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tự tính đi em,giá trị tuyệt đối thì có 2 TH âm và dương nha em.Chúc em học tốt ,maays câu như này thì động não là được nhá
a, -7 + |x-4| = -3
|x-4| = -3-(-7)=4
=> x-4 = 4 hoặc x-4=-4
TH1: x-4=4 => x=8
TH2: x-4=-4 => x=0
Vậy x=8 hoặc x=0
b, 13 - |x+5| = 13
|x+5| = 13 - 13 = 0
=> x+5 = 0
x = 0-5 = -5
Vậy x=-5
c, |x-10| - (-12) = 4
|x-10| = 4 + (-12) = -8 ( vô lý )
=> \(x\in\varnothing\)( Không có trường hợp nào của x thỏa mãn đề bài )
Vậy \(x\varnothing\in\)
Chúc bạn học tốt ^^!!!
a, 43 + ( 9 - 21 ) = 317 - ( x + 317 )
43 + ( -12 ) = 317 - x - 317
43 - 12 = 317 - 317 - x
-x = 31
x = -31
b, (15-x) + (x-12) = 7- (-5 + x)
15-x+x-12 = 7+5-x
15-12 = 12-x
3 = 12-x
x = 12-3
x = 9
c, x - { 57- [42+ (-23 - x)]} = 13- {47+ [25- (32-x)]}
x - [57- (42-23-x)] = 13- [47+ (25-32+x)]
x - [57- (19-x)] = 13- [47+ (x-7)]
x - (57-19+x) = 13- (47+x-7)
x - (38+x) = 13- (40+x)
x-38-x = 13-40-x
x = 13-40+38
x = 11
a) \(43+\left(9-21\right)=317-\left(x+317\right)\\ 43+9-21=317-x-317\\ 52-21=\left(317-317\right)-x\\ 31=-x\\ x=-31\)Vậy x = -31
b) \(\left(15-x\right)+\left(x-12\right)=7-\left(-5+x\right)\\ 15-x+x-12=7+5-x\\ \left(x-x\right)+\left(15-12\right)=12-x\\ 3=12-x\\ x=9\)Vậy x = 9
c) \(x-\left\{57-\left[42+\left(-23-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\\ x-\left\{57-\left[42+\left(-23\right)-x\right]\right\}=13-\left\{47+\left[25-32+x\right]\right\}\\ x-\left\{57-42+23+x\right\}=13-\left\{47+25-32+x\right\}\\ x-57+42-23-x=13-47-25+32-x\\ -57+42-23=-34-25+32-x\\ -15-23=-59+32-x\\ -38=-27-x\\ x=11\)Vậy x = 11
d) \(-7+\left|x-4\right|=-3\\ \left|x-4\right|=4\\ \Rightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\)Vậy \(x\in\left\{8;0\right\}\)
e) \(13-\left|x+5\right|=13\\ \left|x+5\right|=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)Vậy x = -5
g) \(\left|x-10\right|-\left(-12\right)=4\\ \left|x-10\right|=-8\\ \Rightarrow x\in\varnothing\left(\text{vì }\left|x-10\right|\ge0\text{với mọi }x\right)\)Vậy \(x\in\varnothing\)
h) \(\left|x+2\right|< 5\\ 0\le\left|x+2\right|< 5\\ \Rightarrow\left|x+2\right|\in\left\{1;2;3;4\right\}\\ \Rightarrow x+2\in\left\{1;-1;2;-2;3;-3;4;-4\right\}\\ \Rightarrow x\in\left\{-1;-3;0;-4;1;-5;2;-6\right\}\)Vậy \(x\in\left\{-1;-3;0;-4;1;-5;2;-6\right\}\)
a) 43+(9−21)=317−(x+317)43+9−21=317−x−31752−21=(317−317)−x31=−xx=−3143+(9−21)=317−(x+317)43+9−21=317−x−31752−21=(317−317)−x31=−xx=−31Vậy x = -31
b) (15−x)+(x−12)=7−(−5+x)15−x+x−12=7+5−x(x−x)+(15−12)=12−x3=12−xx=9(15−x)+(x−12)=7−(−5+x)15−x+x−12=7+5−x(x−x)+(15−12)=12−x3=12−xx=9Vậy x = 9
c) x−{57−[42+(−23−x)]}=13−{47+[25−(32−x)]}x−{57−[42+(−23)−x]}=13−{47+[25−32+x]}x−{57−42+23+x}=13−{47+25−32+x}x−57+42−23−x=13−47−25+32−x−57+42−23=−34−25+32−x−15−23=−59+32−x−38=−27−xx=11x−{57−[42+(−23−x)]}=13−{47+[25−(32−x)]}x−{57−[42+(−23)−x]}=13−{47+[25−32+x]}x−{57−42+23+x}=13−{47+25−32+x}x−57+42−23−x=13−47−25+32−x−57+42−23=−34−25+32−x−15−23=−59+32−x−38=−27−xx=11Vậy x = 11
d) −7+|x−4|=−3|x−4|=4⇒[x−4=4x−4=−4⇒[x=8x=0−7+|x−4|=−3|x−4|=4⇒[x−4=4x−4=−4⇒[x=8x=0Vậy x∈{8;0}x∈{8;0}
e) 13−|x+5|=13|x+5|=0⇒x+5=0⇒x=−513−|x+5|=13|x+5|=0⇒x+5=0⇒x=−5Vậy x = -5
g) |x−10|−(−12)=4|x−10|=−8⇒x∈∅(vì |x−10|≥0với mọi x)|x−10|−(−12)=4|x−10|=−8⇒x∈∅(vì |x−10|≥0với mọi x)Vậy x∈∅x∈∅
h) |x+2|<50≤|x+2|<5⇒|x+2|∈{1;2;3;4}⇒x+2∈{1;−1;2;−2;3;−3;4;−4}⇒x∈{−1;−3;0;−4;1;−5;2;−6}|x+2|<50≤|x+2|<5⇒|x+2|∈{1;2;3;4}⇒x+2∈{1;−1;2;−2;3;−3;4;−4}⇒x∈{−1;−3;0;−4;1;−5;2;−6}Vậy x∈{−1;−3;0;−4;1;−5;2;−6}
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
a)13-2(x+1)=7
2.(x+1)=13-7
2.(x+1)=6
x+1=6/2
x+1=3
x=3-1
x=2
a, 13 - 2 ( x + 1 ) = 7
2 .( x + 1 ) = 13 - 7
2 .( x + 1 ) = 6
x + 1 = 6 : 2
x + 1 = 3
x = 3 - 1
x = 4
a, => |x+4| = 3-(-7) = 10
=> x+4=10 hoặc x+4=-10
=> x=6 hoặc x=-14
b, => |x+5| = 13 - 13 = 0
=> x+5=0
=> x=0-5 = -5
c, => |x-10| + 12 = 4
=> |x-10| = 4-12 = -8
=> ko tồn tại x tm vì |x-10| > = 0
Tk mk nha
a) -7 + | x - 4 | = 3 b) 13 - | x + 5 | = 13 c) | x - 10 | - (-12) = 4
=> | x - 4 | = 3 - (-7) => | x + 5 | = 13 - 13 =>| x - 10 | + 12 = 4
=> | x - 4 | = 10 => | x + 5 | = 0 =>| x - 10 | = 4 -12
=> x - 4 = 10 hoặc - 10 => x + 5 = 0 =>|x - 10 | = - 8
=> x = 14 hoặc x = -6 => x = 0 - 5 => Với mọi |x - 10| luôn lớn hơn hoặc bằng 0
=> x = - 5 Nên : Không có x thỏa mãn