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7 tháng 7 2018

{ x2 - [ 62 - ( 82 - 9.7)3 - 7.5]3 - 5.3 }3 = 1

{ x2 + [ 36 - (64 - 63)3 - 35]3 - 15}3 = 1

[ x2 - ( 36 - 13 - 35 ) - 15 ]3 = 1

[ x2 - ( 36 - 1 - 35 ) - 15]3 = 1

[ x2 - ( 35 - 35 ) - 15]3 = 1

[ x2 - 0 - 15]3 = 1

( x2 - 15 )3 = 1

<=> ( x2 - 15)3 = 13

=> x2 - 15 = 1

<=> x2 = 16

=> x = 4

17 tháng 8 2015

1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)

2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)

\(2^x=2.2^8=2^9;x=9\)

4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)

\(\Leftrightarrow3^5.x=3^5.5;x=5\)

20 tháng 7 2015

{x2 - [62 - (64 -9 .7)3 -7.5]- 5.3}3=1

{x2 - [62 - (64 - 63)3 -7.5]- 5.3}3=1

{x2 - [62 - 13 -7.5]- 5.3}3=1

{x2 - [36 - 1 -7.5]- 5.3}3=1

{x2 - [36 - 1 - 35]- 5.3}3=1

{x2 - [35 - 35]- 5.3}3=1

{x2 - 0- 5.3}3=1

{x2  - 0 -5.3}3=1

{x2  - 0 - 15}3=1

{x.x - 0 - 15}3=1

{x.x + 0+15}3=1

{x.x +15}3=1

nếu x bằng 1 thì:(1.1+15)3 không bằng 1

---------------2 thì:(2.2+15)3 ------------------1

-------------- 3 thì:(3.3+15)3-------------------1

-------------- 4 thì:(4.4+15)3 bằng 1

Vậy x bằng 4

a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)

\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)

=>x+1=0

hay x=-1

c: |x-2|=13

=>x-2=13 hoặc x-2=-13

=>x=15 hoặc x=-11

d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)

=>7|x-2|=5/3

=>|x-2|=5/21

=>x-2=5/21 hoặc x-2=-5/21

=>x=47/21 hoặc x=37/21

`@` `\text {Ans}`

`\downarrow`

`3^3 * x^2 - 2^4 * x^2 = 8^2 * 5 - 4^2 * 3^2`

`=> x^2 . (3^3 - 2^4) = 2^6 . 5 - 2^4 . 3^2`

`=> x^2 . 11 = 2^4 . (2^2 . 5 - 3^2)`

`=> x^2 . 11 = 2^4 . 11`

`=> x^2 . 11 - 2^4 . 11 = 0`

`=> 11 . (x^2 - 16) = 0`

`=> x^2 - 16 = 0`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = `\(\pm4\)

Vậy, `x \in`\(\left\{4;-4\right\}\)

_____

\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2\cdot2^2=4^2\cdot3\)

`=>`\(\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+\left(3\cdot2\right)^2=48\)

`=>`\(\dfrac{23}{108}\cdot x+6^2=48\)

`=>`\(\dfrac{23}{108}x=48-6^2\)

`=>`\(\dfrac{23}{108}x=48-36\)

`=>`\(\dfrac{23}{108}x=12\)

`=>`\(x=\dfrac{1296}{23}\)

Vậy, `x = `\(\dfrac{1296}{23}\)

13 tháng 7 2023

\(3^3.x^2-2^4.x^2=8^2.5-4^3.3^2\)

\(\Leftrightarrow x^2\left(27-16\right)=2^6.5-2^6.9\)

\(\Leftrightarrow11x^2=2^6.\left(5-9\right)=-4.2^6=-2^8\)

\(\Leftrightarrow x^2=-\dfrac{2^6}{11}< 0\)

\(\Rightarrow x\in\varnothing\)

\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2.2^2=4^2.3\)

\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+36=48\)

\(\Leftrightarrow\dfrac{23}{108}x=12\Leftrightarrow x=\dfrac{12.108}{23}=\dfrac{1296}{23}\)