Найдите значение выражения (4-y)*2-y(y+1) при y=-1/9

8-2у-у²+у=-у²-у+8=1/81+1/9+8=1+9+8=18

х² -2х = х+2-х²

х² - 2х - х +х² - 2 = 0

2х² - 3х - 2 =0

D=(-3)² - 4*2*(-2) = 9+16 = 25 = 5²

x1= (3 - 5)/(2*2) = -2/4 =-0.5

x2 =(3+5)/4 = 8/4 = 2

2) 3х² -8х + 13 = (х-5)²

3х² - 8х + 13 = х² - 10х + 25

3х² - 8х + 13 - х² + 10х - 25 =0

2х² +2х -12 = 0 |÷2

x²+x - 6 =0

D=1² - 4*1*(-6) = 1 +24 = 25 = 5²

x1= (-1-5)/ (2*1) = -6/2 =-3

x2= (-1+5)/2 = 4/2=2

3) (x+1)²=(x-2)²

x²+2x+1 = x² -4x +4

x² +2x + 1 - x² +4x - 4 =0

6x - 3 =0

6x= 3

x=3/6 = 1/2

x=0.5

4)(x-10)² = (1-x)²

x²-20x +100 = 1 -2x+x²

x² -20x +100 -1 +2x -x²=0

18x + 99 =0

x=99/18 = 11/2

x=5.5

∠ABD = ∠ACD = 50°

∠ACB = ∠ADB = x
∠BAC = ∠BDC = y
∠CAD = ∠CBD = z

x:y:z = 5:7:13

∠ABC = ∠ABD + ∠CAD = 50° + z
∠BCD = ∠ACB + ∠ABD = x + 50°
∠CDA = ∠BDC + ∠ADB = y + x
∠DAB = ∠CAD + ∠BAC = z + y

∠ABC + ∠BCD + ∠CDA + ∠BAD =  50 + z + x + 50 + y + x + z + y = 360°

100 + 2z + 2x + 2y = 360
x + z + y = 130
x/y = 5/7
x/z = 5/13
x + 7x/5 + 13x/5 = 130
5x = 130
x = 26
y = 36.4
z = 67.6

∠ABC = 50° + z = 50° + 67.6° = 117.6° 
∠BCD = x + 50° = 26° + 50° = 76°
∠CDA = y + x = 36.4° + 26° = 62.4°
∠DAB = z + y = 67.6° + 36.4° = 104°