Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
a)390-(x-7)=169:13
<=>390-(x-7)=13
<=>x-7=390-13
<=>x-7=377
<=>x=377+7
<=>x=384
b)x-6:2-(48-24):2:6-3=0
<=>x-3-24:2:6-3=0
<=>x-6-12:6=0
<=>x-6-2=0
<=>x-8=0
<=>x=8
a) 390 - ( x - 7 ) = 169 : 13
390 - ( x - 7 ) = 13
x - 7 = 390 - 13
x - 7 = 377
x = 377 + 7
x = 384
b) x - 6 : 2 - ( 48 - 24 ) : 2 : 6 - 3 = 0
x - 3 - ( 48 - 24 ) : 2 : 6 = 0 + 3 = 3
x - 3 - ( 48 - 24 ) : 2 = 3 x 6 = 18
x - 3 - ( 48 - 24 ) = 18 x 2
x - 3 - 24 = 36
x - 3 = 36 + 24
x - 3 = 60
x = 60 + 3
x = 63
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x-2 | 1 | -1 | 3 | -3 |
x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
x-2 | 1 | -1 | 13 | -13 |
x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+7 | 1 | -1 | 2 | -2 |
x | -6 | -8 | -5 | -9 |
390-(x-7)=169:13
390-(x-7)=13
X-7=390-13
X-7=377
X=377+7
X=384
phần a bạn ở dưới làm r nhé
phần b chép sai đề
phần c :
x - 6 : 2 - ( 48 - 24 ) : 2 : 6 - 3 = 0
<=>x - 3 - 24 : 12 - 3 = 0
<=>x - 3 - 2 - 3 = 0
<=>x=8
Vậy x = 8
phần d :
x + 5 . 2 - ( 32 + 16 . 3 : 16 - 15 ) = 0
<=>x + 10 - ( 32 + 3 - 15 ) = 0
<=> x + 10 - 20 = 0
<=>x = 10
Vậy x = 10
hình như bài làm r bạn ạ :0 mình giải r đấy xem lại nhé :V
a) \(390-\left(x-7\right)=169:13\)
\(390-\left(x-7\right)=13\)
\(x-7=390-13\)
\(x-7=377\)
\(x=384\)
b) \(\left(x-110\right):7-3^3-2^3.3\)
Đề sai nhé bạn
c) \(x-6:2-\left(48-24\right):2:6-3=0\)
\(x-3-\left(24\right):2:6=3\)
\(x-3-\left(24\right):2=18\)
\(x-3-\left(24\right)=36\)
\(x-3=36+24\)
\(x-3=60\)
\(x=63\)
d) \(x+5.2-\left(32+16.3:16-15\right)=0\)
\(x+10-\left(32+48:16-15\right)=0\)
\(x+10-\left(32+3-15\right)=0\)
\(x+10-20=0\)
\(x+10=20\)
\(x=10\)
\(a,\Leftrightarrow\left(2-x\right)\left(x^2+4\right)>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\\ b,\Leftrightarrow x+3>0\Leftrightarrow x>-3\\ c,\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>4\end{matrix}\right.\)
\(...\Rightarrow\left[{}\begin{matrix}288:\left(x-3\right)^2-2=0\\x^2-169=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}288:\left(x-3\right)^2=2\\x^2=169\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=288:2\\x^2=13^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=144=12^2\\x^2=13^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\\x=13\\x=-13\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=15\\x=-9\\x=13\\x=-13\end{matrix}\right.\) \(\Rightarrow x\in\left\{-9;15;\pm13\right\}\)