`@` `\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 = `\(\dfrac{1296}{23}\)

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

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

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

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

\(\Rightarrow x\in\varnothing\)

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

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

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

Bạn ghi đề lại câu a.

{ 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

\(\left(x+1\right)^3=2^3.3+3\)

\(\Rightarrow\left(x+1\right)^3=24+3=27=3^3\)

\(\Rightarrow x+1=3\Rightarrow x=2\)

\(5+2.3^x+2=23\)

\(\Rightarrow7+2.3^x=23\)

\(\Rightarrow2.3^x=23-7=16\)

\(\Rightarrow3^x=8.\)

Mà số 3 lũy thừa mấy cũng không thể bằng 8 => x không thỏa mãn.

a) \(x\left(x-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x-1=0\)

+) \(x=0\)

+) \(x-1=0\Rightarrow x=1\)

Vậy \(x\in\left\{0;1\right\}\)

b) \(\left(x+2\right)\left(x-4\right)=0\)

\(\Rightarrow x+2=0\) hoặc \(x-4=0\)

+) \(x+2=0\Rightarrow x=-2\)

+) \(x-4=0\Rightarrow x=4\)

Vậy \(x\in\left\{-2;4\right\}\)

c) \(x^3.x^2=2^8:2^3\)

\(\Rightarrow x^5=2^5\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

d) \(3^{x-3}-3^2=2.3^2\)

\(\Rightarrow3^{x-3}=18+9\)

\(\Rightarrow3^{x-3}=27\)

\(\Rightarrow3^{x-3}=3^3\)

\(\Rightarrow x-3=3\)

\(\Rightarrow x=6\)

Vậy \(x=6\)

a) x(x-1)=0

=> x.x =0+1

=> x.x =1

=> x^2 =1

=> x^2 =1^2

=> x =1

a) \(\Leftrightarrow2.\left(\frac{2.3^x}{3}+3^x.3^2\right)=2.3^6\left(2+3^3\right)\)

\(\Leftrightarrow2.\left(\frac{2.3^x+3.3^x.3^2}{3}\right)=2.3^6.29\)

\(\Leftrightarrow2.\left[\frac{3^x.\left(2+3.3^2\right)}{3}\right]=2.3^6.19\)

\(\Leftrightarrow2.3^{x-1}.29=2.3^6.29\Leftrightarrow3^{x-1}.29=\frac{2.3^6.29}{2}=3^6.29\Leftrightarrow3^{x-1}=\frac{3^6.29}{29}=3^6\)

\(\Leftrightarrow3^{x-1}=3^6\Leftrightarrow x-1=6\Leftrightarrow x=6+1=7\)

vậy x=7 . Chọn mình nha

mấy bài sao tương tự nếu ko biết thì nhắn tin mình chỉ típ nha