Ta có: \(x+y=\frac{1}{2}\) (1)

    \(y+z=\frac{1}{3}\)(2)

  \(x+z=\frac{1}{4}\)(3)

Từ (1), (2) và (3) cộng vế theo vế: 

\(x+y+y+z+x+z=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

<=> \(2\left(x+y+z\right)=\frac{13}{12}\)

<=> \(x+y+z=\frac{13}{24}\)

=> \(\hept{\begin{cases}x=\frac{13}{24}-\left(y+z\right)=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\\y=\frac{13}{24}-\left(x+z\right)=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\\z=\frac{13}{24}-\left(x+y\right)=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\end{cases}}\)

\(\hept{\begin{cases}x+y=\frac{1}{2}\\y+z=\frac{1}{3}\\z+x=\frac{1}{4}\end{cases}}\Rightarrow2\left(x+y+z\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

\(\Leftrightarrow x+y+z=\frac{13}{24}\)

\(x=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)

\(y=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\)

\(z=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)

x=2016

Thanh Huyền giải ra giúp mik với ạ !!!

mk ko biết viết ps sợ bạn khó nhìn

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}< 1\) ( điều phải chứng minh ) 

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 1\left(\text{đ}pcm\right)\)

vậy:\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}< 1\)

k mk bạn nha:)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}< 1\)

\(\Rightarrow\)\(đpcm\)

chúc bạn hok tốt!

từ đề bài ta có \(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=10\)

bạn tính tổng mẫu số đi rồi sẽ làm đc

lm cụ thể giúp mik với

Ta có : \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\Rightarrow\frac{3}{2x+1}+\frac{5.2}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)

=> \(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

=> \(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

=> \(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

=> 2x = 12

=> x = 6

Vậy x = 6

\(\frac{3}{2x+1}+\frac{10}{2\left(2x+1\right)}-\frac{6}{3\left(2x+1\right)}=\frac{6}{13}\)                

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)                

\(\frac{6}{2x+1}=\frac{6}{13}\)  

\(\Rightarrow2x+1=13\left(6=6\right)\)         

\(2x=12\)    

\(x=6\)     

\(a)\)\(\frac{\frac{1}{3}-\frac{1}{7}+\frac{1}{11}}{\frac{2}{3}+\frac{2}{11}-\frac{2}{7}}-\frac{\frac{1}{5}-\frac{1}{3}-\frac{1}{11}}{\frac{2}{3}+\frac{2}{11}-\frac{2}{5}}\)\(=\frac{\frac{1}{3}-\frac{1}{7}+\frac{1}{11}}{2.\left(\frac{1}{3}+\frac{1}{11}-\frac{1}{7}\right)}+\frac{\frac{-1}{5}+\frac{1}{3}+\frac{1}{11}}{2.\left(\frac{1}{3}+\frac{1}{11}-\frac{1}{5}\right)}\)
\(=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1\)

ở câu a : dấu giữa 2 phép tính là cộng chứ không phải trừa nha

\(b)\)\(3^{n+2}+3^n-2^{n+2}-2^n\)

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

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

\(=3^n.10-2^{n-1}.\left(2^3+2\right)\)

\(=3^n.10-2^{n-1}.10\)

\(=10.\left(3^n-2^{n-1}\right)⋮10\)
\(\Rightarrow(3^{n+2}+3^n-2^{n+2}-2^n)⋮10\)

Vì n là số nguyên dương và \((3^{n+2}+3^n-2^{n+2}-2^n)⋮10\) nên \(3^{n+2}+3^n-2^{n+2}-2^n\) có chữ số tận cùng là 0.

mk lười lắm bạn chịu khó nha