Cứu em với ạ em đang cần gấp ạ. Em cảm ơn
CỨU EM CÂU NÀY VỚI Ạ. EM ĐANG CẦN GẤP.
EM CẢM ƠN MỌI NGƯỜI NHIỀU Ạ.
các anh chị ơi . cứu em với , em đang cần gấp lắm ạ , em cảm ơn ạ
Giúp em với ạ , em đang cần gấp . Em cảm ơn ạ ! =.=
\(c,-\dfrac{8}{13}+\left(-\dfrac{7}{5}-x\right)=-\dfrac{1}{2}\\ -\dfrac{7}{5}-x=-\dfrac{1}{2}-\dfrac{8}{13}\\ -\dfrac{7}{5}-x=-\dfrac{29}{26}\\ x=-\dfrac{7}{5}-\left(-\dfrac{29}{26}\right)=-\dfrac{37}{130}\\ d,-1\dfrac{1}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{8}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{8}{7}-\left(-\dfrac{4}{21}\right)\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{20}{21}\\ x-\dfrac{7}{3}=-\dfrac{20}{21}-\left(-\dfrac{5}{3}\right)\\ x-\dfrac{7}{3}=\dfrac{5}{7}\\ x=\dfrac{5}{7}+\dfrac{7}{3}=\dfrac{64}{21}\\ e,-\dfrac{2}{3}-x:\dfrac{1}{2}=\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{2}{3}-\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{16}{15}\\ x=-\dfrac{16}{15}\times\dfrac{1}{2}=-\dfrac{8}{15}\)
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
giúp em với ạ em cảm ơn em đang cần gấp ạ
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 $5$
$\Rightarrow C\not\vdots 6$
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.
giúp em với em đang cần gấp ạ, em cảm ơn nhiều ạ!!
Giúp em với ạ ! Em đang cần gấp , em cảm ơn trước ạ
giúp em với ạ em đang cần gấp cảm ơn ạ
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;...}
\(c,B\left(12\right)=\left\{0;12;24;36;48;...\right\}\\ B\left(18\right)=\left\{0;18;36;54;72;...\right\}\\ BC\left(12;18\right)=B\left(2^2.3^2\right)=B\left(36\right)=\left\{0;36;72;108;144;....\right\}\\ d,Ư\left(16\right)=\left\{1;2;4;8;16\right\}\\ Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\\ ƯC\left(16;24\right)=Ư\left(2^3\right)=Ư\left(8\right)=\left\{1;2;4;8\right\}\\ e,ƯC\left(28;70\right)=Ư\left(2.7\right)=Ư\left(14\right)=\left\{1;2;7;14\right\}\\ BC\left(4;14\right)=B\left(2^2.7\right)=B\left(28\right)=\left\{0;28;56;84;112;140;...\right\}\)
giúp em với ạ em đang cần gấp lắm ạ em cảm ơn trước
(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) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)
\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)
giúp em với ạ em đang cần gấp em cảm ơn ạ 😘😘
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.
$-3^4-2[(-2023)^0+(-5)^2]=-81-2(1+25)=-81-2.26=-81-52$
$=-(81+52)=-133$
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 $x\in \left\{6; 9; 18\right\}$
Mấy anh chị giúp em với ạ, em đang cần gấp ạ, em cảm ơn trước ạ