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Lời giải:
Ta có:
$(x-1)^2\geq 0,\forall x$
$|3-y|\geq 0, \forall y$
$\Rightarrow (x-1)^2+|3-y|\geq 0$
$\Rightarrow (x-1)^2+|3-y|-35\geq -35$
$\Rightarrow P=-[(x-1)^2+|3-y|-35]\leq 35$
Vậy $P_{\max}=35$.
Giá trị này đạt tại $(x-1)^2=|3-y|=0$
$\Leftrightarrow x=1; y=3$
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
\(a,x\in\left\{-4;-3;-2;-1;0;1;2\right\}\)
\(b,x\in\left\{-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11\right\}\)
\(c,x\in\left\{-5;5\right\}\)
\(d;|x|=|-7|\)
<=>\(|x|=7\)
=>\(x\in\left\{-7;7\right\}\)
\(e,|x|=-|6|\)
<=>\(|x|=-6\)
=>\(x\in\varnothing\)
\(g,|-37|-|x|=|-5|\)
<=>\(37-|x|=5\)
<=>\(|x|=32\)
=>\(x\in\left\{-32;32\right\}\)
\(h,3+|x|=9\)
<=>\(|x|=6\)
=>\(x\in\left\{-6;6\right\}\)
\(i,3< |x|< 7\)
=>\(|x|\in\left\{4;5;6\right\}\)
=>\(x\in\left\{-6;-5;-4;4;5;6\right\}\)
a,\(15-\left(-4-x\right)=6\)
\(\Leftrightarrow-4-x=15-6\)
\(\Leftrightarrow-4-x=9\)
\(\Leftrightarrow x=-4-9\)
\(\Leftrightarrow x=-13\)
b,\(-30+\left(25-x\right)=-1\)
\(\Leftrightarrow25-x=-1-\left(-30\right)\)
\(\Leftrightarrow25-x=29\)
\(\Leftrightarrow x=25-29=-4\)
c,\(x-\left(12-25\right)=-8\)
\(\Leftrightarrow x+13=-8\)
\(\Leftrightarrow x=-21\)
e,\(x-5=1\Rightarrow x=6\)
g,\(x+30=-4\)
\(\Leftrightarrow x=-4-30\)
\(\Leftrightarrow x=-34\)
h,\(x-\left(-24\right)=3\)
\(\Leftrightarrow x+24=3\)
\(\Leftrightarrow x=3-24\)
\(\Leftrightarrow x=-21\)
i,\(\left(x+5\right)+\left(x-9\right)=x+2\)
\(\Leftrightarrow x+5+x-9-x-2=0\)
\(\Leftrightarrow x-6=0\)
\(\Leftrightarrow x=6\)
k,\(\left(27-x\right)+\left(15+x\right)=x-24\)
\(\Leftrightarrow27-x+15+x-x+24=0\)
\(\Leftrightarrow-x+66=0\)
\(\Leftrightarrow-x=-66\)
\(\Leftrightarrow x=66\)
a) 15-(-4-x)=6
15 + 4 + x = 6
19 + x = 6
x = 6 - 19
x = 6 + ( -19 )
x = -13
b) -30+( 25- x ) =-1
25- x = -1 - ( - 30 )
25- x = -1 + 30
25- x =29
x = 25 - 29
x = 25 + ( -29 )
x = -4
c) x- ( 12-25) =-8
x - ( - 13 ) = -8
x + 13 = -8
x = -8 - 13
x = -8 + ( -13 )
x = - 21
d) ( x-29 ) - (17- 38 ) = -9
( x-29) - (-21 ) = -9
x -29 = -9 + ( -21 )
x -29 = -30
x = -30 + 29
x = -1
e) x-5=1
x = 1 + 5
x = 6
g) x + 30 = -4
x = -4 - 30
x = -4 + ( -30 )
x = -34
h) x- ( -24 ) = 3
x = 3 + ( -24 )
x = -21
l) 22- ( - x ) = 12
-x = 22 - 12
-x = 10
x = -10
i) ( x + 5 ) + ( x - 9 ) = x + 2
x + 5 + x - 9 = x + 2
x + x + ( 5 - 9 ) = x + 2
x + x + ( -4 ) = x + 2
x + x = x + 2 - ( -4 )
x + x = x + 2 + 4
x + x =x + 6
=> x = 6
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 :
\(\left|x+2\right|+\left|x+5\right|+\left|x+9\right|+\left|x+11\right|\ge0\forall x\)
\(\Rightarrow5x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+2+x+5+x+9+x+11=5x\)
\(\Leftrightarrow4x+27=5x\)
\(\Leftrightarrow5x-4x=27\)
\(\Rightarrow x=27\)