x\(^2\)-y(y+1)(y+2)(y+3)-1

=(x\(^2\)-1)-y(y+3)(y+1)(y+2)

=(x\(^2\)-1)-(y\(^2\)+3y)(y\(^2\)+3y+2)

x\(^{30}\)+x\(^4\)-x\(^{1975}\)+1

=(x\(^{30}\)-1)+[(x\(^{^{ }2}\))\(^2\)-1]-x(x\(^{1974}\)-1)+x+3

=[(x\(^3\))\(^{10}\)-1]+(x\(^2\)-1)(x\(^2\)+1)-x[(x\(^3\))\(^{1974}\)-1]+x+3

(áp dụng hằng đẳng thức mở rộng)

do đó số dư là x+3

bài2

a, x-15=-63-4

=>x-15=-67

=>x=-52

b, -x+3=11

=>x=-11+3

=>x=-8

c,\(|\)x+2\(|\)-4=7

=>\(|\)x+2\(|\)=11

=>\(\left\{{}\begin{matrix}x+2=11\\x+2=-11\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=9\\x=13\end{matrix}\right.\)

bài3

ta có:\(\left|y\right|\)=8

=>\(\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\)

TH1 x=5,y=8

=>x-y=5-8=-3

y-x=8-5=3

TH2x=5 ,y=-8

x-y=5--8=13

y-x=-8-5=-13

có 1 và chỉ 1 đường thẳng đi qua hai đoạn thẳng phân biệt

tk mk , mk tk 

tk mk , mk tk 

\(\dfrac{1}{1-x}\)+\(\dfrac{1}{1+x}\)+\(\dfrac{2}{1+x^2}\)+\(\dfrac{4}{1+x^4}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)

=

=\(\dfrac{4}{1-x^4}\)+\(\dfrac{4}{1+x^4}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)

=\(\dfrac{8}{1-x^8}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)

=\(\dfrac{16}{1-x^{16}}\)+\(\dfrac{16}{1+x^{16}}\)

=\(\dfrac{32}{1-x^{32}}\)

mk k hiểu đầu bài là gì luôn

ta có :C=3x-x\(^2\)

=-(x\(^2\)-3x)

C=3x-x\(^{^{ }2}\)

=-(x\(^2\)-3x)

=-[(x\(^{^{ }2}\)-2x\(\dfrac{3}{2}\)+\(\dfrac{9}{4}\))-\(\dfrac{9}{4}\)]

=-(x-\(\dfrac{3}{2}\))\(^2\)+\(\dfrac{9}{4}\)

vì -(x-\(\dfrac{3}{2}\))\(^{^{ }2}\)\(\le\)0 với mọi x

do đó max C=\(\dfrac{9}{4}\)