Помогите пожалуйста 374;375

Решите задачу:

$374. \\ 1)M+( x^{2} -y)=2 x^{2} +3y-1 \\ M=2 x^{2} +3y-1-( x^{2} -y)=2 x^{2} +3y-1- x^{2} +y= x^{2} +4y-1 \\ 2)M-(b^{2} +3bc)=2b^{2} +c \\ M=2b^{2} +c+b^{2} +3bc=3b^{2} +3bc+c \\ 3)(a ^{2} -ab+b ^{2} )-M=2a ^{2} +2b ^{2} \\ M=a ^{2} -ab+b ^{2} -(2a ^{2} +2b ^{2} )=a ^{2} -ab+b ^{2} -2a ^{2} -2b ^{2} =-a^{2} - \\ -b^{2} -ab \\ \\$

$375. \\ 1)5 x^{2} -(3 x^{2} -( x^{2} +x)+(2x-1)) =5 x^{2} -(3 x^{2} -( x^{2} +x)+ \\ +(2x-1)) 5 x^{2} -(3 x^{2} - x^{2} -x+2x-1)=5 x^{2} -(2 x^{2} +x-1)= \\ =5 x^{2} -2 x^{2} -x+1=3 x^{2} -x+1 \\$

$2)4x-(-(2x-1)+(x-4))+1=4x-(-2x+1+x-4)+1= \\ =4x-(-2x+1+x-4)+1=4x-(-x-3)+1=4x+x+3+1= \\ 5x+4 \\ 3)1-(4ab^{2} -(2a^{2} b-ab)+(ab ^{2} +2a ^{2} b)+ab)= \\ =1-(4ab^{2} -2a^{2} b+ab+ab ^{2} +2a ^{2} b+ab)= \\ =1-(4ab^{2} +ab+ab ^{2} +ab)= 1-(5ab^{2}+2ab )= 1-5ab^{2} -2ab$