Điều kiện: \(x\ge1\)
\(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}=2\\ \Leftrightarrow\sqrt{x-1}=2\sqrt{x}+2\\ \Rightarrow x-1=4x+4+8\sqrt{x}\\ \Leftrightarrow3x+8\sqrt{x}+5=0\\ \Leftrightarrow\left(\sqrt{x}+1\right)\left(3\sqrt{x}+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=-\dfrac{3}{5}\end{matrix}\right.\left(loại\right)} \)
Giải pt sau: \(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\) = \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
giải pt \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\ge\dfrac{3}{2}\)
GIẢI PT
\(\sqrt{x^2+10x+25}=4\)
\(\sqrt{x-2}+3=5\)
\(\sqrt{x^2-x+4}-x^2+x-2=0\)
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{1}{3}\)
Giải PT sau
\(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\)
giải hệ: \(\left\{{}\begin{matrix}\dfrac{1}{x-y}+\dfrac{1}{x+y}=2\\\dfrac{2}{x+y}+\dfrac{3}{x+y}=5\end{matrix}\right.\)
giải pt: \(\sqrt{x^2-4x+7}=\sqrt{x+1}\)
giải pt
x\(\sqrt{y-1}\)+2y\(\sqrt{x-1}\)=\(\dfrac{3xy}{2}\)
giải pt sau
a)\(\sqrt{x^2-6x+9}=3\)
b)\(\sqrt{x+2\sqrt{x-1}}=2\)
c)\(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\)
d)\(\sqrt{x-4}+\sqrt{x+1}=5\)
Help
\(\left\{{}\begin{matrix}\sqrt{3x}.\left(1+\dfrac{1}{x+y}\right)=2\\\sqrt{7y}.\left(1-\dfrac{1}{x-y}\right)=4\sqrt{2}\end{matrix}\right.\)
Giải hệ pt
Tính GTLN của biểu thức A.
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}\)(đk: \(x\ge0,x\ne1,x\ne4\))
B2. Giải pt
\(\sqrt{x-3}+\sqrt{y-5}+\sqrt{z-4}=20-\dfrac{4}{\sqrt{x-3}}-\dfrac{9}{\sqrt{y-5}}-\dfrac{25}{\sqrt{z-4}}\)
