Question

[Phạm Kiều Ngọc Hà]

tìm x biết

a) 1 + 2 + 3 + ...... + x = 500500

b) 2x - 2³ . 3² = 138

c) ( 6x - 3⁴ ) . 7³ = 3 . 7⁴

d) 125 - 5 ( 3x - 1 ) = 5⁵ : 5³

e) 4^x+1 + 4⁰ = 65

j) 17 - 2 ( 2x - 3 ) = 3 

giúp mk nha mk cần gấp

Answer 1

[FL.Hermit]

a) pt <=> \(\frac{x\left(x+1\right)}{2}=500500\)

<=> \(x^2+x=1001000\)

<=> \(x^2-1000x+1001x-1001000=0\)

<=> \(\left(x-1000\right)\left(x+1001\right)=0\)

<=> \(\orbr{\begin{cases}x=1000\\x=-1001\end{cases}}\)

Do \(x>0\)=> \(x=1000\)

b)

<=> \(2x=210\)

<=> \(x=105\)

c) 

<=> \(6x-81=3.7\)

<=> \(x=17\)

d)

<=> \(125-5\left(3x-1\right)=5^2\)

<=> \(5\left(3x-1\right)=100\)

<=> \(3x-1=20\)

<=> \(x=7\)

e)

<=> \(4^{x+1}+1=65\)

<=> \(4^{x+1}=64\)

<=> \(x+1=3\)

<=> \(x=2\)

j) 

<=> \(2\left(2x-3\right)=14\)

<=> \(2x-3=7\)

<=> \(x=5\)

Answer 2

[Phạm Kiều Ngọc Hà]

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