1. В арифметической прогрессии S4=42 и S8=132. Найдите a1 и d. 2. Разложите на множетели: 5aX*x…

1.S4=2 (2a1+3d)=42,S8=4 (2a1+7d)=132, 4a1+6d=42,8a1+28d=132, -8a1-12d=-84,8a1+28d=132, 16d=48,d=3,4a1+18=42,4a1=24,a1=6. 2,5ax^2 — 10ax — bx+2b — x+2=5ax (x-2) -b (x-2) — (x-2)=(x-2) (5ax-b-1). 3,2x — 1/x+1 < 1,2/x+1 > 0, (2x^2-1) /x<0, (2+x) /x>0, x≠0,x (√2x-1) (√2x+1) <0,x (x+2) >0, x (√2x-1) (√2x+1)=0,x1=-1/√2, x2=0, x3=1/√2,x∈ (-∞; -1/√2) U (0; 1/√2) , x (x+2)=0,x1=-2, x2=0,x∈ (-∞; -2) U (0; +∞) , x∈ (-∞; -2) U (0; 1/√2). 4.|a|=sqrt (a_x^2+a_y^2)=11, (|a|) ^2=a_x^2+a_y^2=121,|b|=sqrt (b_x^2+b_y^2)=23, (|b|) ^2=b_x^2+b_y^2=529,|a-b|=sqrt (a_x-b_x) ^2+(a_y-b_y) ^2)=30, (|a-b|) ^2=(a_x-b_x) ^2+(a_y-b_y) ^2=a_x^2-2a_x b_x+b_x^2+a_y^2-2a_y b_y+b_y^2=(|a|) ^2+(|b|) ^2-2a_x b_x-2a_y b_y=900,2a_x b_x+2a_y b_y=(|a|) ^2+(|b|) ^2- (|a-b|) ^2=121+529-900=-250|a+b|=sqrt (a_x+b_x) ^2+(a_y+b_y) ^2) , (|a+b|) ^2=(a_x+b_x) ^2+(a_y+b_y) ^2=a_x^2+2a_x b_x+b_x^2+a_y^2+2a_y b_y+b_y^2=(|a|) ^2+(|b|) ^2+2a_x b_x+2a_y b_y=121+529-250=400,|a+b|=20.