Question

Раскройте формулу квадрата двучлена и решите уровнения a)(x-2)²=x² б)(3y+5)²=9y² в)(c-7)²=c²-7 г)(6-z)²=z²+36 д)(2x-1)²+2=0 е)3(x-2)²=3

Answer 1

а)

$(x - 2)^{2} = {x}^{2} \\ {x}^{2} - 4x + 4 = {x}^{2} \\ - 4x = - 4 \\ x = 1$

б)

$(3y + 5)^{2} = 9 {y}^{2} \\ 9 {y}^{2} + 30y + 25 = 9 {y}^{2} \\ 30y = - 25 \\ y = - \frac{5}{6}$

в)

$(c - 7)^{2} = {c}^{2} - 7 \\ {c}^{2} - 14c + 49 = {c}^{2} - 7 \\ - 14c = - 56 \\ c = 4$

г)

$(6 - z)^{2} = {z}^{2} + 36 \\ 36 - 12z + {z}^{2} = {z}^{2} + 36 \\ - 12z = 0 \\ z = 0$

д)

$(2x - 1)^{2} + 2 = 0 \\ 4 {x}^{2} - 4x + 1 + 2 = 0 \\ 4 {x}^{2} - 4x + 3 = 0 \\ d = {b}^{2} - 4ac = 16 - 4 \times 4 \times 3 = 16 - 48 = - 32$

d<0, поэтому корней нет.

е)

$3(x - 2)^{2} = 3 \: \: \: | \div 3 \\ (x - 2)^{2} = 1 \\ {x}^{2} - 4x + 4 = 1 \\ {x}^{2} - 4x + 3 = 0 \\ d = {b}^{2} - 4ac = 16 - 4 \times 1 \times 3 = 4 \\ x1 = \frac{ - b + \sqrt{d} }{2a} = \frac{4 + 2}{2} = 3 \\ x2 = \frac{ - b - \sqrt{d} }{2a} = \frac{4 - 2}{2} = 1$

Answer 2

a)(x-2)²=x²

x²-2*2х+2²-x²=0

4-4х=0

-4х=-4

х=-4:(-4)

х=1

б)(3y+5)²=9y²

9y² +2*5*3у+5²-9y² =0

30у+25=0

30у=-25

у= -25:30

у= -25\30= - 5\6= - 0,8(3)= -0,83

в)(c-7)²=c²-7

c²-2*7с+ 49-c²+7=0

-14с+56=0

-14с= -56

с= -56: (-14)

с=4

г)(6-z)²=z²+36

36-2*6z+z²-z²-36=0

-12z=0

z= 0: (-12)

z=0

д)(2x-1)²+2=0

4x²-2*2х+1+2=0

4x²-4х+3=0

D=b²-4ac=(-4)²-4*4*3=16-48= -32<0 - нет корней

е)3(x-2)²=3

3(x²-2*2х+4)=3

x²-4х+4=3:3

x²-4х+4=1

x²-4х+4-1=0

x²-4х+3=0

D=b²-4ac=(-4)²-4*1*3= 16-12=4

х1= -b+√D\2а=(4+√4):2=(4+2):2=6:2=3

x2= -b-√D\2а= (4- √4):2= (4-2):2=2:2=1

Пошаговое объяснение: