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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: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
\(\left(X^2-\left(6^2-\left(8^2-9.7\right)-7.5\right)5^3\right)\)\(=1\)
\(\left(X^2-\left(36-\left(64-63\right)^3-35\right)^3-15\right)^3\)\(=1\)
\(\left(X^2-\left(36-1^3-35\right)^3-15\right)^3\)\(=1\)
\(\left(X^2-\left(36-1-35\right)^3-15\right)^3\)\(=1\)
\(\left(X^2-\left(35-35\right)^3-15\right)^3\)\(=1\)
\(\left(X^2-0^3-15\right)^3\)\(=1^3\)
\(\Rightarrow X^2-0-15\)\(=1\)
\(X^2-15\)\(=1\)
\(X^2=15+1\)
\(X^2=16\)
Mà: \(X^2=4^2\)
\(\Rightarrow X=4\)
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
Ta có :
(3x-8):4=7
3x-8 =7*4
3x-8 =28
3x-8 =28+8
3x =36
x =6 (TMĐK)
Vậy x = 6
a/ 8.6+288:(X-3)2=50
48+288:(x-3)2=50
288:(x-3)2=50-48=2
(x-3)2=288:2=144
(x-3)2=122
x-3=12
x=12+3=15
b/ (x2 - (62-(82-9.7)2- 5.3)3= 1
(x2 - (62-(64-9.7)2- 5.3)3= 1
(x2 - (62-(64-63)2- 5.3)3= 1
(x2 - (62-12- 5.3)3= 1
(x2 - (36-1- 15)3= 1
x2 - 203= 1
x2 - 8000= 1
x2=1+8000=8001
x2=................(de sai
b)
{ 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