A\(=\frac{36x85x20}{25x84x34}=\frac{2^2x3^2x5x17x2^2x5}{5^2x2^2x3x7x2x17}=\frac{6}{7}.\)

\(B=\frac{30x63x65x8}{117x200x49}=\frac{\left(2x3x5\right)\left(3^2x7\right)\left(5x13\right)2^3}{\left(3^2x13\right)\left(2^3x5^2\right)7^2}=\frac{6}{7}\)

=>A=B

Lời giải:

Gọi giá trị trên là $A$

$A=\frac{4}{15}+\frac{6}{7}+4+3+\frac{1}{7}$
$=\frac{4}{15}+(4+3)+(\frac{6}{7}+\frac{1}{7})$

$=\frac{4}{15}+7+1=8\frac{4}{15}$

Mình là chủ nhân của câu hỏi này lên các bạn hãy bỏ một phân số 133phaanf 10 ra ngoài

mk ko bt 123

\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}\)

\(\Rightarrow19\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{133}{10}\)

\(\Rightarrow\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{7}{10}\)

\(\frac{7x}{y+z}+\frac{7y}{z+x}+\frac{7z}{x+y}=\frac{133}{10}\)

\(\Rightarrow7\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{133}{10}\)

\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{19}{10}\)

\(\Rightarrow\left(\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\right)=\frac{19}{10}+3\)

\(\Rightarrow\left(\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\right)=\frac{49}{10}\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)=\frac{49}{10}\)

\(\Rightarrow\left(x+y+z\right).\frac{7}{10}=\frac{49}{10}\)

\(\Rightarrow x+y+z=7\)

Vậy x + y + z = 7