\(\frac{1}{1-x^2}+1=\frac{3x}{\sqrt{1-x^2}}\)
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AN
6 tháng 11 2019
a)\(ĐK:-3\le x\le6\)
\(PT\Leftrightarrow\sqrt{x+3}+\sqrt{6-x}=3\)
\(\Leftrightarrow x+3+6-x+2\sqrt{\left(x+3\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\left(tm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 11 2019
b/ ĐKXĐ: \(x\ge7\)
\(\sqrt{3x-2}=1+\sqrt{x-7}\)
\(\Leftrightarrow3x-2=x-6+2\sqrt{x-7}\)
\(\Leftrightarrow x+2=\sqrt{x-7}\)
\(\Leftrightarrow x^2+4x+4=x-7\)
\(\Leftrightarrow x^2+3x+11=0\) (vô nghiệm)
c/ ĐKXĐ: \(x\ge1;x\ne50\)
\(1-\sqrt{3x+1}=\sqrt{x-1}-7\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{3x+1}=8\)
\(\Leftrightarrow4x+2\sqrt{3x^2-2x-1}=64\)
\(\Leftrightarrow\sqrt{3x^2-2x-1}=32-2x\) (\(x\le16\))
\(\Leftrightarrow3x^2-2x-1=\left(32-2x\right)^2\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2019
a/ ĐKXĐ: ...
\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)
Đặt \(\sqrt{x^2-5x-6}=a\ge0\)
\(2a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)
b/ ĐKXĐ: ...
\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)
Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)
\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)
c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)
Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)
\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2019
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)
e/ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)
Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)
f/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)
\(\frac{1}{a}+1+a=3a^2\)
\(\Leftrightarrow3a^3-a^2-a-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)
\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)
TV
17 tháng 10 2019
\((\frac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-2):\frac{1}{x-1}\)
=\(\left(\frac{3x+3\sqrt{x}-3}{(\sqrt{x}+2)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+2}{(\sqrt{x}+2)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(x-1\right)\)
=\(\left(\frac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1-\left(2\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(x-1\right)\)
=\(\left(\frac{3x+5\sqrt{x}-2-\left(2x+4\sqrt{x}-2\sqrt{x}-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(x-1\right)\)
=\(\left(\frac{3x+5\sqrt{x}-2-2x-2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(x-1\right)\)
=\(\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(x-1\right)\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(x-1\right)\)
=\(\frac{\left(\sqrt{x}+1\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)}\)=\(\left(\sqrt{x}+1\right)^2\)
\(\frac{1}{1-x^2}+1=\frac{3x}{\sqrt{1-x^2}}\)
\(\Leftrightarrow1.\sqrt{-x^2+1}+1.\left(-x^2+1\right).\sqrt{-x^2+1}=3x.\left(-x^2+1\right)\)
\(\Leftrightarrow\sqrt{-x^2+1}+\sqrt{-x^2+1}.\left(-x^2+1\right)=3\left(-x^2+1\right)\)
\(\Leftrightarrow-\sqrt{-x^2+1}.x^2+2\sqrt{-x^2+1}=-3x^2+3x\)
\(\Leftrightarrow\sqrt{-x^2+1}.\left(x^2+2\right)=-3x^3+3x\)
\(\Leftrightarrow\sqrt{-x^2+1}=-\frac{3x^2}{-x^2+2}+\frac{3x}{-x^2+2}\)
\(\Leftrightarrow\sqrt{-x^2+1}=\frac{-3x^2+3x}{-x^2+2}\)
\(\Leftrightarrow\left(\sqrt{-x^2+1}\right)^2=\left(\frac{-3x^2+3x}{-x^2+2}\right)^2\)
\(\Leftrightarrow-x^2+1=\frac{9x^6-18x^4+9x^2}{x^4-4x^2+4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{1}{2}}\\x=\frac{2\sqrt{5}}{5}\end{cases}}\)
Vậy nghiệm phương trình là \(\left\{\sqrt{\frac{1}{2}};\frac{2\sqrt{5}}{5}\right\}\)