8.6+288:(x-3)^2=50
dau cham la nhan do
giup voi nha !!!!!!!!!
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8.6+ 288: ( x -3)\(^2\)= 503
48+ 288: (x-3)\(^2\)= 50
288: (x-3)\(^2\)= 50-48
288: (x-3)\(^2\)= 2
(x-3)\(^2\)= 288:2
(x-3)\(^2\)= 144
(x-3)\(^2\)= 12\(^2\)
=>x- 3= 12=> x= 12+ 3x= 15
8*6+288/(x-3)^2=50
48+288/(x-3)^2=50
288/(x-3)^2=50-48
288/(x-3)^2=2
(x-3)^2=2*288=576=24^2
=>x-3=24
x=24+3
x=27
Vậy x=27
toi hoc gioi toan tinh lai thu xem 27 la x hay sao minh tinh miet ko ra
\(8.6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow48+288:\left(x-3\right)^2=50\Leftrightarrow288:\left(x-3\right)^2=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\Leftrightarrow\left(x-3\right)^2=\left(\pm12\right)^2\)
TH1 : \(x-3=12\Leftrightarrow x=15\)
TH2 : \(x-3=-12\Leftrightarrow x=-9\)
Biến đổi được: (x-3)2=144=122=(-12)2
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
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a: Ta có: N nằm trên đường trung trực của AB
nên NA=NB
b: Ta có:M nằm trên đường trung trực của AB
nên MA=MB
Xét ΔMAN và ΔMBN có
MA=MB
AN=BN
MN chung
Do đó: ΔMAN=ΔMBN
Suy ra: \(\widehat{MAN}=\widehat{MBN}=90^0\)
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
\(8.6+288\left(x-3\right)^2=50\)
\(=>48+228\left(x-3\right)^2=50\)
\(=>288\left(x-3\right)^2=2\)
\(=>\left(x-3\right)^2=\frac{1}{144}\)
\(=>x-3=\frac{1}{12}\)
\(=>x=\frac{37}{12}\)