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Chi Blink
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\(c,-\dfrac{8}{13}+\left(-\dfrac{7}{5}-x\right)=-\dfrac{1}{2}\\ -\dfrac{7}{5}-x=-\dfrac{1}{2}-\dfrac{8}{13}\\ -\dfrac{7}{5}-x=-\dfrac{29}{26}\\ x=-\dfrac{7}{5}-\left(-\dfrac{29}{26}\right)=-\dfrac{37}{130}\\ d,-1\dfrac{1}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{8}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{8}{7}-\left(-\dfrac{4}{21}\right)\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{20}{21}\\ x-\dfrac{7}{3}=-\dfrac{20}{21}-\left(-\dfrac{5}{3}\right)\\ x-\dfrac{7}{3}=\dfrac{5}{7}\\ x=\dfrac{5}{7}+\dfrac{7}{3}=\dfrac{64}{21}\\ e,-\dfrac{2}{3}-x:\dfrac{1}{2}=\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{2}{3}-\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{16}{15}\\ x=-\dfrac{16}{15}\times\dfrac{1}{2}=-\dfrac{8}{15}\)

Nguyễn Lê Phước Thịnh
6 tháng 8 2023 lúc 20:17

c: -8/13+(-7/5-x)=-1/2

=>x+7/5+8/13=1/2

=>x=1/2-7/5-8/13=-197/130

d: \(\Leftrightarrow-\dfrac{8}{7}+\dfrac{5}{3}-\left(x-\dfrac{7}{3}\right)=\dfrac{-4}{21}\)

=>\(x-\dfrac{7}{3}=\dfrac{-8}{7}+\dfrac{5}{3}+\dfrac{4}{21}=\dfrac{-24+35+4}{21}=\dfrac{18}{21}=\dfrac{6}{7}\)

=>x=6/7+7/3=18/21+49/21=67/21

e: =>x:1/2=-2/3-2/5=-16/15

=>x=-16/15*1/2=-8/15

f: =>-8/5*x=-1/3+4/9=1/9

=>x=-1/9:8/5=-1/9*5/8=-5/72

g: =>-4/5x-1/4+x=-13/3

=>1/5x=-13/3+1/4=-52/12+3/12=-49/12

=>x=-49/12*5=-245/12

h: =>12/7:x-1/2=0 hoặc 2/5x-3/2=0

=>12/7:x=1/2 hoặc 2/5x=3/2

=>x=12/7:1/2=24/7 hoặc x=3/2:2/5=3/2*5/2=15/4

Hoàng Linh Nguyễn Phan
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Akai Haruma
26 tháng 11 2023 lúc 19:57

1/

$C=5+(5^2+5^3)+(5^4+5^5)+.....+(5^{2022}+5^{2023})$

$=5+5^2(1+5)+5^4(1+5)+....+5^{2022}(1+5)$

$=5+(1+5)(5^2+5^4+....+5^{2022})$
$=5+6(5^2+5^4+....+5^{2022})$

$\Rightarrow C$ chia $6$ dư $5$

$\Rightarrow C\not\vdots 6$

Akai Haruma
26 tháng 11 2023 lúc 19:58

2/

$D=(1+2+2^2)+(2^3+2^4+2^5)+....+(2^{2019}+2^{2020}+2^{2021})$

$=(1+2+2^2)+2^3(1+2+2^2)+....+2^{2019}(1+2+2^2)$

$=(1+2+2^2)(1+2^3+...+2^{2019})$

$=7(1+2^3+...+2^{2019})\vdots 7$ 

Ta có đpcm.

Nguyễn Thanh Ngọc Ánh
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Như Nguyễn
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Xuan Mai
19 tháng 4 2022 lúc 12:06

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Thu Hồng
19 tháng 4 2022 lúc 15:46

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Ha Sino
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Nguyễn Lê Phước Thịnh
24 tháng 10 2021 lúc 20:15

a: Ư(8)={1;2;4;8}

Ư(12)={1;2;3;4;6;12}

UC(8;12)={1;2;4}

b: B(16)={0;16;32;...}

B(24)={0;24;48;...}

BC(16,24)={0;48;96;...}

\(c,B\left(12\right)=\left\{0;12;24;36;48;...\right\}\\ B\left(18\right)=\left\{0;18;36;54;72;...\right\}\\ BC\left(12;18\right)=B\left(2^2.3^2\right)=B\left(36\right)=\left\{0;36;72;108;144;....\right\}\\ d,Ư\left(16\right)=\left\{1;2;4;8;16\right\}\\ Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\\ ƯC\left(16;24\right)=Ư\left(2^3\right)=Ư\left(8\right)=\left\{1;2;4;8\right\}\\ e,ƯC\left(28;70\right)=Ư\left(2.7\right)=Ư\left(14\right)=\left\{1;2;7;14\right\}\\ BC\left(4;14\right)=B\left(2^2.7\right)=B\left(28\right)=\left\{0;28;56;84;112;140;...\right\}\)

LUFFY WANO
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Tô Mì
3 tháng 7 2023 lúc 9:10

(a) \(A=\dfrac{3}{x-2}\in Z\)

\(\Rightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;0;2;4\right\}.\)

 

(b) \(B=-\dfrac{11}{2x-3}\in Z\)

\(\Rightarrow\left(2x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\\2x-3=11\\2x-3=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=7\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;1;2;7\right\}.\)

 

(c) \(C=\dfrac{x+3}{x+1}=\dfrac{\left(x+1\right)+2}{x+1}=1+\dfrac{2}{x+1}\in Z\Rightarrow\dfrac{2}{x+1}\in Z\)

\(\Rightarrow\left(x+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;-2;0;1\right\}.\)

 

(d) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)

\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)

\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)

Noname
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Akai Haruma
29 tháng 12 2023 lúc 23:02

Bài 1:

a. $-27+(-154)-(-27)+54$

$=(-27)-(-27)+(-154)+54=0-154+54=0-(154-54)=0-100=-100$

b.

$-35.127+(-35).(-27)+700$

$=(-35)(127-27)+700=-35.100+700=-3500+700=-2800$

c.

$-3^4-2[(-2023)^0+(-5)^2]=-81-2(1+25)=-81-2.26=-81-52$

$=-(81+52)=-133$

Akai Haruma
29 tháng 12 2023 lúc 23:04

Bài 2: 

a. $-34-2(7-x)=-10$

$2(7-x)=-34-(-10)=-24$

$7-x=-24:2=-12$

$x=7-(-12)=19$
b.

$x=ƯC(36,54,90)$

$\Rightarrow ƯCLN(36,54,90)\vdots x$

$\Rightarrow 18\vdots x$

$\Rightarrow x\in \left\{\pm 1; \pm 2; \pm 3; \pm 6; \pm 9; \pm 18\right\}$

Mà $x>5$ nên $x\in \left\{6; 9; 18\right\}$

Lê Ngân Khánh
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Akai Haruma
19 tháng 6 2023 lúc 18:10

Bạn nên chỉ ra 1 bài bạn thực sự cần thiết. Nếu cần nhiều thì nên tách lẻ ra post riêng chứ chụp cả đề như thế này khả năng bị bỏ qua sẽ cao hơn.

Van Le
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