\(2+\dfrac{\sqrt{19-2x}}{x}=1\)
\(2+\dfrac{\sqrt{19-2x}}{x}=1;x\ne0\)
\(\dfrac{\sqrt{19-2x}}{x}=1-2\)
\(\dfrac{\sqrt{19-2x}}{x}=-1\)
\(\sqrt{19}-2x=-x\)
\(-2x+x=-\sqrt{19}\)
\(x=\sqrt{19};x\ne0\)
\(x=\sqrt{19}\)
\(2+\dfrac{\sqrt{19-2x}}{x}=1\)
\(2+\dfrac{\sqrt{19-2x}}{x}=1;x\ne0\)
\(\dfrac{\sqrt{19-2x}}{x}=1-2\)
\(\dfrac{\sqrt{19-2x}}{x}=-1\)
\(\sqrt{19}-2x=-x\)
\(-2x+x=-\sqrt{19}\)
\(x=\sqrt{19};x\ne0\)
\(x=\sqrt{19}\)
Giải phương trình:
1, \(4\sqrt{x+3}+\sqrt{19-3x}=x^2+2x+9\)
2, \(\sqrt{3x-8}-\sqrt{x+1}=\dfrac{2x-11}{5}\)
3, \(\sqrt{x+\dfrac{3}{x}}=\dfrac{x^2+7}{2\left(x+1\right)}\)
Cho x=\(\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\). Tính A=(4x5+4x4-x3+1)19+\(\sqrt{4x^5+4x^4-5x^3+5x}\)+\(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2019}\)
\(\sqrt{2x-1}+\sqrt{19-3x}\) = \(\dfrac{6}{-x^2+10x-24}\)
Giải phương trình:
1. \(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)
2. \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\)
3. \(\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{6+2x}{\sqrt{5+x}}=\dfrac{8}{3}\)
4. \(x^2+1-\left(x+1\right)\sqrt{x^2-2x+3}=0\)
5. \(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)
6. \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
1) \(\dfrac{x-3x^2}{2}+\sqrt{2x^4-x^3+7x^2-3x+3}=2\)
2) \(1+\sqrt{\dfrac{x-2}{1-x}}=\dfrac{2x^2-2x+1}{x^2-2x+2}\)
3) \(x+y+z+\dfrac{3}{x-1}+\dfrac{3}{y-1}+\dfrac{3}{z-1}=2\left(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2}\right)\) với x ,y ,z > 1
4) \(\sqrt[3]{x+6}+x^2=7-\sqrt{x-1}\)
5) \(x^4-2x^3+x-\sqrt{2\left(x^2-x\right)}=0\)
Cho biểu thức: \(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\). Tính giá trị của A tại x=\(\sqrt{2}\)
Cho \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2.\left(\sqrt{3}+1\right)}}\). Tính: \(A=\dfrac{4.\left(x+1\right).x^{2013}-2.x^{2012}+2x+1}{2x^2+3x}\)
rút gọn bt
A= \(\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right):\dfrac{2\sqrt{x}-1}{\sqrt{x}-x}\)
Rút gọn biểu thức A= \(\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right):\dfrac{2\sqrt{x}-1}{\sqrt{x}-x}\)