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O
0
27 tháng 11 2021
\(a,ĐK:1\le x\le3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{3-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(PT\Leftrightarrow a+b-ab=1\Leftrightarrow a+b-ab-1=0\\ \Leftrightarrow\left(a-1\right)\left(1-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\3-x=1\end{matrix}\right.\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:0\le x\le9\\ PT\Leftrightarrow9+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\\ \Leftrightarrow2\sqrt{-x^2+9x}-\left(-x^2+9x\right)=0\\ \Leftrightarrow\sqrt{-x^2+9x}\left(2-\sqrt{-x^2+9x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x^2+9x=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\\x^2-9x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=9\left(n\right)\\x=\dfrac{9+\sqrt{65}}{2}\left(n\right)\\x=\dfrac{9-\sqrt{65}}{2}\left(n\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 11 2019
a/ ĐKXĐ: \(-2\le x\le5\)
\(\sqrt{x+2}+\sqrt{5-x}+\sqrt{\left(x+2\right)\left(5-x\right)}-4=0\)
Đặt \(\sqrt{x+2}+\sqrt{5-x}=a>0\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}\)
\(\Rightarrow a+\frac{a^2-7}{2}-4=0\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}=1\)
\(\Leftrightarrow-x^2+3x+10=1\)
\(\Leftrightarrow x^2-3x-9=0\)
b/ \(\Leftrightarrow\sqrt{x+1}-\sqrt{4-x}+2\left(5+2\sqrt{\left(x+1\right)\left(4-x\right)}\right)=17\)
Đặt \(\sqrt{x+1}-\sqrt{4-x}=a\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{5-a^2}{2}\)
\(a+2\left(5+5-a^2\right)=17\)
\(\Leftrightarrow-2a^2+a+3=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}-\sqrt{4-x}=-1\\\sqrt{x+1}-\sqrt{4-x}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}+1=\sqrt{4-x}\\2\sqrt{x+1}=2\sqrt{4-x}+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2+2\sqrt{x+1}=4-x\\4x+4=25-4x+12\sqrt{4-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1-x\left(x\le1\right)\\12\sqrt{4-x}=8x-21\left(x\ge\frac{21}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\left(1-x\right)^2\\144\left(4-x\right)=\left(8x-21\right)^2\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 11 2019
c/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)
\(a^2-1=3\left(a-1\right)\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-x^2}=\frac{a^2-1}{2}=0\\\sqrt{x-x^2}=\frac{a^2-1}{2}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x^2=0\\x-x^2=\frac{9}{4}\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{5+2x}=a\ge0\\\sqrt{5-2x}=b\ge0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}\left(3a-1\right)\left(3b-1\right)=16\\a^2+b^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3ab-\left(a+b\right)=5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=3ab-5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\)
\(\Rightarrow\left(3ab-5\right)^2-2ab=10\)
\(\Leftrightarrow9\left(ab\right)^2-32ab+15=0\Rightarrow\left[{}\begin{matrix}ab=3\\ab=\frac{5}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(ab\right)^2=9\\\left(ab\right)^2=\frac{25}{81}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}25-4x^2=9\\25-4x^2=\frac{25}{81}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=4\\x^2=\frac{500}{81}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
VT
1
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 11 2019
ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1-x^2}=a\ge0\) ta được:
\(\left\{{}\begin{matrix}x^2+a^2=1\\x^3+a^3=\sqrt{2}ax\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+a^2=1\\\left(x+a\right)\left(x^2+a^2-ax\right)=\sqrt{2}ax\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+a^2=1\\\left(x+a\right)\left(1-ax\right)=\sqrt{2}ax\end{matrix}\right.\)
Đặt \(x+a=t\Rightarrow x^2+a^2+2ax=t^2\Rightarrow ax=\frac{t^2-1}{2}\)
\(\Rightarrow t\left(1-\frac{t^2-1}{2}\right)=\sqrt{2}\left(\frac{t^2-1}{2}\right)\)
\(\Leftrightarrow t^3+\sqrt{2}t^2-3t-\sqrt{2}=0\)
\(\Leftrightarrow\left(t-\sqrt{2}\right)\left(t^2+2\sqrt{2}t+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{2}\\t=1-\sqrt{2}\\t=-1-\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\sqrt{1-x^2}=\sqrt{2}\\x+\sqrt{1-x^2}=-1-\sqrt{2}\left(l\right)\\x+\sqrt{1-x^2}=1-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1-x^2}=\sqrt{2}-x\\\sqrt{1-x^2}=1-\sqrt{2}-x\left(x\le1-\sqrt{2}\right)\\\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x^2=\left(\sqrt{2}-x\right)^2\\1-x^2=\left(1-\sqrt{2}-x\right)^2\end{matrix}\right.\) \(\Leftrightarrow...\)
LT
1
HN
27 tháng 12 2017
\(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\)
\(\Leftrightarrow\dfrac{x^2+\sqrt{\left(x+3\right)\left(x+1\right)}}{\sqrt{x+3}+\sqrt{x+1}}=x\)
\(\Leftrightarrow\left(x-\sqrt{x+1}\right)\left(x-\sqrt{x+3}\right)=0\)
\(\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}\)
\(Đk\left[{}\begin{matrix}x\ge1\\x=0\\x\le-2\end{matrix}\right.\)
\(Pt:\Leftrightarrow x^2-x+2\sqrt{x^2\left(x-1\right)\left(x+2\right)}+x^2+2x=4x^2\)
\(\Leftrightarrow2\sqrt{x^2\left(x-1\right)\left(x+2\right)}=2x^2-x\)
\(\Leftrightarrow4x^2\left(x-1\right)\left(x+2\right)=4x^4-4x^3+x^2\)
\(\Leftrightarrow4x^4+4x^3-8x^2=4x^4-4x^3+x^2\)
\(\Leftrightarrow8x^3-9x^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{9}{8}\end{matrix}\right.\left(tm\right)\)