\(x^2-4x+9y^2+6y+10\\ =\left(x^2-4x+4\right)+\left(9y^2+6y+1\right)+5\\ =\left(x-2\right)^2+\left(3y+1\right)^2+5\ge5>0\)

có ai giúp mình câu này với mai phải nộp rồi

Ta có:

\(5^2E=5^2+5^4+5^6+5^8+...+5^{110}+5^{112}\)

\(\Rightarrow25E-E=5^{112}-1\Leftrightarrow E=\dfrac{5^{112}-1}{24}\)

     \(x^2+2x+4⋮x+1\)

\(\Leftrightarrow\left(x^2+x\right)+\left(x+1\right)+3⋮x+1\)

\(\Leftrightarrow x\left(x+1\right)+\left(x+1\right)+3⋮x+1\)

\(\Leftrightarrow3⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow x\in\left\{0;-2;2;-4\right\}\)

Ta có: \(x^2+2x+4\)

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

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

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

Để \(x^2+2x+4\) chia hết cho x + 1 thì 3 phải chia hết cho x + 1

\(\Rightarrow\left(x+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow x\in\left\{-4;-2;0;2\right\}\)

\(2^{14}:2^{10}-3x+3^2=4^3-\left(-5\right)\\ 2^2-3x+9=64-\left(-5\right)\\ 4-3x+9=69\\ 4-3x=69-9\\ 4-3x=60\\ 3x=4-60\\ 3x=-56\\ x=-56;3\\ x=\dfrac{-56}{3}\)

tìm x \(\in\) Z  biết :             

2¹⁴ :  2¹⁰ - 3x + 3² = 4³ -(-5)

       2⁴  - 3x   + 9   = 64 + 5

      16 + 9 - 3x      = 69

         25 - 3x         =  69

                 3x        = 25 - 69

                 3x        = - 44

                  x        = - \(\dfrac{44}{3}\)  (loại vì x \(\in\) Z)

     Kết luận : x \(\in\) \(\varnothing\)

A = ( - 5)^32 . 20^43 / ( -8 )^29 . 125^25

   = 5^32 . 20^43 / ( -8 )^29 . 125^25

   = 5^32 . 20^43 / ( -8 )^29 . ( 5^3 )^25

   = 5^32 . 20^43 / ( -8 )^29 . 5^75

   = 20^43 / ( -8 )^29 . 5^43

   = ( 4 . 5 )^43 / ( -8 )^29 . 5^43

   = 4^43 . 5^43 / ( -8 )^29 . 5^43

   = 4^43 / ( -8 )^29

   = ( 2^2 )^43 / ( -2^3 )^29

   = 2^86 / -2^87

AI GIÚP MÒNH VỚI !

k rồi giúp

`@` `\text {Ans}`

`\downarrow`

`a)`

`13/50 + 9% + 41/100 + 0,24`

`= 0,26 + 0,09 + 0,41 + 0,24`

`= (0,26 + 0,24) + (0,09 + 0,41)`

`= 0,5 + 0,5`

`= 1`

`b)`

`2018 \times 2020 - 1/2017 + 2018 \times 2019`

`= 2018 \times (2020 + 2019) - 1/2017`

`= 2018 \times 4039 - 1/2017`

`= 8150702`

`c)`

`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`

`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)

`=`\(1-\dfrac{1}{7}\)

`= 6/7`

\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)

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