1)giải pt: 1+\(\dfrac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
2)giải pt: \(\dfrac{x^2}{\sqrt{3x-2}}-\sqrt{3x-2}=1-x\)
giải pt :
a, \(\dfrac{\sqrt{x-3}}{\sqrt{2x-1}-1}=\dfrac{1}{\sqrt{x+3}-\sqrt{x-3}}\)
b, \(\left(\sqrt{x^2+x+1}+\sqrt{4x^2+x+1}\right)\left(\sqrt{5x^2+1}-\sqrt{2x^2+1}\right)=3x^2\)
Giải PT: \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right).\sqrt{\dfrac{x-1}{x-2}}=3\)
giải pt :
a, \(4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{16x^4+4x^2+1}=0\)
b, \(x^2-3x+1+\dfrac{1}{\sqrt{3}}\sqrt{x^4+x^2+1}=0\)
a.
\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:
\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)
\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)
\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)
\(\Leftrightarrow3a^2=b^2\)
\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)
\(\Leftrightarrow...\)
b.
\(x^2-3x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)
Lặp lại cách làm câu a
giải pt :
a, (x+5)(2-x)=3\(\sqrt{x^2+3x}\)
b, \(\sqrt[3]{\dfrac{2x}{x+1}}+\sqrt[3]{\dfrac{1}{2}+\dfrac{1}{2x}}=2\)
c,\(\sqrt[5]{\dfrac{16x}{x-1}}+\sqrt[5]{\dfrac{x-1}{16x}}=\dfrac{5}{2}\)
d, \(\sqrt{5x^2+10x+1}=7-2x-x^2\)
e, \(\sqrt{2x^2+4x+1}=1-2x-x^2\)
giải pt :
a,\(2x^2-11x+21=3\sqrt[3]{4x-4}\)
b,\(\dfrac{\sqrt{x-3}}{\sqrt{2x-1}-1}=\dfrac{1}{\sqrt{x+3}-\sqrt{x-3}}\)
c,\(\left(\sqrt{x^2+x+1}+\sqrt{4x^2+x+1}\right)\left(\sqrt{5x^2+1}-\sqrt{2x^2+1}\right)=3x^2\)
giải pt sau :
\(2\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2-\dfrac{1}{4}+\sqrt{...+\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2+x+\dfrac{1}{4}}}}}}=2x^3+3x^2+3x+1\)
trong đó biểu thức ở vế trái có 2010 dấu căn thức bậc 2
Giải pt sau
a,\(^{x^2-6x+26=6\sqrt{2x+1}}\)
b,\(x^2+2x\sqrt{x-\dfrac{1}{x}}=3x+1\)
giải hệ pt :
\(\left\{{}\begin{matrix}\dfrac{\sqrt{x}}{1+\sqrt{1-x}}-\dfrac{\sqrt{y}}{1+\sqrt{y}}+x+y=1\\8x^2+7x+20y-13=\left(1+\dfrac{1}{1-y}\right)\sqrt[3]{3x^2-2}\end{matrix}\right.\)
1,
A=\(3\sqrt{8}-\sqrt{50}-\sqrt{\sqrt{2}-1}\)
B=2\(\dfrac{2}{x-1}\sqrt{\dfrac{x^2-2x+1}{4x^2}}\) với 0<x<1
2,Giải pt
\(\sqrt{x^2-3x+2}+\sqrt{x+2}=\sqrt{x+2}+\sqrt{x^2+2x+3}\)
\(A=3\sqrt{8}-\sqrt{50}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow6\sqrt{2}-5\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(B=2.\dfrac{2}{x-1}.\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{x^2-2x+1}}{2x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{\left(x-1\right)^2}}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{x-1}{x}\)
\(\Leftrightarrow\)\(2.\dfrac{1}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x}\)