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\(x=\dfrac{7}{25}+\dfrac{-1}{5}=\dfrac{7}{25}-\dfrac{1}{5}=\dfrac{2}{25}.\\ x=\dfrac{5}{11}+\dfrac{4}{-9}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{1}{99}.\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\Leftrightarrow\dfrac{5}{9}+x=-\dfrac{1}{3}.\Leftrightarrow x=-\dfrac{8}{9}.\)
\(x=\dfrac{7}{25}+-\dfrac{1}{5}=>\dfrac{7}{25}+-\dfrac{5}{25}=>x=\dfrac{2}{25}\)
\(x=\dfrac{5}{11}+\dfrac{4}{-9}=>\dfrac{-45}{-99}+\dfrac{44}{-99}=>x=\dfrac{-1}{-99}=\dfrac{1}{99}\)
\(\dfrac{5}{9}-\dfrac{x}{-1}=-\dfrac{1}{3}=>-\dfrac{1}{3}-\dfrac{5}{9}=>\dfrac{x}{-1}=-\dfrac{8}{9}=>x=-\dfrac{8}{9}\)
a, bổ sung đề
\(\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\right)=0\Leftrightarrow x=50\)
a. Tìm x thuộc N sao cho : 2x + 1 thuộc Ư ( 2x + 10)
(2x + 10) ⋮ (2x + 1)
Ta có (2x + 10) = (2x + 1 + 9)
Mà (2x + 10) ⋮ (2x + 1)
Nên 9 ⋮ (2x + 1)
Do đó ta có (2x + 1) ∈ Ư (9) = {-1; 1; -3; 3; -9; 9}
2x + 1 | -1 | 1 | -3 | 3 | -9 | 9 |
2x | -2 | 0 | -4 | 2 | -10 | 8 |
x | -1 | 0 | -2 | 1 | -5 | 4 |
Vậy x = {-1; 0; -2; 1; -5; 4}
b. A = 3 - 5 + 13 - 15 + 23 - 25 + ....... + 93 - 95 + 2020
A = (3 - 5) + (13 - 15) + (23 - 25) +.......+ (93 - 95) + 2020
A = (-2) + (-2) + (-2) + ......... + (-2) + 2020
Có 10 số (-2)
A = (-2) . 10 + 2020
A = (-20) + 2020
A = 2000
c. 2( x + 1) - x - 2 = (-5) - 3
2x + 2 - 1x - 2 = (-8)
2x - 1x + 2 - 2 = (-8)
2x - 1x + 0 = (-8)
2x - 1x = (-8)
(2 - 1)x = (-8)
1 . x = (-8) : 1
x = (-8)
Bài 1:
a) 17-25+55-17
=(17-17)+(-25+55)
=0+30
=30
b.(-37+101)-(91+37)
= -37+101-91-37
=(-37-37)+(101-91)
= -74 + 10
=-64
c.(-4-6):(-5)
=(-10) : (-5)
=2
d)35-(-95)+372-(372+95)
=35+95+372-372-95
=35+(95-95) + ( 372-372)
=35
A, (X+1)+(x+4)+(x+7)+...+(x+25)+(x+28)=155
b,(x+9)+(x-2)+(x+7)
+(x-4)+(x+5)+(x-6)+(x+3)+(x-8)+(x+1)=95
A,(x+1)+(x+4)+(x+7)+...+(x+25)+(x+28)=155
x+1+x+4+x+7+...+x+25+x+28=155
x.[(28-1):3+1]+(1+4+7+...+25+28)=155
10.x+[(28+1).10:2]=155
10.x+145=155
A
<=>x+1+x+4+x+7+...+x+25+x+28=155
<=>10x.(1+4+7+10+13+16+19+22+25+28)=155
<=>10x.145 =155
<=>10x=155:145=\(\frac{31}{29}\)
x=\(\frac{31}{29}:10=\frac{31}{290}\)
Vậy x=\(\frac{31}{290}\)
\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)
\(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+1}{99}+1\right)=-4+4\)
\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)
\(\left(x+100\right).\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)
=> \(\orbr{\begin{cases}x+100=0\\\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}=0\end{cases}}\)
Mà \(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\ne0\)
=> x + 100 = 0
=> x = -100
Vậy x = -100
Câu b trừ mỗi số đi 1 tức là trừ cả cụm đó cho 3 rùi lm tương tự câu a
1) 70 - 5( x - 3 ) = 45
5( x - 3 ) = 70 - 45
5( x - 3 ) = 25
x - 3 = 25 : 5
x - 3 = 5
x = 5 + 3
x = 8
2) 12 + ( 5 + x ) = 20
5 + x = 20 - 12
5 + x = 8
x = 8 - 5
x = 3
3) 130 - ( 100 + x ) = 25
100 + x = 130 - 25
100 + x = 105
x = 105 - 100
x = 5
4) 175 + ( 30 - x ) = 200
30 - x = 200 - 175
30 - x = 25
x = 30 - 25
x = 5
5) 5( x + 12 ) + 22 = 92
5( x + 12 ) = 92 - 22
5( x + 12 ) = 70
x + 12 = 70 : 5
x + 12 = 14
x = 14 - 12
x = 2
6) 95 - 5( x + 2 ) = 45
5( x + 2 ) = 95 - 45
5( x + 2 ) = 50
x + 2 = 50 : 5
x + 2 = 10
x = 10 - 2
x = 8
7) 14x + 54 = 82
14x = 82 - 54
14x = 28
x = 28 : 14
x = 2
8) 15x - 133 = 17
15x = 17 + 133
15x = 150
x = 150 : 15
x = 10
9) 155 - 10( x + 1 ) = 55
10( x + 1 ) = 155 - 55
10( x + 1 ) = 100
x + 1 = 100 : 10
x + 1 = 10
x = 10 - 1
x = 9
a, ( -31 ) + ( -95 ) + 131 + ( -5 )
= <( -31 ) + 131> + <( -95 ) + ( -5 )>
= 100 + ( -100 )
= 0
Dấu < > là ngoặc vuông nha
95-5(5+x)=25
5(5+x)=95-25
5(5+x)=70
5+x=70:5
5+x=14
x=14-5
x=9
vậy x=9
tick cho tớ nhie