Các câu hỏi tương tự
\(\frac{\text{1}}{5x8}\) + \(\frac{\text{1}}{8x\text{1}\text{1}}\)+ \(\frac{\text{1}}{\text{1}\text{1}x\text{1}4}\) + ..... + \(\frac{\text{1}}{x\left(x+3\right)}\)= \(\frac{\text{1}0\text{1}}{\text{1}540}\)
cho các số x,y,z khác 0 va thoả mãn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.t\text{ính}gi\text{á}tr\text{ị}bi\text{ểu}th\text{ức}P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
( Câu siêu khó dành cho học sinh đội tuyển Toán )( Giúp với )Bài 10 :b) frac{left(text{13}frac{text{1}}{text{4}}-text{2}frac{text{5}}{text{27}}-text{10}frac{text{5}}{text{6}}right).text{230}frac{text{1}}{text{25}}+text{46}frac{text{3}}{text{4}}}{left(text{1}frac{text{3}}{text{7}}+frac{text{10}}{text{3}}right):left(text{12}frac{text{1}}{text{3}}-text{14}frac{text{2}}{text{7}}right)}
Đọc tiếp
( Câu siêu khó dành cho học sinh đội tuyển Toán )
( Giúp với )
Bài 10 :
b) \(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}\)
\(\frac{\text{1}}{6}\)x + \(\frac{\text{1}}{\text{1}0}\) x - \(\frac{4}{5}\) x + 1 = 0
\(\frac{y+\text{z}+1}{x}=\frac{x+\text{z}+2}{y}\frac{x+y-3}{\text{z}}=\frac{1}{x+y+\text{z}}\)
\(x\text{(\frac{1}{5} +\frac{1}{4})-\text{( \frac{1}{7}-\frac{1}{8})=0}}\)
Tìm x;y biết rằng : \(\text{|}x-5\text{|}+\text{|}1-x\text{|}=\frac{12}{\text{|}y+1\text{|}+3}\)
Tìm các cặp số nguyên (x;y) thõa mãn
\(\text{|}x-5\text{|}+\text{|}1-x\text{|}=\frac{12}{\text{|}y+1\text{|}+3}\)
Cho \(\frac{x^{\text{4}}}{a}+\frac{y^{\text{4}}}{b}=\frac{1}{a+b};x^2+y^2=1\)
Chứng minh rằng:\(\frac{x^{200\text{4}}}{a^{1002}}+\frac{y^{200\text{4}}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)