Bài 1:

a: Ta có: \(48751-\left(10425+y\right)=3828:12\)

\(\Leftrightarrow y+10425=48751-319=48432\)

hay y=38007

b: Ta có: \(\left(2367-y\right)-\left(2^{10}-7\right)=15^2-20\)

\(\Leftrightarrow2367-y=1222\)

hay y=1145

Bài 2: 

Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

<=> [x²-(36-1-35)³-15]³=1

<=> (x²-15)³=1

<=> x²-15=1

<=> x²=16

<=>\(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

{ x² - [ 6² - ( 8² -9.7 )³ -7.5 ]³ -5.3 }³ =1 ... {x² - [6² - 1³ -7.5]³ - 5.3}³=1. {x² - [36 - 1 -7.5]³ - 5.3}³=1. {x² - [36 - 1 - 35]³ - 5.3}³=1. {x² - [35 - 35]³ -5.3}³=1. {x² - 0³ - 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.

Là sao

{x^2−[6^2−(8^2−9⋅7)^3−7⋅5]^3−5⋅3}^3=1

⇒x^2−[36−(64−63)^3−35]^3−15=1

⇒x^2−[36−35−1^3]^3=16

⇒x^2−0^3=16

⇒x^2=16

⇒x=±4

Hok tốt

a,8.6+288:(x-3)^2=50

48+288:(x-3)^2=50

288:(x-3)^2=2

(x-3^2=288:2

(x-3)^2=144=12^2

x=12+3

=>x=15

x=100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

{ 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

<=>{x²-[6²(64-63)³-35]³-15}³=1

<=>[x²-(6².1³-35)³-15]³=1

<=>[x²-(36-35)³-15]³=1

<=>(x²-1-15)³=1

<=>(x²-16)=\(\sqrt{1}=1\)

=>x²=17=>x=\(\sqrt{17}\)