Violympic toán 6

Vương Hà Thu

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

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

Bình luận (0)
 Khách vãng lai đã xóa
Nguyễn Huy Hoàng Sơn
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}

Bình luận (0)
 Khách vãng lai đã xóa

Các câu hỏi tương tự
nguyen viet anh
Xem chi tiết
Hoàng Thị Trâm
Xem chi tiết
nguyen ngoc son
Xem chi tiết
Chung Diem Ngoc Ha
Xem chi tiết
Vũ Hương Giang
Xem chi tiết
Phươngg Thùyy
Xem chi tiết
Xem chi tiết
Linda Linda
Xem chi tiết
nguyen ngoc son
Xem chi tiết