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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.
{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=\(\sqrt{17}\)
\(\left\{x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3-5.3\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-\left[6^2-\left(8^2-\left(8+1\right).\left(8-1\right)\right)^3-\left(6+1\right).\left(6-1\right)\right]^3-5.3\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-\left[6^2-\left(8^2-8^2+1\right)^3-6^2+1\right]^3-5.3\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-\left[6^2-1-6^2+1\right]^3-5.3\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-5.3\right\}^3=1\)
\(\Leftrightarrow x^2-15=1\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
\(\left\{x^2-\left[6^2-\left(8^2-9.7\right)-7.5\right]^3-5.3\right\}^3=1\)
\(\Rightarrow x^2-\left[6^2-\left(64-63\right)^3-35\right]^3-5.3=1\)
\(\Rightarrow x^2-\left(36-1^3-35\right)^3-15=1\)
\(\Rightarrow x^2-0^3-15=1\)
\(\Rightarrow x^2=1+15\)
\(\Rightarrow x^2=16\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)