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![](https://rs.olm.vn/images/avt/0.png?1311)
a) x = \(\dfrac{1}{5}+\dfrac{2}{11}\)
x = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)
x = \(\dfrac{21}{55}\)
b) \(\dfrac{x}{15}\) = \(\dfrac{3}{5}+\dfrac{-2}{3}\)
\(\dfrac{x}{15}=\dfrac{9}{15}+\dfrac{-10}{15}\)
\(\dfrac{x}{15}=\dfrac{-1}{15}\)
\(x=\dfrac{-1}{15}\cdot15\)
x = -1
c) \(\dfrac{11}{8}+\dfrac{13}{6}=\dfrac{85}{x}\)
\(\dfrac{33}{24}+\dfrac{52}{24}=\dfrac{85}{x}\)
\(\dfrac{85}{24}=\dfrac{85}{x}\)
\(\dfrac{85}{x}=\dfrac{85}{24}\)
\(x=85:\dfrac{85}{24}\)
\(x=85\cdot\dfrac{24}{85}\)
\(x=24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ( \(\dfrac{1}{4}\) + \(\dfrac{-5}{13}\)) +( \(\dfrac{2}{11}\) + \(\dfrac{-8}{13}\) + \(\dfrac{3}{4}\))
= \(\dfrac{1}{4}\) - \(\dfrac{5}{13}\) + \(\dfrac{2}{11}\) - \(\dfrac{8}{13}\) + \(\dfrac{3}{4}\)
= ( \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) + \(\dfrac{2}{11}\)
= 1 - 1 + \(\dfrac{2}{11}\)
= \(\dfrac{2}{11}\)
b, ( \(\dfrac{21}{31}\) + \(\dfrac{-16}{7}\)) +( \(\dfrac{44}{53}\) + \(\dfrac{10}{31}\)) + \(\dfrac{9}{53}\)
= \(\dfrac{21}{31}-\dfrac{16}{7}+\dfrac{44}{53}+\dfrac{10}{31}+\dfrac{9}{53}\)
= ( \(\dfrac{21}{31}\) + \(\dfrac{10}{31}\)) + ( \(\dfrac{44}{53}\) + \(\dfrac{9}{53}\)) - \(\dfrac{16}{7}\)
= 1 + 1 - \(\dfrac{16}{7}\)
= \(\dfrac{14}{7}-\dfrac{16}{7}\)
= - \(\dfrac{2}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
e) \(24=2^3.3\)
\(84=2^2.3.7\)
\(180=2^2.3^2.5\)
\(\RightarrowƯCLN\left(24;84;180\right)=2^2.3=12\)
b) \(24=2^2.3\)
\(36=2^2.3^2\)
\(\RightarrowƯCLN\left(24;36\right)=2^2.3=12\)
g) \(56=2^3.7\)
\(140=2^2.5.7\)
\(\RightarrowƯCLN\left(56;140\right)=2^2.7=28\)
h) \(12=2^2.3\)
\(14=2.7\)
\(8=2^3\)
\(20=2^2.5\)
\(\RightarrowƯCLN\left(12;14;8;20\right)=2\)
d) \(6=2.3\)
\(8=2^3\)
\(18=2.3^2\)
\(\RightarrowƯCLN\left(6;8;18\right)=2\)
k) \(7=7\)
\(9=3^2\)
\(12=2^2.3\)
\(21=3.7\)
\(\RightarrowƯCLN\left(7;9;12;21\right)=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x - 5)4 = (x - 5)6
⇒ (x - 5)6 - (x - 5)4 = 0
⇒ (x - 5)4.[(x - 5)2 - 1]
⇒ (x - 5)4 (x - 6)(x - 4) = 0
⇒ x - 5 = 0 hoặc x - 6 = 0 hoặc x - 4 =0
⇒ x ϵ {5;6;4}
nếu có sai thì báo mình nhé, đây là đề ccuar cô chứ mình ko biết đau
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 2n+15 chia hết cho 2n+7
=> (2n+7)+8 chia hết cho 2n+7
Mà 2n+7 chia hết cho 2n+7
=>8 chia hết cho 2n+7
=>2n+7 thuộc Ư(8)=1;-1;2;-2;4;-4;8;-8
Với 2n+7=1=>2n=-6
=>n= -3
rồi mấy trường hợp kia bn tự làm nhé :)
Ta có: 2n+15 chia hết cho 2n+7
=> (2n+7)+8 chia hết cho 2n+7
Mà 2n+7 chia hết cho 2n+7
=>8 chia hết cho 2n+7
=>2n+7 thuộc Ư(8)=1;-1;2;-2;4;-4;8;-8
Với 2n+7=1=>2n=-6
=>n= -3
Rồi mấy trường hợp kia bn tự làm nhé !
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\overline{11a10b}\) : 5 dư 1 nên b = 1; 6
\(\overline{11a10b}\) ⋮ 9 ⇔ 1 + 1 + a + 1 + 0 + b ⋮ 9 ⇔ 3 + a + b ⋮ 9 ⇒ a + b = 6; 15
Lập bảng ta có:
a+b | 6 6 | 15 15 |
b | 1 6 | 1 6 |
a | 5 0 | 14 9 |
Vậy ta có các cặp a; b thỏa mãn đề bài lần lượt là:
(a; b) =( 5; 1); ( 0; 6); ( 9; 6)
b, \(\overline{23a4b}\) : 2 dư 1; ⋮ 5; ⋮ 9
vì \(\overline{23a4b}\) : 2 dư 1 và ⋮ 5 nên b = 5
\(\overline{23a4b}\) ⋮ 9 ⇔ 2 + 3 + a + 4 + b ⋮ 9 ⇔ 9 + a + b ⋮ 9
⇔ a + b ⋮ 9 ⇔ a + 5 ⋮ 9 ⇔ a = 4
Vậy a = 4; b = 5
a) 7/20 + 3/5 + (-1/4)
= 7/20 + 12/20 - 5/20
= 14/20
= 7/10
b) 1/4 + 1/5 + 1/12
= 15/60 + 12/60 + 5/60
= 32/60
= 8/15
c) 1/4 + 1/6 + (-1/12)
= 3/12 + 2/12 - 1/12
= 4/12
= 1/3
d) 3/4 + 7/10 + 11/20
= 15/20 + 14/20 + 11/20
= 40/20
= 2
a) \(\dfrac{7}{20}\) + \(\dfrac{3}{5}\) + \(\dfrac{-1}{4}\)
= \(\dfrac{7}{20}+\dfrac{12}{20}+\dfrac{-5}{20}\)
= \(\dfrac{14}{20}\)
= \(\dfrac{7}{10}\)
b) \(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{12}\)
= \(\dfrac{15}{60}+\dfrac{12}{60}+\dfrac{5}{60}\)
= \(\dfrac{32}{60}\)
= \(\dfrac{8}{15}\)