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3 tháng 9 2018

\(a,8^7-2^{18}=2^{21}-2^{18}=2^{17}\left(2^4-2\right)=14.2^7⋮14\)

\(b,10^6-5^7=2^6.5^6-5^7=5^6\left(2^6-5\right)=59.5^6⋮59\)

c ko nói chia hết cho số nào

\(d,3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

\(=3^{n+1}\left(3^2+3\right)+2^{n+1}\left(2^2+2\right)\)

\(=3^{n+1}.12+2^{n+1}.6\)

\(=6.\left(3^{n+1}.2+2^{n+1}\right)⋮6\)

3 tháng 9 2018

Gấp quá nên viết thiếu đề câu c, viết sai đề câu d

Bài 2: 

1: \(5^n+5^{n+2}=650\)

\(\Leftrightarrow5^n\cdot26=650\)

\(\Leftrightarrow5^n=25\)

hay x=2

2: \(32^{-n}\cdot16^n=1024\)

\(\Leftrightarrow\dfrac{1}{32^n}\cdot16^n=1024\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^n=1024\)

hay n=-10

13: \(9\cdot27^n=3^5\)

\(\Leftrightarrow3^{3n}=3^5:3^2=3^3\)

=>3n=3

hay n=1

28 tháng 5 2018

\(a,7^6+7^5-7^4=7^4\left(7^2+7-1\right)\\ =7^4\cdot55\\ \Rightarrow7^6+7^5-7^4⋮55\)

\(b,3^{n+2}-2^{n+2}+3^n-2^n\\ =3^n\cdot3^2+3^n-2^n\cdot2^2-2^n\\ =3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\\ =3^n\cdot10-2^{n-1}\cdot2\cdot5\\ =10\cdot\left(3^n-2^{n-1}\right)\\ \Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)

\(c,8^7-2^{18}=8^7-\left(2^3\right)^6\\ =8^7-8^6\\ =8^6\cdot\left(8-1\right)\\ =8^5\cdot8\cdot7\\ =8^5\cdot4\cdot14\\ \Rightarrow8^7-2^{18}⋮14\)

20 tháng 8 2017

1.Tính

a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)

b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)

c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)

d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)

e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)

Bài 2

a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)

\(x=\dfrac{13}{49}\)

b.\(\left|x-1,5\right|=2\)

Xảy ra 2 trường hợp

TH1

\(x-1,5=2\)

\(x=3,5\)

TH2

\(x-1,5=-2\)

\(x=-0,5\)

Vậy \(x=3,5\) hoặc \(x=-0,5\) .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

20 tháng 8 2017

Ths bn nhé

29 tháng 8 2017

a) ta có : \(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)

\(5^3.21=5^3.3.7⋮7\) (đpcm)

b) ta có : \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.\left(49+7-1\right)\)

\(=7^4.55=7^4.5.11⋮11\) (đpcm)

c) ta có : \(3^{x+2}-2^{x+3}+3^x-2^{x+1}=3^{x+2}+3^x-2^{x+3}-2^{x+1}\)

\(=3^x\left(3^2+1\right)-2^x\left(2^3+2\right)=3^x.\left(9+1\right)-2^x.\left(8+2\right)\)

\(=3^x.10-2^x.10=10\left(3^x-2^x\right)⋮10\) (đpcm)

d) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}=3^x.\left(3^3+3\right)+2^x.\left(2^3+2^2\right)\)

\(=3^x.\left(27+3\right)+2^x\left(8+4\right)=3^x.30+2^x.12=6.\left(3^x.5+2^x.2\right)⋮6\) (đpcm)

29 tháng 8 2017

a)Ta có:\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)(vì 21 chia hết cho 7)

\(\)\(\RightarrowĐPCM\)

b)Ta có: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)

\(\Rightarrowđpcm\)

28 tháng 11 2019

Bài 2:

1) \(\frac{x}{12}-\frac{5}{6}=\frac{1}{12}\)

\(\Rightarrow\frac{x}{12}=\frac{1}{12}+\frac{5}{6}\)

\(\Rightarrow\frac{x}{12}=\frac{11}{12}\)

\(\Rightarrow x.12=11.12\)

\(\Rightarrow x.12=132\)

\(\Rightarrow x=132:12\)

