-25-(x+5) = 415 + 5(x-83)

=> -25-x-5=415+5x-415

=> -30-x-5x = 0

=> -30=6x

=> x = -5 

`-25-(x+5)=415+5(x-83)`

`<=> -25-x-5=415+5x-415`

`<=> -x-5x=415-415+25+5`

`<=> -6x=30`

`<=> x=-5`

-25 - ( x+5 ) = 415 + 5(x- 83)

\(-25-x-5-415-5x+415=0\)

\(-30-6x=0\)

\(6x=-30\)

\(x=-30:6=-5\)

Vậy x=-5

=>\(\dfrac{3}{x-5}-\dfrac{y}{3}=\dfrac{1}{6}\)

=>\(\dfrac{9-y\left(x-5\right)}{3\left(x-5\right)}=\dfrac{1}{6}\)

=>9-y(x-5)=1/2(x-5)

=>(x-5)(1/2+y)=9

=>(x-5)(2y+1)=18

=>\(\left(x-5;2y+1\right)\in\left\{\left(18;1\right);\left(-18;-1\right);\left(2;9\right);\left(-2;-9\right);\left(6;3\right);\left(-6;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(23;0\right);\left(-13;-1\right);\left(7;4\right);\left(3;-5\right);\left(11;1\right);\left(-1;-2\right)\right\}\)

\(3\left(x+2\right)^2-5^2=2.5^2\)

\(\Rightarrow3\left(x+2\right)^2=2.5^2+5^2\)

\(\Rightarrow3\left(x+2\right)^2=5^2\left(2+1\right)\)

\(\Rightarrow3\left(x+2\right)^2=5^2.3\)

\(\Rightarrow\left(x+2\right)^2=5^2\)

\(\Rightarrow\left[{}\begin{matrix}x+2=5\\x+2=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\) \(\Rightarrow x=3\left(x\inℕ\right)\)

3(x + 2)² - 5² = 2.5²

3(x + 2)² - 25 = 50

3(x + 2)² = 50 + 25

3(x + 2)² = 75

(x + 2)² = 75 : 3

(x + 2)² = 25

x + 2 = 5 hoặc x + 2 = -5

*) x + 2 = 5

x = 5 - 2

x = 3 (nhận)

*) x + 2 = -5

x = -5 - 2

x = -7 (loại)

Vậy x = 3

\(\dfrac{20142014}{20152015}\times x+7986=1+3+5+...+199\)

Vì các số ở vế 2 đều cách nhau 2 đơn vị

=> Số số hạng của vế 2 là \(\left(199-1\right)\div2+1=100\) ( số hạng )

=> Tổng của vế 2 là \(\left(199+1\right)\times100\div2=10000\) 

   Thay vào biểu thức, ta có:

\(\dfrac{20142014}{20152015}\times x+7986=10000\)

                \(\dfrac{2014}{2015}\times x=10000-7986=2014\) 

                              \(x=2014\div\dfrac{2014}{2015}\) 

                               \(x=2015\)

\(\dfrac{20142014}{20152015}\)\(x\)+ 7986 = 1 + 3 + 5 + ...+ 197 + 199

\(\dfrac{2014}{2015}\)\(x\) + 7986 = (199 + 3){ (199 -1): 2 + 1}: 2

\(\dfrac{2014}{2015}\)\(x\) + 7986 = 202. 100: 2

\(\dfrac{2014}{2015}x\)             = 10000

\(\dfrac{2014}{2015}\)\(x\)             =  10000 -  7986

\(\dfrac{2014}{2015}\)\(x\)             = 2014

         \(x\)            = 2014 : \(\dfrac{2014}{2015}\)

         \(x\)            = 2015

\(\dfrac{20142014}{20152015}.x+7986=1+3+5+...+197+199\)

\(\Rightarrow\dfrac{20142014}{20152015}.x+7986=\left[\left(199-1\right):2+1\right]\left(1+199\right):2\)

\(\Rightarrow\dfrac{20142014}{20152015}.x+7986=100.200:2\)

\(\Rightarrow\dfrac{20142014}{20152015}.x=10000-7986\)

\(\Rightarrow\dfrac{20142014}{20152015}.x=2014\)

\(\Rightarrow x=2014:\dfrac{20142014}{20152015}=2014.\dfrac{20152015}{20142014}=\dfrac{2015.10001}{10001}\)

\(\Rightarrow x=2015\)

\(5^x-2-3^2=2^4-\left(2^8x2^4-2^{10}x2^2\right)\)

\(5^x-2-3^2=2^4-\left(2^{8+4}-2^{10+2}\right)\)

\(5^x-2-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(5^x-2-3^2=2^4-0\)

\(5^x-2-3^2=2^4\)

\(5^x-2-9=16\)

\(5^x-2=16+9\)

\(5^x-2=25\)

\(5^x=25+2\)

\(5^x=27\)

Bởi vì 27 không phân tích được 1 số có số mũ là 2

\(\Rightarrow\) Không tồn tại x

\(5^{x-2}-9=16-\left(256.16-1024.4\right)\)

\(\Rightarrow5^{x-2}-9=16-\left(4096-4096\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}-9=16\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow x-2=25:5\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=5\)

\(5^{x-2}-3^2=2^4-\left(2^8\cdot2^4-2^{10}\cdot2^2\right)\)

\(\Rightarrow5^{x-2}-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}=16+9\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow5^{x-2}=5^2\)

\(\Rightarrow x-2=2\)

\(\Rightarrow x=2+2\)

\(\Rightarrow x=4\)

Vậy: x=4

=> 1/x = 5/6 - y/3

1/x = 5-2y/6

=> x(5-2y) = 1.6 = 6

Do x ∈ N => x >= 0

Mà 6>0 => 5-2y > 0

Vì y ∈ N => 5-2y ∈ N*

Ta có bảng:

Do x,y ∈ N => (x,y) = (6,2) (thử lại thỏa mãn)

Vậy x=6; y = 2

a) \(A=\left|x-5\right|+\left|x-7\right|=\left|x-5\right|+\left|7-x\right|\ge\left|x-5+7-x\right|=\left|2\right|=2\)

\(minA=2\Leftrightarrow\)\(7\ge x\ge5\)

b) \(B=\left|2x+1\right|+\left|2x-2\right|=\left|2x+1\right|+\left|2-2x\right|\ge\left|2x+1+2-2x\right|=\left|3\right|=3\)

\(minB=3\Leftrightarrow1\ge x\ge-\dfrac{1}{2}\)