Ta đặt

  \(A=1\times3+3\times5+...+61\times63\)

\(6A=1\times3\times6+3\times5\times6+....+61\times63\times6\)

\(6A=1\times3\times6+3\times5\times\left(7-1\right)+...+61\times63\times\left(65-59\right)\)

\(6A=1\times3\times6+3\times5\times7-1\times3\times5+...+61\times63\times65-59\times61\times63\)

\(6A=1\times3\times6-1\times3\times5+61\times63\times65\)

\(6A=3+61\times63\times65\)

\(6A=3\times\left(1+61\times21\times65\right)\)

\(2A=83266\)

\(A=83266\div2=41633\)

`@` `\text {Ans}`

`\downarrow`

`a,`

` F(x)=3x^2-7+5x-6x^2-4x^2+8`

`= (3x^2 - 6x^2 - 4x^2) + 5x + (-7 + 8)`

`= -7x^2 + 5x + 1`

Bậc của đa thức: `2`

`G(x)=x^4+2x-1+2x^4+3x^3+2-x`

`= (x^4 + 2x^4) + 3x^3 + (2x - x) + (-1+2)`

`= 3x^4 + 3x^3 + x + 1`

Bậc của đa thức: `4`

`b,`

`F(x) + G(x) = (-7x^2 + 5x + 1)+(3x^4 + 3x^3 + x + 1)`

`= -7x^2 + 5x + 1+3x^4 + 3x^3 + x + 1`

`= 3x^4 + 3x^3 - 7x^2 + (5x + x) + (1+1)`

`= 3x^4 + 3x^3 - 7x^2 + 6x + 2`

`F(x) - G(x) = (-7x^2 + 5x + 1) - (3x^4 + 3x^3 + x + 1)`

`= -7x^2 + 5x + 1 - 3x^4 - 3x^3 - x - 1`

`= -3x^4 - 3x^3 - 7x^2 + (5x - x) + (1-1)`

`= -3x^4 - 3x^3 - 7x^2 + 4x`

a/

\(F\left(x\right)=\left(3-6-4\right)x^2+5x+\left(-7+8\right)=-7x^2+5x+1\) -> Đa thức bậc 2

\(G\left(x\right)=\left(1+2\right)x^4+3x^3+\left(2-1\right)x+\left(-1+2\right)=3x^4+3x^3+x+1\) -> Đa thức bậc 4

b/

\(F\left(x\right)+G\left(x\right)=-7x^2+5x+1+3x^4+3x^3+x+1\\ =3x^4+3x^3-7x^2+6x+2\)

\(F\left(x\right)-G\left(x\right)=-7x^2+5x+1-3x^4-3x^3-x-1\\ =-3x^4-3x^3-7x^2+4x\)

Lũy thừa có thể hiểu là tích số của một số với chính nó nhiều lần. Luỹ thừa ký hiệu là \(a^b\) , đọc là lũy thừa bậc b của a hay a mũ b , số a gọi là cơ số, số b gọi là số mũ. Ngoài ra, ta cần biết rằng, phép toán ngược với phép tính lũy thừa là phép khai căn.

 Đ/N:Lũy thừa được viết dưới dạng aⁿ, gồm cơ số a và số mũ là n.

C/T : aⁿ = a.a.............a ( n thừa số a ) ( n khác 0 ).

⇒ ( 4x + 28 ) . 3 + 55  = 35 . 5

 ⇒ ( 4x + 28 ) . 3 + 55 = 175

  ⇒ ( 4x + 28 ) . 3 = 175 - 55

  ⇒ ( 4x + 28 ) .3 = 120

   ⇒  4x + 28 = 120 : 3

    ⇒ 4x + 28 = 40

   ⇒  4x = 40 -  28

    ⇒ 4x = 12

     ⇒ x = 12: 4

⇔ x = 3.

Vậy x= 3.

\(\left[\left(4x+28\right)\times3+55\right]\div5=35\)

\(\left(4x+28\right)\times3+55=175\)

\(12x+84+55=175\)

\(12x=175-55-84\)

\(12x=36\)

\(x=3\)

\(a,x\left(y-z\right)+y\left(z-x\right)+z\left(x-y\right)\\ =xy-xz+yz-xy+xz-yz\\ =\left(xy-xy\right)+\left(xz-xz\right)+\left(yz-yz\right)\\ =0+0+0\\ =0\left(dpcm\right)\)

\(b,x\left(y+z-yz\right)-y\left(z+x-zx\right)+z\left(y-x\right)\\ =xy+xz-xyz-yz-xy+xyz+yz-xz\\ =\left(xy-xy\right)+\left(xz-xz\right)+\left(xyz-xyz\right)+\left(yz-yz\right)\\ =0+0+0+0\\ =0\left(dpcm\right)\)

a) 

\(P=\left(x^{14}-9x^{13}\right)-\left(x^{13}-9x^{12}\right)+\left(x^{12}-9x^{11}\right)-...+\left(x^2-9x\right)-\left(x-9\right)+1\)

\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+x^{11}\left(x-9\right)+...+x\left(x-9\right)-\left(x-9\right)+1\)

\(P\left(9\right)=1\)

b)

\(Q=\left(x^{15}-7x^{14}\right)-\left(x^{14}-7x^{13}\right)+\left(x^{13}-7x^{12}\right)-...-\left(x^2-7x\right)+\left(x-7\right)+2\)

\(=x^{14}\left(x-7\right)-x^{13}\left(x-7\right)+x^{12}\left(x-7\right)-...-x\left(x-7\right)+\left(x-7\right)+2\)

\(Q\left(7\right)=2\)

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

\(\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)

Vì \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\) 

=> \(x+2014=0\) 

                  \(x=0-2014\) 

                  \(x=-2014\)

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)\(\left(x+1\right)\times\dfrac{1}{10}+\left(x+1\right)\times\dfrac{1}{11}+\left(x+1\right)\times\dfrac{1}{12}-\left(x+1\right)\times\dfrac{1}{13}-\left(x+1\right)\times\dfrac{1}{14}=0\)

\(\left(x+1\right)\times\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0\) 

 => \(x+1=0\)

             \(x=0-1\)

             \(x=-1\)

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\\ \Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\\ \Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\\ \Rightarrow x+1=0\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\\ \Rightarrow x=-1\)