\ m2 = ( - 12 + \sqrt{24} ) \div 6$

ж)

4p²+12p+9-(p²-2p+1)=-8

4p²+12p+9-p²+2p-1+8=0

3p²+14p+16=0

D=196-4×3×16=4

x1=(-14-2)/6=-16/6=-8/3

x2=(-14+2)/6=-12/6=-2

Answer 2

Ответ:

$1)\ \ x\, (5x-3)=3(2x-1\frac{1}{3})\\\\5x^2-3x=6x-4\ \ ,\ \ 5x^2-9x+4=0\ \ ,\ \ D=1\ ,\\\\x_1=\dfrac{9-1}{10}=\dfrac{4}{5}\ \ ,\ \ x_2=\dfrac{9+1}{10}=1\\\\\\2)\ \ (3x-2)(3x+2)=4x(x-1)\\\\9x^2-4=4x^2-4x\ \ ,\ \ 5x^2+4x-4=0\ ,\ \ D/4=24\ ,\\\\x_1=\dfrac{-2-2\sqrt6}{5}\ \ ,\ \ x_2=\dfrac{-2+2\sqrt6}{5}$

$3)\ \ (2m+3)^2=(m-1)(m+1)\\\\4m^2+12m+9=m^2-1\ \ ,\ \ 3m^2+12m+10=0\ \ ,\ \ D/4=6\ ,\\\\m_1=\dfrac{-6-\sqrt6}{3}\ \ ,\ \ m_2=\dfrac{-6+\sqrt6}{3}\\\\\\4)\ \ (2p+3)^2-(p-1)^2=-8\\\\4p^2+12p+9-(p^2-2p+1)+8=0\ \ ,\ \ 3p^2+14p+16=0\ \ ,\ \ D/4=1\ ,\\\\p_1=\dfrac{-7-1}{3}=-\dfrac{8}{3}\ \ ,\ \ p_2=\dfrac{-7+1}{3} =-2$