a, 2.(x+15)=40

2x+30=40

2x = 10

x = 5

Vậy x=5

b, 9.x-33=\(3^{2015}:3^{2014}\)

9.x-33=3

9.x =36

x = 4

Vậy x= 4

A. 2.(x+15)=40

(x+15)=40:2

x+15=20

x =20-15

x =5

Lời giải:

Vì $x=9$ nên $x-9=0$
Ta có:

$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$

$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$

$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$

$=x-10=9-10=-1$

\(x^2+2y^2-2xy+4y+3< 0\)

\(\Rightarrow x^2-2xy+y^2+y^2+4y+4-1< 0\)  

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)-1< 0\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Mà: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\) 

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1\ge-1\forall x,y\)

Mặt khác: \(\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\)

\(\Rightarrow x=y=-2\)

Vậy: .... 

Áp dụng công thức : a = bq + r , ta có

x : 4 = 16 dư 3

x = 16 . 4 + 3

x = 67

x : 5 = 22 dư 1

x = 22 . 5 + 1

x = 111

x : 4 = 16 (dư 3)

      x =16 x 4 + 3

      x = 67.

x :5 = 22 (dư 1 )

    x = 22 x 5 +1

     x= 111

\(\frac{2x-4}{2014}+\frac{2x-2}{2016}\) và \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)

VT = \(\frac{2x-4}{2014}+\frac{2x-2}{2016}\)

= \(\frac{2x-4}{2014}+1+\frac{2x-2}{2016}+1\)

= \(\frac{2x-2018}{2014}+\frac{2x-2018}{2016}\)

VP = \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)

= \(\frac{2x-1}{2017}+1+\frac{2x-3}{2015}+1\)

= \(\frac{2x-2018}{2017}+\frac{2x-2018}{2015}\)

Mà \(\frac{2x-2018}{2014}>\frac{2x-2018}{2015}\) và \(\frac{2x-2018}{2016}>\frac{2x-2018}{2017}\)

nên \(\frac{2x-4}{2014}+\frac{2x-2}{2016}\) > \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)

Chúc bn học tốt!!