Question

помогите решить уравнения умалаю буду очееень благодарен вам очень:
$4 - 5(3x + 2.5) = 3x + 9.5$
$6(10 - x)(3x + 9) = 0$
$x(2x + 1) = 3x + 4$
$9 {x}^{2} - 6x = - 1$
$2 - 3x = - 5 {x}^{2}$

$(6 - 2x) {}^{2} = (2x - 8) {}^{2}$
$\frac{x + 7}{6} + 2 = \frac{x}{3}$
$10 {x}^{2} + 5x = 0$
$12 - {x}^{2} = 11$
$\frac{1}{4} {x}^{2} - 4 = 0$

$\frac{x - 1}{2} + 2 = \frac{x}{3}$
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Answer 1

$4-5(3x+2.5)=3x+9.5\\ 4-15x-12.5=3x+9.5\\-8.5-15x-3x=9.5\\-18x=9.5+8.5\\-18x=18\\x=-1$

$6(10-x)(3x+9)=0 \\(60-6x)(3x+9)=0\\180x+540-18x ^2-54x=0\\-18x^2+126x+540=0\\18(10-x)(x+3)=0\\10-x=0\\x=10\\x+3=0\\x=-3$

$x(2x+1)=3x+1 \\ 2x ^ { 2 } +x=3x+4 \\ 2x ^ { 2 } +x-3x-4=0\\ 2x ^ { 2 } -2x-4=0\\x ^ { 2 } -x-2=0\\\sqrt{D} =\sqrt{9} =3\\ x= \dfrac{ 1+3 }{ 2 } =2\\ x= \dfrac{ 1-3 }{ 2 } =-1$

$9x ^ { 2 } -6x=-1 \\9x ^ { 2 } -6x+1=0\\\sqrt{D} =\sqrt{36-4*9*1} =\sqrt{0} =0\\ x= \dfrac{ 6+0 }{ 18 } =\dfrac{1}{3}$

$2-3x = -5 { x }^{ 2 } \\5 { x }^{ 2 } -3x+2 = 0 \\\sqrt{D}=\sqrt{9-4*5*2} =\sqrt{-31}$

$(6-2x) ^ { 2 } =(2x-8) ^ { 2 } \\36-24x+4x^ 2 =4x^ 2 -32x+64\\36-24x+4x^ 2 -4x^ 2 +32x=64\\36+8x=64\\8x=28\\x= 3 \dfrac { 1 } { 2 }$

$\dfrac{ x+7 }{ 6 } +2= \dfrac{ x }{ 3 } \\x+7+12=2x\\x+19=2x\\x+19-2x=0\\-x+19=0\\-x=-19\\x=19$

$10x ^ { 2 } +5x=0 \\x(10x+5)=0\\x=0\\10x+5=0\\10x=-5\\x=-\dfrac{1}{2}$

$12-x ^ { 2 } =11 \\-x ^ { 2 } =11-12\\-x ^ { 2 }=-1\\x^2=1\\x=1\\x=-1$

$\dfrac{ 1 }{ 4 } x ^ { 2 } -4=0 \\x^{2} -16=0\\x^{2} =16\\x=-4\\x=4$

$\dfrac{ x-1 }{ 2 } -2= \dfrac{ x }{ 3 } \\3(x-1)-12=2x\\3x-15-2x=0\\x-15=0\\x=0$

Объяснение: