Đáp án A

2 x 7 = 75 35 2 . 7 + x 7 = 15 7 14 + x = 15 x = 15 - 14 x = 1

Đáp án A

2 x 7 = 75 35 2 . 7 + x 7 = 15 7 14 + x = 15 x = 15 - 14 x = 1

a/\(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{18}{5}\)\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot11}{5\cdot7}=\frac{22}{35}\)

b/\(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11^2\cdot5^3\cdot13^3\cdot2\cdot3}{2^3\cdot3^4\cdot5\cdot11^2\cdot13}=\frac{5^2\cdot13^2}{2^2\cdot3^3}=\frac{4225}{108}\)

c/\(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}=\frac{2^2\cdot3^3\cdot37\cdot5\cdot199+2\cdot3^2\cdot13\cdot17}{2^3\cdot3\cdot83\cdot11\cdot181-2^4\cdot3\cdot83}=\frac{2\cdot3^2\cdot11\cdot20101}{2^3\cdot3^3\cdot13\cdot17\cdot83}=\frac{11\cdot20101}{2^2\cdot3\cdot13\cdot17\cdot83}\)

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a) Ta có:  x 2 = 2 2  nên x = 2.

b) Ta có: x 2 = 5 2 nên x = 5.

c) Ta có:  3 x 5 = 3  nên  x 5 = 1 . Do đó x = 1.

d) Ta có:  6 x 3 = 48  nên  x 3 = 8 . Do đó x = 2.

e) Ta có:  x - 1 2 = 2 2  nên  x - 1 = 2 . Do đó x = 3.

f) Ta có:  x + 1 2 = 5 2  nên x +1 = 5. Do đó x = 4.

g) Ta có:  x - 1 3 = 3 3  nên  x - 1 = 3 . Do đó x = 4.

h) Ta có:  x + 1 3 = 4 3 nên x +1 = 4. Do đó x = 3

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+3\right)}=\frac{20}{41}\)

\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}\right)=2.\frac{20}{41}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

\(1-\frac{1}{x+2}=\frac{40}{41}\)

\(\frac{1}{x+2}=1-\frac{40}{41}\)

\(\frac{1}{x+2}=\frac{1}{41}\)

=> x + 2 = 41

=> x = 41 - 2

=> x = 39

Vẫy x = 39

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

=> \(1-\frac{1}{x+2}=\frac{40}{41}\)

=> \(\frac{1}{x+2}=\frac{1}{41}\)

=> x + 2 = 41

=> x = 39

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

\(\Leftrightarrow2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}\right)=2.\frac{8}{17}\)

\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{16}{17}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)

\(\Rightarrow x+2=17\Rightarrow x=15\)

x là số lẻ vậy x có thể là: 1 ; 3 ; 5 ; 7 ; 9

  Còn lại bạn tự giải nha! Cứ dùng phương pháp loại suy thử với từng số là ra! dễ mà

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{8}{17}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{8}{17}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{16}{17}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{17}\)

\(\Rightarrow x+2=17\Rightarrow x=15\)

\(3\cdot5^x=60\cdot5^8+15\cdot25^4\)

\(3\cdot5^x=60\cdot5^8+15\cdot5^8\)

\(3\cdot5^x=5^8\left(60+15\right)\)

\(3\cdot5^x=5^8\cdot75\)

\(5^x=5^8.75:3\)

\(5^x=5^8.25=5^8.5^2=5^{10}\)

\(\Rightarrow x=10\)

P/s: Hoq chắc :<