Bài 1:

\(a.5^5-5^4+5^3\)

\(=5^3.5^2-5^3.5+5^3.1\)

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

\(=5^3.21\)

\(=5^3.3.7⋮7\)

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Bài 2:

\(a.32< 2^n< 128\)

\(\Rightarrow2^5< 2^n< 2^7\)

\(\Rightarrow n=2\)

\(b.9.27\le3^n\le243\)

\(\Rightarrow3^2.3^3\le3^n\le3^5\)

\(\Rightarrow3^5\le3^n\le3^5\)

\(\Rightarrow n=5\)

\(4^{a.b.c.d}=\left(4^a\right)^{bcd}=5^{bcd}=\left(5^b\right)^{cd}=6^{cd}=\left(6^c\right)^d=7^d=8\)

=> \(2^{2abcd}=8=2^3\Rightarrow2abcd=3\Rightarrow abcd=\frac{3}{2}\)

\(TDB:\)

   \(4^a=8\Leftrightarrow a=1,5\)

   \(5,5^b=8\Rightarrow b=1,219\)

  \(6,6^c=8\Rightarrow c=1,101\)

  \(7,7^d=8\Rightarrow d=1,018\)

\(\Rightarrow a.b.c.d=1,5\times1,219\times1,101\times1,018=2,049\)

a: |x-2|<=3

=>x-2>=-3 và x-2<=3

=>-1<=x<=5

mà x thuộc A

nên \(x\in\left\{-1;0;1;2;3;4;5\right\}\)

b: |x-3|>5

=>x-3<-5 hoặc x-3>5

=>x>8 hoặc x<-2

mà x thuộc A

 nên \(x\in\left\{-10;-9;...;-3;9;10\right\}\)

a, \(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\frac{9}{2}=288\)

\(\Rightarrow2^n=64\)

\(\Rightarrow n=6\)

\(KL....\)

b, đề hơi sai pn ạ

c, \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\)chia hết cho 55

d, \(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)

\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\)

\(\Rightarrow5A-A=5^{51}-1\)

\(\Rightarrow A=\frac{5^{51}-1}{4}\)

a, 2−1.2n+4.2n=9.25

⇒2n.92 =288

⇒2n=64

⇒n=6

KL....

b, đề hơi sai pn ạ

c, 76+75−74=74(72+7−1)=74.55chia hết cho 55

d, A=1+5+52+53+...+549+550

⇒5A=5+52+53+54+...+550+551

⇒5A−A=551−1

⇒A=551−14 

Thế muốn giải thích thì liệt kê đau đầu =(

\(\frac{3}{\sqrt{7}-5}-\frac{3}{\sqrt{7+5}}=\frac{-10}{9}\inℚ\)

\(\frac{\sqrt{7}+5}{\sqrt{7}-5}+\frac{\sqrt{7}-5}{\sqrt{7}+5}=12\inℚ\)

Đây là TH là số hữu tỉ còn lại.....

\(\frac{4}{2-\sqrt{3}}-\frac{4}{2+\sqrt{3}}=8\sqrt{3}\notinℚ\)

\(\frac{\sqrt{3}}{\sqrt{7}-2}-2\sqrt{7}=2-\sqrt{7}\notinℚ\)

a) \(\frac{5}{6}=\frac{-1}{x}\) \(\Leftrightarrow\) 5x = -1 . 6 \(\Leftrightarrow\) x = \(\frac{-6}{5}\) = -1,2

b)\(\frac{-3}{7}=\frac{x}{14}\Leftrightarrow\) -3 . 14 = 7x \(\Leftrightarrow\) x = \(\frac{-3.14}{7}\) = -6

c)\(\frac{x+1}{4}=\frac{-3}{2}\) \(\Leftrightarrow\) 2(x+1) = -3.4\(\Leftrightarrow\) \(x+1=\frac{-3.4}{2}\) = -6 \(\Leftrightarrow\) x = -6 - 1= -7

d) \(\frac{2x-3}{5}=\frac{-6}{7}\) \(\Leftrightarrow\) 2x - 3 = \(\frac{-6.5}{7}\) = \(\frac{-30}{7}\) \(\Leftrightarrow\) 2x = \(\frac{-30}{7}+3\) = \(\frac{-9}{7}\)

\(\Leftrightarrow\) x = \(\frac{-9}{7}:2\) = \(\frac{-9}{14}\)

e) \(\frac{3-5x}{4}=\frac{5}{6}\) \(\Leftrightarrow\) 3-5x = \(\frac{5.4}{6}\) = \(\frac{10}{3}\) \(\Leftrightarrow\) 5x = 3-\(\frac{10}{3}\) = \(\frac{-1}{3}\) \(\Leftrightarrow\) x = \(\frac{-1}{15}\)

f)\(\frac{12}{x}=\frac{-6}{5}\) \(\Leftrightarrow\) x = \(\frac{12.5}{-6}\) = -10

Đặt \(m=3^{4^4},n=4^{\frac{5^6-1}{4}}=2^{\frac{5^6-1}{2}}\)

Khi đó ta có \(m^4=\left(3^{4^4}\right)^4=3^4^{^5};4n^4=4\left(4^{\frac{5^6-1}{4}}\right)^4=4\cdot4^{5^6-1}=4^{5^6}\)

Ta có \(A=m^4+4n^4=\left(m^4+4m^2n^2+4n^4\right)-4m^2n^2=\left(m^2+2n^2\right)^2-\left(2mn\right)^2\)

\(A=\left(m^2+2n^2-2mn\right)\left(m^2+2n^2+2mn\right)\)

\(m^2+2n^2+2mn>m^2+2n^2-2mn\)

\(=\left(m-n\right)^2+n^2\ge n^2=2^{5^6-1}>2^{8064}=\left(2^4\right)^{2016}>10^{2016}\)

Vậy bài toán được chứng minh

a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)

⇔ 3(x - 3) = 90 - 5(1 - 2x)

⇔ 3x - 9 = 90 - 5 + 10x

⇔ 3x - 10x = 90 - 5 + 9

⇔ -7x = 94

⇔ x = \(\frac{-94}{7}\)

S = { \(\frac{-94}{7}\) }

b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)

⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)

⇔ 6x - 4 - 60 = 9 - 6x - 42

⇔ 6x + 6x = 9 - 42 + 60 + 4

⇔ 12x = 31

⇔ x = \(\frac{31}{12}\)

S = { \(\frac{31}{12}\) }

c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7

⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210

⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210

⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40

⇔ 13x = 150

⇔ x = \(\frac{150}{13}\)

S = { \(\frac{150}{13}\) }

d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)

⇔ 21x - 120(x - 9) = 4(2x + 1,5)

⇔ 21x - 120x + 1080 = 8x + 6

⇔ 21x - 120x - 8x = 6 - 1080

⇔ -107x = -1074

⇔ x = \(\frac{1074}{107}\)

S = { \(\frac{1074}{107}\) }

e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5

⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840

⇔ 140x -140+56 -294x+42= 96x+48 -840

⇔ 140x -294x -96x = 48 -840 -42 -56+140

⇔ -250x = -750

⇔ x = 3

S = { 3 }

f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)

⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x

⇔ 4x+4+18x+9 = 4x+6x+6+7+12x

⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4

⇔ 0x = 0

S = R

Chúc bạn học tốt !