a, \(\frac{3}{5}.x-\frac{1}{2}=\frac{1}{7}\)
b, \(\frac{1}{4}+\frac{1}{3}:3x=-5\)
c, \(\frac{1}{3}.x+\frac{2}{5}\left(x+1\right)=0\)
d, \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
Cho biểu thức P=\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
Cho x la nghiệm của pt: x2 + 3x + 1 =0
tính giá trị của biểu thức:
B= (\(x+\frac{1}{x}\))2 + \(\left(x^2+\frac{1}{x^2}\right)^2+\left(x^3+\frac{1}{x^3}\right)^2+....+\left(x^6+\frac{1}{x^6}\right)^2\)
Giai phương trình:
\(x^{3^{ }}+\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}-2=0\)
Giải phương trình sau :
\(x^3+\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}-2=0\)
Giải hệ phương trình :
1, \(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\)
2 , \(\left\{{}\begin{matrix}\frac{2}{2x-y}-\frac{1}{x+y}=0\\\frac{3}{2x-y}-\frac{6}{x+y}=-1\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-2y\right)-15\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}2x+y=7\\-x+4y=10\end{matrix}\right.\)
Giải PT và HPT:
1)\(\left\{{}\begin{matrix}xy+x+y=3\\\frac{1}{x^2+2x}+\frac{1}{y^2+2y}=\frac{2}{3}\end{matrix}\right.\)
2)\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)
3)\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\9xy\left(3x-y\right)+6=26x^3-2y^3\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^2-2xy+x-2y+3=0\\y^2-x^2+2xy+2x-2=0\end{matrix}\right.\)
1) Giaỉ hệ \(\left\{{}\begin{matrix}\left(x+y\right)^2\left(8x^2+8y^2+4xy-13\right)+5=0\\2x+\frac{1}{x+y}=1\end{matrix}\right.\)
2) Tính P\(=\frac{4\left(x+1\right)x^{2018}-2x^{2017}+2x+1}{2x^2+3x}\)Với x\(=\sqrt{\frac{1}{2\sqrt{3}-2}-\frac{3}{2\sqrt{3}+2}}\)
Giải phương trình: \(2\left[3x\right]=\left[x+\frac{2}{3}\right]+\left[x+\frac{1}{3}\right]+\left[x\right]+1\)
