Giải phương trình: \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
Cho \(\left(x-\dfrac{1}{x}\right):\left(x+\dfrac{1}{x}\right)\)\(=\dfrac{1}{2}\). Tính \(\left(x^2-\dfrac{1}{x^2}\right):\left(x^2+\dfrac{1}{x^{2.}}\right)\)
Cho a + b + c + d = 0. Tính \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right);N=\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{a^2+c^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
cho \(\dfrac{x^2-3x}{x.\left(1-3y\right)}=\dfrac{y^2-3x}{y.\left(1-3x\right)}\).CM \(\dfrac{8}{3}+x+y=\dfrac{1}{x}+\dfrac{1}{y}\)
Cho bieu thuc: \(Q=\left(\dfrac{x^2-2x}{2x^2+8}+\dfrac{2x^2}{x^2.\left(x-2\right)}\right).\left(\dfrac{x^2-x-2}{x^2}\right)\)
a, Rut gon bieu thuc Q
b, Tim gia tri ca x de Q co gia tri bang \(\dfrac{1}{4}\)
Cho \(\left[{}\begin{matrix}x,y,z\ne0\\x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\\x^3+y^3+z^3=1\end{matrix}\right.\).Tính A=\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
Rút gọn biểu thức và tìm điều kiện xác định
\(A=\left(\dfrac{1}{x+2}+\dfrac{x}{x^2-4}+\dfrac{2}{2-x}\right):\left(1-\dfrac{x}{x+2}\right)\)
E=\(\left(\dfrac{2}{1+2x}+\dfrac{4x^2}{4x^2-1}-\dfrac{1}{1-2x}\right):\left(\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)\)
a,Rút gọn
b,tính x để E>3
c,Tính Min B
Cho a+b=1. CM \(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{2.\left(ab-2\right)}{a^2b^2+3}\)