58-(x+34)=15(tìm x)x_x 18+(x24)=50(tìm x)x_x
18x-18=0 (tìm x) x_x
17(x-15)=0(tìm x) x_x
18x-18=0
18x=0+18
18x=18
x=18:18
x=1
17(x-15)=0
(x-15)=0:17
(x-15)=0
x=15+0
x=15
(x-42)-17=127 (tìm x) x_x
23(x+1)=69(tìm x)x_x
2.x+5=120:2(tìm x) x_x
5.x-2=613(tìm x) x_x
(x-42) - 17 = 127
=> x - 42 = 127 + 17 = 144
=> x = 144 + 42 = 186
23(x+1) = 69
=> x + 1 = 69 : 23 = 3
x = 3 - 1 = 2
2x + 5 = 120 : 2 = 60
=> 2x = 60 - 5 = 55
x = 55 : 2 = 27,5
5x - 2 = 613
=> 5x = 613 + 2 = 615
x = 615 : 5 = 123
a)(x-42)-17=127
(x-42)=127+17
(x-42)=144
x=144+42
x=186
b)23(x+1)=69
(x+1)=69:23
(x+1)=3
x=3-1
x=2
c)2.x+5=120:2
2.x+5=60
2.x=60-5
2.x=55
x=55:2
x=27,5
d)5.x-2=613
5.x=613+2
5.x=615
x=615:5
x=123
(x-42)-17=127(tim x)xx
23(x+1)=69(timx)xx
2x+5=120:2(Timx)xx
5x-2=613(timx)xx
(15+x)-24=246 (tìm x) x_x (298-x)-42=252(tìm x)x_x
(15 + x) - 24 = 246
=> 15 + x = 246 + 24 = 270
x = 270 - 15 = 255
(298 - x) - 42 = 252
=> (298 - x) = 252 + 42 = 294
=> x = 298 - 2984 = 4
2.x+13=35 (tìm x)x_x
2x + 13 = 35
=> 2x = 35 - 13
2x = 22
x = 22 : 2
x = 11
Tìm X
125*X_X*47= 25350
\(125\times x-x\times47=25350\\x\times(125-47)=25350\\x\times78=25350\\x=25350:78\\x=325\)
x+78*x+25*x_x*4=78,6+121,4
\(x\) + 78 x \(x\)+ 25 x \(x\)- \(x\)x 4 = 78,6 + 121,4
\(x\)x 1 + 78 x \(x\)+ 25 x \(x\)- \(x\)x 4 = 200
\(x\)x ( 1 + 78 + 25 - 4 ) = 200
\(x\)x 100 = 200
\(x\)= 200 : 100
\(x\)= 2
X + 78 * X + 25 * X - X * 4 = 78,6 + 121,4
X * ( 1 + 78 + 25 - 4 ) = 200
X * 100 = 200
X = 200 : 100
X = 2
Vậy X = 2
Tìm x ∈ Z biết 5 6 + − 7 8 ≤ x 24 ≤ − 5 12 + 5 8
A. x∈{0;1;2;3;4}
B. x∈{−1;0;1;2;3;4;5}
C. x∈{−1;0;1;2;3;4}
D. x∈{0;1;2;3;4;5}
Đáp án cần chọn là: B
x∈{−1;0;1;2;3;4;5}
5 6 + − 7 8 ≤ x 24 ≤ − 5 12 + 5 8 − 1 24 ≤ x 24 ≤ 5 24 − 1 ≤ x ≤ 5
Tìm x ∈ Z, biết:
b) - 7 8 + 5 6 < x 24 ≤ 5 8 + - 5 12
- 7 8 + 5 6 < x 24 ≤ 5 8 + - 5 12
⇔ - 21 24 + 20 24 ≤ x 24 ≤ 15 24 + - 10 24 ⇔ - 1 24 ≤ x 24 ≤ 5 24
-1 ≤ x ≤ 5 mà x ∈ Z. Do đó: x ∈ { -1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5 }
Tìm x ∈ Z, biết:
a ) - 5 19 + 3 19 < x 19 ≤ 13 19 + - 11 19 b ) - 7 8 + 5 6 < x 24 ≤ 5 8 + - 5 12
-1 ≤ x ≤ 5 mà x ∈ Z. Do đó: x ∈ { -1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5 }