\(\Rightarrow x=11\)

Vậy \(x=11.\)

2) \(\frac{2}{3}-1\frac{4}{15}x=\frac{-3}{5}\)

\(\Rightarrow\frac{2}{3}-\frac{19}{15}x=\frac{-3}{5}\)

\(\Rightarrow\frac{19}{15}x=\frac{2}{3}+\frac{3}{5}\)

\(\Rightarrow\frac{19}{15}x=\frac{19}{15}\)

\(\Rightarrow x=\frac{19}{15}:\frac{19}{15}\)

\(\Rightarrow x=1\)

Vậy \(x=1.\)

3) \(\left(-2\right)^3+0,5x=1,5\)

\(\Rightarrow-8+0,5x=1,5\)

\(\Rightarrow0,5x=1,5+8\)

\(\Rightarrow0,5x=9,5\)

\(\Rightarrow x=9,5:0,5\)

\(\Rightarrow x=19\)

Vậy \(x=19.\)

Chúc bạn học tốt!

22 tháng 2 2020

1.

a) \(\frac{3}{7}+\frac{-5}{2}+\frac{-3}{5}\\ =\frac{30}{70}+\frac{-175}{70}+\frac{-42}{70}\\ =\frac{30-175-42}{70}\\ =\frac{-187}{70}\)

b) \(\frac{-8}{18}-\frac{15}{27}\\ =\frac{-4}{9}-\frac{5}{9}\\ =\frac{-9}{9}=-1\)

c) \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\\ =\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\\ =\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\\ =\frac{56+20-49}{70}\\ =\frac{27}{70}\)

2.

a) \(x+\frac{1}{4}=\frac{4}{3}\\ x=\frac{4}{3}-\frac{1}{4}\\ x=\frac{16}{12}-\frac{3}{12}\\ x=\frac{13}{12}\)

Vậy \(x=\frac{13}{12}\)

b) \(-x-\frac{2}{3}=\frac{-6}{7}\\ -x=\frac{-6}{7}+\frac{2}{3}\\ -x=\frac{-18}{21}+\frac{14}{21}\\ -x=-\frac{4}{21}\\ x=\frac{4}{21}\)

Vậy \(x=\frac{4}{21}\)

c) \(x^2=16\\ x^2=4^2=\left(-4\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

Vậy \(x\in\left\{4;-4\right\}\)

d) \(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\\ \frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\\ \frac{5}{3}x=\frac{15}{21}-\frac{14}{21}\\ \frac{5}{3}x=\frac{1}{21}\\ x=\frac{1}{21}:\frac{5}{3}\\ x=\frac{1}{21}\cdot\frac{3}{5}\\ x=\frac{1}{35}\)

Vậy \(x=\frac{1}{35}\)

3.

a) Xét △AKB và △AKC có:

AB = AC

KB = KC

AK: cạnh chung

\(\Rightarrow\text{△AKB = △AKC (c.c.c) }\)

b) \(\text{△AKB = △AKC }\)

\(\Rightarrow\widehat{AKB}=\widehat{AKC}\) (2 góc tương ứng)

\(\widehat{AKB}+\widehat{AKC}=180^o\) (2 góc kề bù)

\(\Rightarrow\widehat{AKB}=\widehat{AKC}=90^o\\ \Rightarrow AK\perp BC\)

22 tháng 2 2020

Câu 3:

a/ Xét ΔAKB và ΔAKC có:

AB = AC (GT)

\(\widehat{BAK}=\widehat{CAK}\left(GT\right)\)

AK: cạnh chung

=> ΔAKB = ΔAKC (c.g.c)

b/ VìΔAKB = ΔAKC (câu a)

\(\widehat{AKB}=\widehat{AKC}\) (2 góc tương ứng)

Mà 2 góc này lại là hai góc kề bù

=> \(\widehat{AKB}=\widehat{AKC}=180^0:2=90^0\)

=> AK BC

Cau 2:

a) \(x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

b) \(-x-\frac{2}{3}=-\frac{6}{7}\)

=> \(-x=-\frac{6}{7}+\frac{2}{3}=-\frac{4}{21}\)

=> \(x=\frac{4}{21}\)

c) x2 = 16

=> x = 4 hoặc x =-4