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Answer:
a, \(\left|x\right|=5\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
b, \(\left|x\right|< 2\Rightarrow\orbr{\begin{cases}x=\pm1\\x=0\end{cases}}\)
c, \(\left|x\right|=-1\Rightarrow x\in\varnothing\)
d, \(\left|x\right|=\left|-5\right|\Rightarrow\left|x\right|=5\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
e, \(\left|x+3\right|=0\Rightarrow x+3=0\Rightarrow x=-3\)
f, \(\left|x-1\right|=4\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)
g, \(\left|x-5\right|=10\Rightarrow\orbr{\begin{cases}x-5=10\\x-5=-10\end{cases}}\Rightarrow\orbr{\begin{cases}x=15\\x=-5\end{cases}}\)
h, \(\left|x+1\right|=-2\Rightarrow x\in\varnothing\)
j, \(\left|x+4\right|=5-\left(-1\right)\Rightarrow\left|x+4\right|=5+1=6\Rightarrow\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)
k, \(\left|x-1\right|=-10-3\Rightarrow\left|x-1\right|=-13\Rightarrow x\in\varnothing\)
l, \(\left|x+2\right|=12+\left(-3\right)+\left|-4\right|\Rightarrow\left|x+2\right|=12-3+4=13\Rightarrow\orbr{\begin{cases}x+2=13\\x+2=-13\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-15\end{cases}}\)
m, \(\left|x+2\right|-12=-1\Rightarrow\left|x+2\right|=11\Rightarrow\orbr{\begin{cases}x+2=11\\x+2=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=-13\end{cases}}\)
n, \(135-\left|9-x\right|=35\Rightarrow\left|9-x\right|=100\Rightarrow\orbr{\begin{cases}9-x=100\\9-x=-100\end{cases}}\Rightarrow\orbr{\begin{cases}x=-91\\x=109\end{cases}}\)
o, \(\left|2x+3\right|=5\Rightarrow\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)
\(i,3\left(x+4\right)=105\)
\(x+4=35\)
\(x=31\)
\(k,\left(4x+28\right).3+55=7\)
\(\left(4x+28\right).3=-48\)
\(4x+28=-16\)
\(4x=-44\)
\(x=-11\)
\(l,\left(3x-16\right).343=4802\)
\(3x-16=14\)
\(3x=30\)
\(x=10\)
\(m,43+\left(9-21\right)-317+\left(x+317\right)=0\)
\(-244+\left(x+317\right)=0\)
\(x+317=244\)
\(x=-73\)
\(a,\left(x-10\right)=22:11\)
\(x-10=2\)
\(x=2+10=>x=12\)
\(b.2x=-17-15\)
\(2x=-32\)
\(x=-32:2\)
\(x=-16\)
\(c,305+x=-765-100\)
\(305+x=-865\)
\(x=-865-305\)
\(x=-1170\)
\(d,2^x=16:4\)
\(2^x=4=>2^2=4=>x=2\)
Bài 7:
a; 3\(^{x}\).3 = 243
3\(^{x+1}\) = 3\(^5\)
\(x+1\) = 5
\(x=5-1\)
\(x=4\)
Vậy \(x=4\)
b; 2\(^{x}\).162 = 1024
\(2^{x}\) = 1024 : 162
2\(^{x}\) = \(\frac{512}{81}\) (loại vì 512/81 không phải là số tự nhiên)
Không có số tự nhiên nào của x thỏa mãn đề bài.
Vậy \(x\in\) ∅
c; 64.4\(^{x}\) = 168
4\(^{x}\) = 168 : 64
4\(^{x}\) = \(\frac{21}{8}\) (loại)
vì 21/8 không phải là số tự nhiên
Vậy \(x\in\) ∅
d; \(2^{x}\) = 16
\(2^{x}\) = 2\(^4\)
\(x=4\)
Vậy \(x=4\)
Bài 8:
a; (\(2^{17}\) + 17\(^2\)).(9\(^{15}\) - 3\(^{15}\)).(2\(^4\) - 4\(^2\))
= (\(2^{17}\) + 17\(^2\)).(9\(^{15}\) - 3\(^{15}\)).(16 - 16)
= (\(2^{17}\) + 17\(^2\)).(9\(^{15}\) - 3\(^{15}\)).0
= 0
b; (\(8^{2017}\) - 8\(^{2015}\)):(8\(^{2014}\).8\(\))
= (\(8^{2017}\) - 8\(^{2015}\)): \(8^{2015}\)
= \(8^{2017}:8^{2015}\) - \(8^{2015}\) : 8\(^{2015}\)
= 8\(^2\) - 1
= 64 - 1
= 63
c; (1\(^3\) + 2\(^3\) + 3\(^4\) + 4\(^5\)).(\(3^8-81^2\))
= (1\(^3\) + 2\(^3\) + 3\(^4\) + 4\(^5\)).(3\(^8\) - 3\(^8\))
= (1\(^3\) + 2\(^3\) + 3\(^4\) + 4\(^5\)).0
= 0
d; (2\(^8\) + 8\(^3\)) : (2\(^5\).2\(^3\))
= (2\(^8\) + 2\(^9\)):(2\(^8\))
= 2\(^8\) : 2\(^8\) + 2\(^9\) : 2\(^8\)
= 1 + 2
= 3
2x + 12 = 3(x - 7)
\(\Rightarrow\) 2x +12 = 3x - 3 . 7
\(\Rightarrow\) 2x + 12 = 2x + (x - 21)
\(\Rightarrow\) 12 = x - 21
\(\Rightarrow\) x = 12 + 21
\(\Rightarrow\) x = 33
Vậy: x = 33
\(C=\frac{5}{12}\div\frac{1}{12\%}+0,04\)
\(C=\frac{5}{12}\div\frac{25}{3}+0,04\)
\(C=\frac{5}{12}\times\frac{3}{25}+0,04\)
\(C=\frac{1}{4}.\frac{1}{5}+0,04\)
\(C=\frac{1}{20}+0,04\)
\(C=0,5+0,04=0,54\)
2 . x + 12 = 3 . ( x - 7)
2 . x + 12 = 3 .x - 21
21 + 12 = 3 . x - 2 . x
33 = x
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