Giúp em với. Em cần gấp. Em cảm ơn ạ
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1 am going to send
2 will get
3 will do
4 will be
5 are going to visit
6 will win
7 am going to take
8 are going
9 will go
10 is going to defeat
11 is going to have
12 will never lie
13 will fly
14 won't tell
15 will like
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\(c,-\dfrac{8}{13}+\left(-\dfrac{7}{5}-x\right)=-\dfrac{1}{2}\\ -\dfrac{7}{5}-x=-\dfrac{1}{2}-\dfrac{8}{13}\\ -\dfrac{7}{5}-x=-\dfrac{29}{26}\\ x=-\dfrac{7}{5}-\left(-\dfrac{29}{26}\right)=-\dfrac{37}{130}\\ d,-1\dfrac{1}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{8}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{8}{7}-\left(-\dfrac{4}{21}\right)\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{20}{21}\\ x-\dfrac{7}{3}=-\dfrac{20}{21}-\left(-\dfrac{5}{3}\right)\\ x-\dfrac{7}{3}=\dfrac{5}{7}\\ x=\dfrac{5}{7}+\dfrac{7}{3}=\dfrac{64}{21}\\ e,-\dfrac{2}{3}-x:\dfrac{1}{2}=\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{2}{3}-\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{16}{15}\\ x=-\dfrac{16}{15}\times\dfrac{1}{2}=-\dfrac{8}{15}\)
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
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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.
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\(=>Qthu1=0,2.340000=68000J\)
\(=>Qthu2=2100.0,2.20=8400J\)
\(=>Qtoa=2.4200.25=210000J\)
\(=>Qthu1+Qthu2< Qtoa\)=>đá nóng chảy hoàn toàn
\(=>0,2.2100.20+0,2.340000+0,2.4200.tcb=2.4200\left(25-tcb\right)\)
\(=>tcb=14,5^oC\)
Cho em hỏi ngu tí ạ vậy tcb ở nhưng phép tính trên vứt đi đâu ạ
Câu 56: A
Câu 57: C
Câu 58: A
Câu 59: A