Violympic toán 6

VT

a , 43 + ( 9 - 21 ) = 317 - ( x + 317 )

b , ( 15 - x ) + ( x - 12 ) = 7 - ( -5 + x )

c , x - { 57 - [ 42 + ( -23 - x )]} = 13 - { 47 + [ 25 - ( 32 - x )]}

d , -7 + | x - 4 | = -3

e , 13 - | x + 5 | = 13

g , | x - 10 | - ( -12 ) = 4

h , | x + 2 | < 5

H24
4 tháng 2 2020 lúc 21:21

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 \(x\in\left\{-1;-3;0;-4;1;-5;2;-6\right\}\)

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NS
5 tháng 2 2020 lúc 9:42

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}

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