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1)
25+27+x=21+l-22l
=>25+27+x=21+22
=>25+27+x=43
=>52+x=43
=>x=43-52
=>x=-9
2)
l-5l+l-7l=x+3
=>5+7=x+3
=>12=x+3
=>x=12-3
=>x=9
3)
8+lxl=l-8l+15
=>8+lxl=8+15
=>8+lxl=23
=>lxl=23-8
=>lxl=15
=>x=15 hoặc x=-15
4)
lxl+15=-7
=>lxl=-7-15
=>lxl=-22
=>x ko tồn tại
1)25+27+x=21+22 3)8+x=8+15
52+x=43 8+x=23
x=52-43=9 x=23-8=15
2)5+7=x+3 x=15 hoặc x=-15
12=x+3 4) x+15=-7
x=12-3=9 x=(-7)-15=-22
x=22 hoặc x=-22
-(-2009+79)-74.(-18)+74.(-118)-2009-3
=2009-79-74.[(-18)+118]-2009-3
= (2009-2009)-(79+3)-(74.100)
= 0-81-7400
= -7481
-1+3 -5+7 -... - 97+99
câu này sai đề nha bạn chỉnh sửa đi mik làm
\(3,1+5^2+5^4+...+5^{26}\)
\(=\left(1+5^2\right)+\left(5^4+5^6\right)+...+\left(5^{24}+5^{26}\right)\)
\(=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{24}\left(1+5^2\right)\)
\(=26+5^4.26+...+5^{24}.26\)
\(=26\left(5^4+...+5^{24}\right)\)
Vì \(26⋮26\)
\(\Rightarrow26\left(5^4+...+5^{24}\right)⋮26\)
\(\Rightarrow1+5^2+5^4+...+5^{26}⋮26\)
\(4,1+2^2+2^4+...+2^{100}\)
\(=\left(1+2^2+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=\left(1+2^2+2^4\right)+....+2^{98}\left(1+2^2+2^4\right)\)
\(=21+2^6.21...+2^{98}.21\)
\(=21\left(2^6+...+2^{98}\right)\)
Có : \(21\left(2^6+...+2^{98}\right)⋮21\)
\(\Rightarrow1+2^2+2^4+...+2^{100}⋮21\)
a) \(\left|x\right|-2=5\Leftrightarrow\left|x\right|=7\Leftrightarrow x=\pm7\)
b) \(\left|x-2\right|=5\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-5\end{cases}}}\)
c) \(2\left(x+7\right)=-16\Leftrightarrow x+7=8\Leftrightarrow x=1\)
d) \(\left(x-5\right)\left(x+7\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
\(a,x\in\left\{-4;-3;-2;-1;0;1;2\right\}\)
\(b,x\in\left\{-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11\right\}\)
\(c,x\in\left\{-5;5\right\}\)
\(d;|x|=|-7|\)
<=>\(|x|=7\)
=>\(x\in\left\{-7;7\right\}\)
\(e,|x|=-|6|\)
<=>\(|x|=-6\)
=>\(x\in\varnothing\)
\(g,|-37|-|x|=|-5|\)
<=>\(37-|x|=5\)
<=>\(|x|=32\)
=>\(x\in\left\{-32;32\right\}\)
\(h,3+|x|=9\)
<=>\(|x|=6\)
=>\(x\in\left\{-6;6\right\}\)
\(i,3< |x|< 7\)
=>\(|x|\in\left\{4;5;6\right\}\)
=>\(x\in\left\{-6;-5;-4;4;5;6\right\}\)
Ta thấy :
\(\left|x+2\right|\ge0\forall x\)
\(\left|x+5\right|\ge0\forall x\)
\(\left|x+9\right|\ge0\forall x\)
\(\left|x+11\right|\ge0\forall x\)
Cộng vế với vế ta được :
\(\left|x+2\right|+\left|x+5\right|+\left|x+9\right|+\left|x+11\right|\ge0\forall x\)
\(\Rightarrow5x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+2+x+5+x+9+x+11=5x\)
\(\Leftrightarrow4x+27=5x\)
\(\Leftrightarrow5x-4x=27\)
\(\Rightarrow x=27\)
= 14 + 5 +7 - 18 = 8