A(x)+B(x)=2x-3x³+2x²+1+4x³+2x²-5

               = x³+4x²+2x-4

thay x=1 vào B(x) ta được

B(x)=4.1³+2.1³-5

      =4+2-5

     =1

\(A\left(x\right)+B\left(x\right)=\left(x+2\right)\left(x^2+2x-2\right)\)

thay x=1 \(=>A\left(1\right)+B\left(1\right)=3\left(1+2-2\right)=3\)

Ta có:

A(x)+B(x)=2x-3x³+2x²+1+4x³+2x²-5

                 =(-3x³+4x³)+(2x²+2x²)+2x+(1-5)

                 =x³+4x²+2x-4

Thay x=1 vào B(x) ta có:

B(1)=4*1³+2*1²-5

       =4+2-5=1

a: \(f\left(x\right)+g\left(x\right)=2x^3-2x^2+4x\)

b: \(f\left(x\right)-g\left(x\right)=-2x^2+2x+2\)

a)\(f\left(x\right)+g\left(x\right)=\left(x^3+x^3\right)-2x^2+\left(3x+x\right)+\left(1-1\right)\)

\(f\left(x\right)+g\left(x\right)=2x^3-2x^2+4x\)

b)\(f\left(x\right)-g\left(x\right)=x^3-2x^2+3x+1-x^3-x+1\)

\(f\left(x\right)-g\left(x\right)=-2x^2+2x+2\)

\(a,=3abc\left(5b+7c\right)\\ b,=\left(x+1\right)\left(9x^2-3x\right)=3x\left(3x-1\right)\left(x+1\right)\\ c,=2x\left(x+3\right)\)

a) \(=3abc\left(5b+7c\right)\)

b) \(=3x\left(x+1\right)\left(3x-1\right)\)

c) \(=2x\left(x+3\right)\)

chịu :)))))))))

Ko biết đc 

a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)

\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)

\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)

\(M=1-\frac{1}{49}\)

\(M=\frac{48}{49}\)

b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)

=  \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)

\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)

\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)

\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=2\times\frac{9}{22}\)

\(=\frac{9}{11}\)

Mình trả lời câu a nha  M= 4/1*5+4/5*9+4/9*13+...+4/45*49   M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49    M=1-1/49=48/49

mình na

B = 1x1 + 2x2 + 3x3 + ...+ 1999x1999 

B = 1 x (2-1) + 2 x (3 - 1 ) + 3 x ( 4 - 1 ) + .....+ 1999x(2000 -1)

B = (1 x 2 +2 x 3 + .... + 1999 x 2000 ) - ( 1 + 2 + 3 +  .... + 1999 )

B =                        P                                  -             Q

P x 3 = 1x2x3 + 2x3x3 + ..... + 1999 x 2000 x 3

P x 3 =1x2x(3-0) + 2x3x(4-1) +....+ 1999 x 2000 x ( 2001 - 1998)

P x 3 = 1x2x3 + 2x3x4 - 1x2x3 +......+ 1999 x 2000 x 2001 -  1998 x 1999 x 2000

P x 3 = 1999 x 2000 x 2001

P = 1999 x 2000 x 2001 : 3  = 2666666000 

Q = 1 + 2 + 3 + ..... + 1999 ( Có 1999 số )

Q = ( 1999 + 1 ) x 1999 : 2 = 1999000 

P - Q = B = 2666666000 - 1999000 = 2664667000 

Số cặp trong dãy trên là:

1 + 1999 = 2000

vì các cạp có số giống nhau nên:

2000 x 2 = 4000

        Đáp sô:4000

Mình k chắc

a) \(2x^2-16x=0\)

\(\Rightarrow2x\left(x-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)

b) \(\left(2x-1\right)^2-25=0\)

\(\Rightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)

\(\Rightarrow4\left(x-3\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

\(b.\left(2x-1\right)^2-25=0\)

<=>\(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

<=>\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

\(a.2x^2-16x=0< =>2x\left(x-8\right)=0\)

\(< =>\left[{}\begin{matrix}2x=0\\x-8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)