M=( x^2y^3 + x^3y^2 - x^2 + y^2 + 5) - (x^2y^3 + x^3y^2 + 2xy^2 -1) 

M= x^2y^3 + x^3y^2 - x^2 + y^2 + 5 - x^2y^3 - x^3y^2 - 2xy^2 +1

M= y^2 - x^2- 2xy^2 +6

b)\(x^6+2x^3+2=x^6+x^3+x^3+1+1\)

\(=x^3\left(x^3+1\right)+\left(x^3+1\right)+1=\left(x^3+1\right)\left(x^3+1\right)+1=\left(x^3+1\right)^2+1\)

Vì \(\left(x^3+1\right)^2\ge0\Rightarrow\left(x^3+1\right)^2+1\ge0+1>0\) với mọi x \(\in\) R

=>vô nghiệm

Vậy...............

a) f(x)=x²+bx-a

Ta có: f(2)=2²+b.2-a=4+2b-a

Mà f(2)=5 =>4a+2b-a=5

=>4a+2b=5+a=>2(2a+b)=5+a (*)

Ta có: f(1)=1²+b.1-a=1+b-a

Mà f(1)=0=>1+b-a=0=>b-a=-1=>a=b-(-1)=b+1

Thay a=b+1 vào (*) =>2.[2.(b+1)+b]=5+(b+1)

=>2.(2b+2+b)=b+6

=>2.(3b+2)=b+6

=>6b+4=b+6

=>6b-b=6-4

=>5a=2=>a=2/5

Khi đó a=b+1 =>b=a-1=>b=2/5-1=-3/5

Vậy..................

S = 1+1/2.(1+2)+1/3.(1+2+3)+...+1/100.(1+2+3+...+100)

   = 1+1/3.(1+2+3)+1/5.(1+2+3+4+5)+...+1/99(1+2+3+...+99) +  1/2.(1+2)+1/4.(1+2+3+4)+...+1/100.(1+2+3+...+100)

   =  (1+2+3+...+50)+(3/2+5/2+7/2+...+101/2)

   =  1275+1300

   =       2575

làm giùm bn í đi mọi người ........ mk cx k cho ......

F(x) = ax³ + bx² + cx + d 

F(x) = a.x.x.x + b.x.x + c.x + d

F(x) = x.x ( ax + bx + c ) + d

F(x) = x ( ax + bx + c + d )

F(1) = 1 ( a + b + c + d )

Muốn x = 1 là nghiệm 

=)) 1 ( a + b + c + d ) =0

=) a + b + c + d = 0 

a+b+c+d=0 

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)

Ta có: \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\) với mọi n \(\ne\) 0

\(A=1+\frac{1}{2}.\left(\frac{2.3}{2}\right)+\frac{1}{3}.\left(\frac{3.4}{2}\right)+.....+\frac{1}{20}.\left(\frac{20.21}{2}\right)\)

\(A=1+\frac{3}{2}+\frac{4}{2}+.....+\frac{21}{2}\)

\(A=\frac{1}{2}.\left(2+3+....+21\right)\)

Tổng trong ngoặc có:(21-2)+1=20(số hạng)

=>\(2+3+...+21=\frac{\left(21+2\right).20}{2}=230\)

Khi đó \(A=\frac{1}{2}.230=115\)

Vậy..............

giúp tui vs nào .......