Đặt \(t=6x+1\)và \(h=\sqrt{x^2+3}\)
\(\frac{1}{4}\cdot t^2+h^2-\frac{9}{4}=th\)
\(\Leftrightarrow\left(t-2h\right)^2=9\)
\(\Leftrightarrow t-2h=\pm3\)
Với \(t-2h=3\)ta có
\(6x+1-2\sqrt{x^2+3}=3\)
\(\Leftrightarrow3x-1=\sqrt{x^2+3}\)
\(\Leftrightarrow\hept{\begin{cases}3x-1\ge0\\x^2+3=\left(3x+2\right)^2\end{cases}\Leftrightarrow x=\frac{\sqrt{7}-3}{4}}\)
Vậy pt có nghiệm là \(x=1;x=\frac{\sqrt{7}-3}{4}\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
Giải phương trình: \(10x^2\)\(+3x+1=\left(6x+1\right)\sqrt{x^2+3}\)
Giải phương trình:
a.\(\left(17-6x\right)\sqrt{3x-5}+\left(6x-7\right)\sqrt{7-3x}=2+8\sqrt{36x-9x^2-35}\)
b.\(\sqrt{x^2-3x+2}=\sqrt{10x-20}-\sqrt{x-3}\)
giải pt:
1) \(4\sqrt{\frac{x^2}{3}+4}=1+\frac{3x}{2}+\sqrt{6x}\)
2) \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
3) \(\sqrt{1+x}+\sqrt{1-x}+\frac{x^2}{4}=2\)
giải pt \(\sqrt{3x^2-6x-6}=3\sqrt{\left(2-x\right)^5}+\left(7x-19\right)\sqrt{2-x}\)
giải các pt vô tỉ sau:
1, \(x^2\)-\(5x\)+\(4\)=\(\left(2x-1\right)\sqrt{x^2-3x+4}\)
2, \(2\sqrt{\left(x+2\right)^3}=6x+3x^2-x^3\)
3, \(x=\left(\sqrt{x}+2\right)\left(1-\sqrt{1-\sqrt{x}}\right)^2\)
Cho \(x=\dfrac{3+\sqrt{5}}{2}\). Tình \(P=\left(10x^2-30x+11\right)^2+\dfrac{\left(2x^2-6x+3\right)^{10}}{x^5-3x^4+x^3-1}\)
1)giải pt \(x^3-9x^2+6x-6-3\sqrt[3]{6x^2+2}=0\)
2) giải hệ pt \(\int^{\sqrt{x}\left(1+\frac{3}{x+3y}\right)=2}_{\sqrt{7y}\left(1-\frac{3}{x+3y}\right)=4\sqrt{2}}\)
Giải pt: \(\sqrt[4]{\left(x-2\right)\left(4-x\right)}+\sqrt[4]{x-2}+\sqrt[4]{4-x}+6x\sqrt{3x}=x^3+30\)
