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HN
17 tháng 8 2017
b/ \(\sqrt{12-\dfrac{12}{x^2}}+\sqrt{x^2-\dfrac{12}{x^2}}=x^2\)
\(\Leftrightarrow x-\sqrt{12-\dfrac{12}{x^2}}=\sqrt{x^2-\dfrac{12}{x^2}}\)
Bình phương 2 vế rút gọn
\(\Leftrightarrow x^4-x^2-4\sqrt{3\left(x^4-x^2\right)}+12=0\)
Đặt \(\sqrt{x^4-x^2}=a\)
\(\Rightarrow a^2-4\sqrt{3}a+12=0\)
\(\Leftrightarrow a=2\sqrt{3}\)
\(\Leftrightarrow x^4-x^2=12\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
LP
1
3 tháng 10 2018
https://diendantoanhoc.net/topic/163051-x-fracxsqrtx2-1-frac3512/
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 \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
11 tháng 3 2021
1) \(\Leftrightarrow4-4\sqrt{\dfrac{x+2}{x-3}}=x+7\)
\(\Leftrightarrow-4\sqrt{\dfrac{x+2}{x-3}}=x+3\)
\(\Leftrightarrow16\dfrac{x+2}{x-3}=x^2+6x+9\)
\(\Leftrightarrow16x+3=x^3+6x^2+9x-3x^2-18x-27\)
\(\Leftrightarrow x^3+3x^2-25x-59=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4,79\\x=-2,2\\x=-5,58\end{matrix}\right.\)
Vậy tập nghiệm....
HP
28 tháng 3 2021
ĐK: \(-1\le x< 0;x\ge1\)
TH1: \(-1\le x< 0\Rightarrow VP< 0;VT\ge0\Rightarrow\) vô nghiệm
TH2: \(x\ge1\)
\(pt\Leftrightarrow x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}\)
\(\Leftrightarrow x^2+1-\dfrac{1}{x}-2x\sqrt{1-\dfrac{1}{x}}=x-\dfrac{1}{x}\)
\(\Leftrightarrow x^2-x+1-2\sqrt{x^2-x}=0\)
\(\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x^2-x}=1\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{1+\sqrt{5}}{2}\)
Vậy ...
DY
1
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2022
ĐKXĐ: \(x>0\)
\(3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}+1\right)-9\)
\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-9\)
Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a>0\)
\(\Rightarrow3a< 2a^2-9\Rightarrow2a^2-3a-9>0\)
\(\Rightarrow\left(a-3\right)\left(2a+3\right)>0\)
\(\Rightarrow a-3>0\Rightarrow a>3\)
\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>3\Leftrightarrow2x+1>6\sqrt{x}\)
\(\Leftrightarrow2x-6\sqrt{x}+1>0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{3+\sqrt{7}}{2}\\0\le\sqrt{x}< \dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>\dfrac{8+3\sqrt{7}}{2}\\0\le x< \dfrac{8-3\sqrt{7}}{2}\end{matrix}\right.\)
G
3
D
8 tháng 2 2023
`[x+35]/1984-[x+30]/1989+[x+19]/2000+[x+23]/[1996=-2`
`<=>[x+35]/1984+1-[x+30]/1989-1+[x+19]/2000+1+[x+23]/1996+1=0`
`<=>[x+2019]/1984-[x+2019]/1989+[x+2019]/2000+[x+2019]/1996=0`
`<=>(x+2019)(1/1984-1/1989+1/2000+1/1996)=0`
`=>x+2019=0`
`<=>x=-2019`
8 tháng 2 2023
\(\dfrac{x+35}{1984}-\dfrac{x+30}{1989}+\dfrac{x+19}{2000}+\dfrac{x+23}{1996}\text{=}-2\)
\(\Leftrightarrow\dfrac{x+35}{1984}-\dfrac{x+30}{1989}+\dfrac{x+19}{2000}+\dfrac{x+23}{1996}+3-1\text{=}0\)
\(\Leftrightarrow\left(\dfrac{x+35}{1984}+1\right)-\left(\dfrac{x+30}{1989}+1\right)+\left(\dfrac{x+19}{2000}+1\right)+\left(\dfrac{x+23}{1996}+1\right)\text{=}0\)
\(\Leftrightarrow\dfrac{x+2019}{1984}-\dfrac{x+2019}{1989}+\dfrac{x+2019}{2000}+\dfrac{x+2019}{1996}\text{=}0\)
\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{1984}-\dfrac{1}{1989}+\dfrac{1}{2000}+\dfrac{1}{1996}\right)\text{=}0\)
\(\Leftrightarrow\left(x+2019\right)\text{=}0\)
\(\Leftrightarrow x\text{=}-2019\)
NN
0