bài 1:giải phương trình
a, x=\(\sqrt{x}\)
b, \(\frac{4}{x-1}+\frac{2}{x}=\frac{3x+4}{x^2-x}\)
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HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a)
\(\sin \left( {2x + \frac{\pi }{4}} \right) = \sin x \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{4} = x + k2\pi \\2x + \frac{\pi }{4} = \pi - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\3x = \pi - \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{4} + \frac{{k2\pi }}{3}\end{array} \right.;k \in Z\)
b)
\(\begin{array}{l}\sin 2x = \cos 3x\\ \Leftrightarrow \cos 3x = \cos \left( {\frac{\pi }{2} - 2x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} - 2x + k2\pi \\3x = - \left( {\frac{\pi }{2} - 2x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\end{array}\)
c)
\(\begin{array}{l}{\cos ^2}2x = {\cos ^2}\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = - \cos \left( {x + \frac{\pi }{6}} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\pi - \left( {x + \frac{\pi }{6}} \right)} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right)\end{array} \right.\end{array}\)
Với \(\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}2x = - \left( {x + \frac{\pi }{6}} \right) + k2\pi \\2x = x + \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{18}} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{6} + k2\pi \end{array} \right.\)
Với \(\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right) \Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{6} - x + k2\pi \\2x = - \left( {\frac{{5\pi }}{6} - x} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = \frac{{5\pi }}{6} + k2\pi \\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}\\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right.\)
TN
4 tháng 7 2017
b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)
\(\Leftrightarrow\left(\frac{4}{x}-x\right)+\frac{x-\frac{1}{x}-\frac{3}{2}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}=\frac{2x-\frac{5}{x}-\frac{3}{2}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\)
\(\Leftrightarrow\frac{-\left(x-2\right)\left(x+2\right)}{x}+\frac{\frac{\left(x-2\right)\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(x-2\right)\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{-\left(x+2\right)}{x}+\frac{\frac{\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\right)=0\)
Pt trong ngoặc VN suy ra x=2
TN
4 tháng 7 2017
a)\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)
\(\Leftrightarrow x^2+3\sqrt{x^2-1}-1=\sqrt{x^4-x^2+1}-1\)
\(\Leftrightarrow\frac{x^2\left(3\sqrt{x^2-1}+1\right)}{3\sqrt{x^2-1}+1}+\frac{9\left(x^2-1\right)-1}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2+1-1}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)
pt trong căn vô nghiệm
suy ra x=1; x=-1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 4 2019
a/ ĐK: \(3x^2-10x+6\ge0\)
Nhận thấy \(x=0\) không phải nghiệm
\(\Leftrightarrow2\left(x^2+4\right)=\left(3x^2-10x+6\right)^2\)
\(\Leftrightarrow2\left(x^2+\frac{4}{x^2}\right)=\left(3x-10+\frac{6}{x}\right)^2=\left(3\left(x+\frac{2}{x}\right)-10\right)^2\)
Đặt \(x+\frac{2}{x}=a\Rightarrow x^2+\frac{4}{x^2}=a^2-4\)
\(\Leftrightarrow2\left(a^2-4\right)=\left(3a-10\right)^2\)
\(\Leftrightarrow7a^2-60a+108=0\Rightarrow\left[{}\begin{matrix}a=6\\a=\frac{18}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{2}{x}=6\\x+\frac{2}{x}=\frac{18}{7}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+2=0\\7x^2-18x+14=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3+\sqrt{7}\\x=3-\sqrt{7}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 4 2019
b/ \(x\ge-\frac{1}{4}\)
Đặt \(\sqrt{x+\frac{1}{4}}=a\ge0\Rightarrow x=a^2-\frac{1}{4}\)
\(\Leftrightarrow a^2-\frac{1}{4}+\sqrt{a^2-\frac{1}{4}+\frac{1}{2}+a}=2\)
\(\Leftrightarrow a^2-\frac{1}{4}+\sqrt{\frac{1}{4}\left(4a^2+4a+1\right)}=2\)
\(\Leftrightarrow a^2-\frac{1}{4}+\frac{1}{2}\left(2a+1\right)=2\)
\(\Leftrightarrow4a^2+4a-7=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{-1+2\sqrt{2}}{2}\\a=\frac{-1-2\sqrt{2}}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+\frac{1}{4}}=\frac{-1+2\sqrt{2}}{2}\Rightarrow x=2-\sqrt{2}\)
L
4 tháng 7 2020
1.a) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}.\left(\sqrt{x+2}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\sqrt{x+2}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x+2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Vậy x=2 hoặc x=-1
a) \(ĐKXĐ:x\ge0\)
\(x=\sqrt{x}\)\(\Leftrightarrow x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}.\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)( thỏa mãn )
Vậy \(x=0\)hoặc \(x=1\)
b) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
\(\frac{4}{x-1}+\frac{2}{x}=\frac{3x+4}{x^2-x}\)\(\Leftrightarrow\frac{4x}{x\left(x-1\right)}+\frac{2\left(x-1\right)}{x\left(x-1\right)}=\frac{3x+4}{x\left(x-1\right)}\)
\(\Rightarrow4x+2\left(x-1\right)=3x+4\)
\(\Leftrightarrow4x+2x-2=3x+4\)
\(\Leftrightarrow4x+2x-3x=4+2\)
\(\Leftrightarrow3x=6\)\(\Leftrightarrow x=2\)( thỏa mãn )
Vậy \(x=2\)