\(\text{Giải phương trình: }2x+\frac{x-1}{x}=\sqrt{1-\frac{1}{x}}+3\sqrt{x-\frac{1}{x}}\left(x>0\right)\)
giải phương trình
a) \(\left(x+\frac{5-x}{\sqrt{x}+1}\right)^2+\frac{16\sqrt{x}\left(5-x\right)}{\sqrt{x}+1}-16\)\(=0\)
b) \(\sqrt{2x-\frac{3}{x}}+\sqrt{\frac{6}{x}-2x}=1+\frac{3}{2x}\)
c) \(\sqrt{2x+1}+\frac{2x-1}{x+3}-\left(2x-1\right)\sqrt{x^2+4}-\sqrt{2}=0\)
d) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
Tìm điều kiện xác định và giải các phương trình sau
a) \(\frac{3}{x-5}.\frac{\sqrt{\left(5-x\right)^2.\left(x-1\right)}}{\sqrt{\left(x-1\right)^2}}-\frac{1}{x+1}\)
b) \(\sqrt{\frac{1+x}{2x}}:\sqrt{\frac{\left(x+1\right)^3}{8x}}-\sqrt{x^2-4x+4}=0\)
Giải các phương trình sau:
a) \(\sin x = \frac{{\sqrt 3 }}{2}\);
b) \(2\cos x = - \sqrt 2 \);
c) \(\sqrt 3 \tan \left( {\frac{x}{2} + {{15}^0}} \right) = 1\);
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\)
a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)
b) \(2\cos x = - \sqrt 2 \;\; \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x = - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)
c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)
\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
giải phương trình \(2x+\frac{x-1}{x}=\sqrt{1-\frac{1}{x}}+3\sqrt{x-\frac{1}{x}}\left(x>0\right)\)
Giải phương trình: \(2x+\frac{x-1}{x}=\sqrt{1-\frac{1}{x}}+3\sqrt{x-\frac{1}{x}}\left(x>0\right)\)
Giải Phương trình sau : \(\sqrt{x}-x\left(x-\frac{1}{2}\right)=\frac{1}{2x^3}-\frac{1}{2x\sqrt{x}}\)
giải hệ phương trình sau :\(\hept{\begin{cases}\sqrt{4x-2y}-2\sqrt{x-2y}=-1\\\sqrt{x-2y}+7\left(2x-y\right)=37\end{cases}}\)
Giải phương trình \(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\Rightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{2x^3}{2}+\frac{x^2}{2}+\frac{2x}{2}+\frac{1}{2}\)
\(\Rightarrow\sqrt{x^2+x+\frac{1}{2}-\frac{1}{4}}=\sqrt{x^2+x+\frac{1}{4}}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)
\(\Rightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=x+\frac{1}{2}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)
\(\Rightarrow x^3+\frac{x^2}{2}+x+\frac{1}{2}-x-\frac{1}{2}=x^3+\frac{x^2}{2}=0\Rightarrow\frac{2x^3+x^2}{2}=0\)
\(\Rightarrow2x^3+x^2=0\Rightarrow x^2\left(2x+1\right)=0\Rightarrow\hept{\begin{cases}x^2=0\Rightarrow x=0\\2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\end{cases}}\)
vậy x=0 và x=-1/2
giải phương trình:
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\) (*) (ĐKXĐ: \(\forall x\in R\))
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+\left(2x+1\right)\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
+) Xét \(x+\frac{1}{2}\ge0\Leftrightarrow x\ge-\frac{1}{2}\). Khi đó pt (*) trở thành:
\(\sqrt{x^2-\frac{1}{4}+x+\frac{1}{2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\) (Do \(x\ge\frac{1}{2}\))
\(\Leftrightarrow\frac{\left(2x+1\right)\left(x^2+1\right)-\left(2x+1\right)}{2}=0\)
\(\Leftrightarrow x^2\left(2x+1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\) (t/m ĐKXĐ)
+) Xét \(x+\frac{1}{2}< 0\Leftrightarrow x< -\frac{1}{2}\). Khi đó: \(2x+1< 0\)
Ta thấy: \(2x+1< 0;x^2+1>0;\frac{1}{2}>0\Rightarrow\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)< 0\)
Mà \(\sqrt{x^2-\frac{1}{4}+\left|x+\frac{1}{2}\right|}\ge0\) nên Vô lí ---> Loại TH này.
Vậy tập nghiệm của pt (*) là \(S=\left\{0;-\frac{1}{2}\right\}.\)
rthgsdgdh olweikehgf
Giải phương trình:
a) \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\)
b) \(2\sqrt[3]{\frac{2x-3}{1-x}}+\sqrt[3]{\frac{1-x}{2x-3}}=3\)
c) \(x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6=0\)
a) \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\)
Ta có: \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}\ge2\sqrt{\sqrt{\frac{2x-1}{x+1}}\cdot\sqrt{\frac{x+1}{2x-1}}}=2\) (BĐT Cô-si)
Mà \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\) (theo đề bài)
Suy ra dấu bằng phải xảy ra \(\Rightarrow\sqrt{\frac{2x-1}{x+1}}=\sqrt{\frac{x+1}{2x-1}}\) \(\Leftrightarrow\frac{2x-1}{x+1}=\frac{x+1}{2x-1}\) \(\Leftrightarrow\left(2x-1\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+1\\2x-1=-x-1\end{matrix}\right.\Leftrightarrow\) \(x=2\) (tmđkxđ) hoặc \(x=0\) (không tmđkxđ)
Vậy \(S=\left\{2\right\}\).
Bạn đừng quên tự tìm ĐKXĐ cho câu a nhé bạn.
c) \(x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6=0\) ĐKXĐ: \(x>0\)
Vì \(x>0\Rightarrow x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6>0\)
Vậy \(S=\varnothing\).