Ta có \(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21=5^3.3.7\)

Vì 5³.3 là số nguyên nên \(5^3.3.7⋮7\)

Vậy  \(5^5-5^4+5^3⋮7\)

c) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}\)

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

\(=3^x\left(3^2+3\right)+2^x\left(2^2+2\right)\)

\(=3^x.12+2^x.6\)

\(=6\left(2.3^x+2^x\right)\)

Vì \(2.3^x+2^x\in Z\)

Nên : \(6\left(2.3^x+2^x\right)⋮6\)

Vậy \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}⋮6\)

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)

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

\(\Rightarrowđpcm\)

2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)

1: \(\Leftrightarrow\left(x+1\right)^2=4\)

=>x+1=2 hoặc x+1=-2

=>x=1 hoặc x=-3

2: \(\Leftrightarrow7x-21=5x+25\)

=>2x=46

=>x=23

3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)

=>4,5x+2=4x+3

=>x=1

1: =>1/3:x=3/5-2/3=9/15-10/15=-1/15

=>x=-1/3:1/15=5

2: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)

=>x*2/3=-1

=>x=-3/2

3: \(\Leftrightarrow\dfrac{8}{3}:x=\dfrac{25}{12}:\dfrac{-3}{50}=\dfrac{25}{12}\cdot\dfrac{-50}{3}\)

hay x=-48/625

9: =>x=-2*3/1,5=-4

8: =>2/3:x=5/2:-3/10=5/2*(-10)/3=-50/6=-25/3

=>x=-2/3:25/3=-2/3*3/25=-2/25

Bài 3: 

 \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}=\dfrac{2}{3}+\dfrac{1}{3}=1\)

1) \(\frac{x-1}{5}=\frac{x+2}{7}\)

\(\Rightarrow7x-7=5x+10\)

=> 7x - 5x = 10 + 7

2x = 17

x = 8,5

b) \(\frac{x-1}{7}=\frac{5}{x+1}\)

=> x² - 1 = 35

x² = 34

\(\Rightarrow x=\sqrt{34};x=-\sqrt{34}\)

c) \(\frac{x+1}{x-2}=\frac{x+2}{x+3}\)

=> x² + 4x + 3 = x² - 4

=> x² - x² + 4x + 3 = -4

4x + 3 = - 4

4x = -7

x = -7/4