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chỉ cần giải cho mình câu C bài 1 và câu B,C bài 2 thôi nhé
(139139 . 133 - 133133 . 139) : (2 + 4+ 6 + ... + 2002)
= (139 . 1001 . 133 - 133 . 1001 . 139) : (2 + 4 + 6 + ... + 2002)
= 0 : (2 + 4 + 6 + ... + 2002)
= 0
a) 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
b) 10 + 2.x = 4^5 : 4^3
10 + 2.x=4^2
10 + 2.x=16
2.x=16-10
2.x=6
x=6:2
x=3
c) 2x - 138 = 2^3 . 3^2
2x - 138 = 8.9
2x - 138 = 72
2x=72+138
2x=210
x=210:2
x=105
d) 231 - (x - 6) = 133 : 13
(đề sai)
a: \(A=\left[111\cdot\left(38\cdot3-133\right)\right]\cdot2018\)
\(=111\cdot\left(-19\right)\cdot2018=-4255962\)
b: \(B=\left[25\left(1+3\cdot3-5\cdot2\right)\right]:20182019\)
\(=25\left(1+9-10\right):20182019=0\)
c: \(C=\dfrac{\left[244\left(395-195\right)\right]}{122\cdot200}=\dfrac{244}{122}=2\)
d: \(D=\dfrac{355\cdot45\cdot55}{11\cdot9\cdot7}=\dfrac{355\cdot5\cdot5}{7}=\dfrac{8875}{7}\)
Ta có: \(1-\left(\frac{49}{9}+x-\frac{133}{18}\right):\frac{11}{2}=\frac{4}{9}\)
\(\Rightarrow\left(\frac{49}{9}+x-\frac{133}{18}\right):\frac{11}{2}=1-\frac{4}{9}=\frac{5}{9}\)
\(\frac{49}{8}+x-\frac{133}{19}=\frac{5}{9}.\frac{11}{2}=\frac{55}{18}\)
\(\frac{49}{8}+x=\frac{55}{18}+\frac{133}{19}=\frac{181}{18}\)
\(\Rightarrow x=\frac{181}{18}-\frac{49}{8}=\frac{283}{72}\)
Vậy \(x=\frac{283}{72}\)
1 - ( 49/9 + x -133/18 ):11/2 = 4/9
1 - ( 49/9 + x - 133/18 ) = 4/9 . 11/2
1 - ( 49/9 + x - 133/18 ) = 22/9
49/9 + x - 133/18 = 1 - 22/9
49/9 + x - 133/18 = - 13/9
49/9 + x = -13/9 - 133/18
49/9 + x = -53/6
x = -53/6 -49/9
x = -257/18
a) \(10+2x=4^5:4^3\)
\(\Leftrightarrow10+2x=4^2\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(14x+54=82\)
\(\Leftrightarrow14x=28\)
\(\Leftrightarrow x=2\)
c) \(15x-133=17\)
\(\Leftrightarrow15x=150\)
\(\Leftrightarrow x=10\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`x + 10 = 20`
`=> x = 20 -10`
`=> x = 10`
Vậy, `x = 10`
`b)`
`2 * x + 15 = 35`
`=> 2x = 35 - 15`
`=> 2x = 20`
`=> x = 20 \div 2`
`=> x = 10`
Vậy, `x = 10`
`c)`
`3 * ( x + 2 ) = 15`
`=> x + 2 = 15 \div 3`
`=> x + 2 = 5`
`=> x = 5 - 2`
`=> x = 3`
Vậy, `x = 3`
`d)`
`10 * x + 15 * 11 = 20 * 10`
`=> 10x + 165 = 200`
`=> 10x = 200 - 165`
`=> 10x = 35`
`=> x = 35 \div 10`
`=> x = 3,5`
Vậy,` x = 3,5`
`e)`
`4 * ( x + 2 ) = 3 * 4`
`=> x + 2 = 12 \div 4`
`=> x + 2 = 3`
`=> x = 3 - 2`
`=> x = 1`
Vậy,` x = 1`
`f)`
`33 x + 135 = 26 * 9`
`=> 33x + 135 = 234`
`=> 33x = 234 - 135`
`=> 33x = 99`
`=> x = 99 \div 33`
`=> x = 3`
Vậy, `x = 3`
`g)`
`2 * x + 15 + 16 + 17 = 100`
`=> 2x + 48 = 100`
`=> 2x = 100 - 48`
`=> 2x = 52`
`=> x = 52 \div 2`
`=> x =26`
`h)`
`2 * (x + 9 + 10 + 11) = 4 . 12 . 25`
`=> 2 * (x + 9 + 10 + 11) = 4*25*12`
`=> 2 * (x + 9 + 10 + 11) = 100*12`
`=> x + 9 + 10 + 11 = 100*12 \div 2`
`=> x + 30 = 600`
`=> x = 600 - 30`
`=> x = 570`
Vậy, `x = 570.`
a) \(x+10=20\Leftrightarrow x=10\)
b) \(2x+15=35\Leftrightarrow2x=20\Leftrightarrow x=10\)
c) \(3.\left(x+2\right)=15\Leftrightarrow x+2=5\Leftrightarrow x=3\)
d) \(10x+15.11=20.10\Leftrightarrow10x+165=200\Leftrightarrow10x=35\Leftrightarrow x=\dfrac{35}{10}=\dfrac{7}{2}\)
e) \(4.\left(x+2\right)=3.4\Leftrightarrow x+2=3\Leftrightarrow x=1\)
f) \(35x+135=26.9\Leftrightarrow35x=234-135\Leftrightarrow35x=99\Leftrightarrow x=\dfrac{99}{35}\)
g) \(2x+15+16+17=100\Leftrightarrow2x+48=100\Leftrightarrow2x=52\Leftrightarrow x=26\)
h) \(2.\left(x+9+10+11\right)=4.12.25\)
\(\Leftrightarrow x+30=2.12.25\)
\(\Leftrightarrow x=600-30\)
\(\Leftrightarrow x=570\)