\(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)\)
\(\Leftrightarrow x^4+x^3+4x^2+3x+1=0\)
\(\Leftrightarrow\left(x^4+x^3+\frac{x^2}{4}\right)+\left(\frac{9x^2}{4}+3x+1\right)+\frac{3x^2}{2}=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{2}\right)^2+\left(\frac{3x}{2}+1\right)^2+\frac{3x^2}{2}=0\)
Tới đây thì thấy nó vô nghiệm rồi nhé
Giải phương trình :
\(\left|x+\frac{1}{1\times5}\right|+\left|x+\frac{1}{5\times9}\right|+\left|x+\frac{1}{9\times13}\right|+...+\left|x+\frac{1}{397\times401}\right|=101\times x\)
giải phương trình
1. \(\frac{2x+3}{5x-3}\)-\(\frac{3}{4x-6}\)=\(\frac{2}{5}\)
2. \(\frac{1}{x-1}\)-\(\frac{7}{x-2}\)=\(\frac{1}{\left(x-1\right)\left(2-x\right)}\)
3. x+\(\frac{x-1}{x-2}\)=3+\(\frac{1}{x-2}\)
4. \(\frac{x+1}{x-3}\)-\(\frac{1}{x-1}\)=\(\frac{2}{\left(x-1\right)\left(x-3\right)}\)
5. \(\frac{2x-1}{x+1}+\frac{x}{\left(x-1\right)\left(x-2\right)}=\frac{6x-2}{x-2}\)
Tính:
\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}+\frac{1}{\left(x+3\right).\left(x+4\right)}+\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{x+5}\)
Giải phương trình:
a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
b) \(\frac{3}{10}\left(1,2-x\right)-\frac{5+7x}{4}=\frac{1}{20}\left(9x+0,2\right)-\frac{12,5x+4,5}{5}\)
Giải phương trình sau : 20\(\left(\frac{x-2}{x+1}\right)^2\)-5\(\left(\frac{x+2}{x-1}\right)^2\)+48\(\frac{x^2-4}{x^2-1}\)=0
GIẢI PHƯƠNG TRÌNH SAU
A) \(\frac{X^2+2X+1}{X^2+2X+2}+\frac{X^2+2X+2}{X^2+2X+3}=\frac{7}{6}\)
B) \(\frac{\left(X^2-3X-4\right)^4}{\left(X-3\right)^5\left(X+2\right)^3}+\frac{\left(X^2+4X+3\right)^6}{\left(X-3\right)^3\left(X+2\right)^5}=0\)
Rút gọn \(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^5}.\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}.\left(\frac{1}{x}+\frac{1}{y}\right)\)
giải jùm vs
Rút gọn Pthức:
A= \(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2+x+1}\right)\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
Tính tổng: \(A=\frac{x^4-\left(x-1\right)^2}{\left(x^2+1\right)^2-x^2}+\frac{x^2-\left(x^2-1\right)^2}{x^2\left(x+1\right)^2-1}+\frac{x^2\left(x-1\right)^2-1}{x^4-\left(x+1\right)^2}\)
