xin lỗi em mới học lớp 6 

Ta có : \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{2}{3}+\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{2\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}\right)}=\frac{1}{2}\)

Lại có : \(\frac{\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right).2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=\frac{0.2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=0\)

Khi đó \(B=\frac{1}{2}+0=\frac{1}{2}\)

H = 2012 - 1 - ( \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+99}\))
   = 2011 - ( \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{\left(99+1\right).\left[\left(99-1\right):1+1\right]:2}\)
   = 2011 - ( \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\))
   = 2011 - 2.( \(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\))
   = 2011 - 2.(\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)) 
   = 2011 - 2.( \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\))
   = 2011 - 2.(\(\frac{1}{2}-\frac{1}{100}\)) = 2011 - 2.\(\frac{49}{100}\)= 2011 - \(\frac{49}{50}\)= \(\frac{100501}{50}\)

\(H=2012-\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}\right)\)

\(=2012-\left(1+\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{99\left(99+1\right):2}\right)\)

\(=2012-\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\right)\)

\(=2012-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2}{99.100}\right)\)

\(=2012-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2012-2\left(1-\frac{1}{100}\right)\)

\(=2012-2\cdot\frac{99}{100}\)

\(=2012-\frac{99}{50}\)

\(=\frac{100501}{50}\)

= 1 là đúng

Bạn có thể cho mình biết cách giải được không vậy bạn.

A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)

\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+6-4}=\frac{2x-2+3y-6-z+3}{4+6-4}\)

\(=\frac{\left(2x+3y-z\right)+\left(-2+6+3\right)}{6}=\frac{50+\left(-5\right)}{6}=\frac{45}{6}=7,5\)

\(\frac{x-1}{2}=7,5\Rightarrow x-1=15\Rightarrow x=16\)

\(\frac{y-2}{3}=7,5\Rightarrow y-2=24,5\Rightarrow y=20,5\)

\(\frac{z-3}{4}=7,5\Rightarrow z-3=30\Rightarrow z=33\)

Áp dngj tính chất dãy các tỉ số bằng nhau. ta có:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x+3y-z}{2.2+3.3-4}=\frac{50}{9}\)\(\frac{50}{9}\)

\(\frac{x-1}{2}=\frac{50}{9}\Rightarrow x-1=\frac{50}{9}.2=\frac{100}{9}\)

                                       \(x=\frac{100}{9}+1=\frac{109}{9}\)

\(\frac{y-2}{3}=\frac{50}{9}\Rightarrow y-2=\frac{50}{9}3=\frac{50}{3}\)

                                        \(y=\frac{50}{3}+2=\frac{56}{3}\)

\(\frac{z-3}{4}=\frac{50}{9}\Rightarrow z-3=\frac{50}{9}.4=\frac{200}{9}\)

                                        \(z=\frac{200}{9}+3=\frac{227}{9}\)      

Chúc bạn học tốt

\(\)

cái đoạn có hai phân số \(\frac{50}{9}\)bạn bớt đi một cái nha cái đó mik ghi nhầm

4Ta Co   x-1/2=y-2/3=z-3/4=2x-2/4=3y-6/9=2x-2+3y-6-z-3/4+9-4=2x+3y-z-5/9=50-5/9=45/9=5[tc day ti so bang nhau]

vay  x-1/2=5;y-2/3=5;z-3/4=5

nen   x=11;y=17;z=23

Mình không bày bn cách giải, nhưng sẽ gợi ý:

2 bài tương tự nhau, mẫu gấp nhau 3 lần nhé