hình như bạn chép sai đề thì phải !!

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

 Giải

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

\(\frac{2\left(x+y+z.\right)}{x+y+z}=\frac{1}{x+y+z}\)

\(2=\frac{1}{x+y+z}\)

=> \(x+y+z=\frac{1}{2}=0,5\)

=> \(\frac{0,5-x+1}{x}=\frac{0,5-y+2}{y}=\frac{0,5-z-3}{z}=2\)

+ \(\frac{15-x}{x}=2\Rightarrow2x+x=15\Rightarrow3x=15\Rightarrow x=5\)

+ \(\frac{2,5-y}{y}=2\Rightarrow2y+y=2,5\Rightarrow3y=2,5\Rightarrow y=\frac{5}{6}\)

+ \(\frac{-2,5-z}{z}=2\Rightarrow2z+z=\left(-2,5\right)\Rightarrow3z=\left(-2,5\right)\Rightarrow z=\frac{-5}{6}\)

S1 = 1 + 2 + 2^2 + 2^3 + … + 2^63

=> 2S1 = 2 + 2^2 + 2^3 + ... + 2^64

=> 2S1 - S1 = S^64  - 1 = S1

2S = 2 + 2^2 + 2^3 + 2^4 + ... + 2^64

2S - S = 2 + 2^2 + 2^3 + 2^4 + ... + 2^64 - 1 - 2 - 2^2 - 2^3 - .... - 2^63

S = 2^64 - 1

Trả lời:

\(\left(2x-4\right)^2:\frac{3}{4}-\frac{1}{3}=1\)

\(\Leftrightarrow\left(2x-4\right)^2:\frac{3}{4}=\frac{4}{3}\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x-4=1\\2x-4=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=5\\2x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}}\)

Vậy x = 5/2; x = 3/2

\(\left(2x-4\right)^2\div\frac{3}{4}=\frac{4}{3}\)

\(\left(2x-4\right)^2=1\)

\(\Rightarrow\)\(\orbr{\begin{cases}2x-4=-1\\2x-4=1\end{cases}\Rightarrow\orbr{\begin{cases}2x=3\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{5}{2}\end{cases}}}\)

\(D=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{1979\cdot1982}\)

\(D=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{1979\cdot1982}\right)\)

\(D=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{1979}-\frac{1}{1982}\right)\)

\(D=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{1982}\right)=\frac{165}{991}\)

gọi d là ƯC(12n + 1; 30n + 2)

\(\Rightarrow\hept{\begin{cases}12n+1⋮d\\30n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}5\left(12n+1\right)⋮d\\2\left(30n+2\right)⋮d\end{cases}}\Rightarrow\hept{\begin{cases}60n+5⋮d\\60n+4⋮d\end{cases}}}\)

=> 60n + 5 - 60n - 4 ⋮ d

=> 1 ⋮ d

=> d = +-1

=> ps đó tối giản

thanks nha kb ko

Không mấy tính tổng quát, giải sử x=<y=<z

=>  1/x+1/x+1/x >=1/x +1/y +1/z = 3/5

=>   3/x>=3/5

=>   X=<5

Có 1/x< 3/5; do 1/x +1/y +1/z = 3/5

=>   X>5/3 => x=2,3,4,5

Xét các trường hợp ta thấy chỉ có x=y=z=5 thỏa

TL: 

X=y=z=5