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ĐK: \(x\ge\dfrac{1}{2}\)
\(pt\Leftrightarrow\sqrt{x}-1+\sqrt{2x-1}-1+x^2+x-2=0\)
\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{2x-2}{\sqrt{2x-1}+1}+\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2\right)\left(x-1\right)=0\)
Vì \(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2>0\) nên \(x-1=0\Leftrightarrow x=1\left(tm\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-2>=0\\4-x>=0\\x+1< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2< =x< =4\\x< >-1\end{matrix}\right.\Leftrightarrow x\in\left[2;4\right]\)
\(\dfrac{x-2}{x+1}-\dfrac{3}{x+2}>0.\left(x\ne-1;-2\right).\\ \Leftrightarrow\dfrac{x^2-4-3x-3}{\left(x+1\right)\left(x+2\right)}>0.\\ \Leftrightarrow\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)
Đặt \(f\left(x\right)=\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)
Ta có: \(x^2-3x-7=0.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{37}}{2}.\\x=\dfrac{3-\sqrt{37}}{2}.\end{matrix}\right.\)
\(x+1=0.\Leftrightarrow x=-1.\\ x+2=0.\Leftrightarrow x=-2.\)
Bảng xét dấu:
\(\Rightarrow f\left(x\right)>0\Leftrightarrow x\in\left(-\infty-2\right)\cup\left(\dfrac{3-\sqrt{37}}{2};-1\right)\cup\left(\dfrac{3+\sqrt{37}}{2};+\infty\right).\)
\(\sqrt{x^2-3x+2}\ge3.\\ \Leftrightarrow x^2-3x+2\ge9.\\ \Leftrightarrow x^2-3x-7\ge0.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{37}}{2}.\\x=\dfrac{3+\sqrt{37}}{2}.\end{matrix}\right.\)
Đặt \(f\left(x\right)=x^2-3x-7.\)
\(f\left(x\right)=x^2-3x-7.\)
\(\Rightarrow f\left(x\right)\ge0\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)
\(\Rightarrow\sqrt{x^2-3x+2}\ge3\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)
1:
ĐKXĐ: x<>3
\(\dfrac{x-1}{x-3}>1\)
=>\(\dfrac{x-1-\left(x-3\right)}{x-3}>0\)
=>\(\dfrac{x-1-x+3}{x-3}>0\)
=>\(\dfrac{2}{x-3}>0\)
=>x-3>0
=>x>3
2: ĐKXĐ: \(\left[{}\begin{matrix}x>=3\\x< =-4\end{matrix}\right.\)
\(\sqrt{x^2+x-12}< 8-x\)
=>\(\left\{{}\begin{matrix}8-x>=0\\x^2+x-12< \left(8-x\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =8\\x^2+x-12-x^2+16x-64< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =8\\17x-76< 0\end{matrix}\right.\)
=>\(x< \dfrac{76}{17}\)
Kết hợp ĐKXĐ, ta được: \(\left[{}\begin{matrix}3< =x< \dfrac{76}{17}\\x< =-4\end{matrix}\right.\)
\(sinx+cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(=\sqrt{2}cos\left(\frac{\pi}{2}-\left(x+\frac{\pi}{4}\right)\right)=\sqrt{2}cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)
\(sinx-cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx-\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
\(=-\sqrt{2}sin\left(\frac{\pi}{4}-x\right)=-\sqrt{2}cos\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+sin2x\)
\(=sin^2x-cos^2x+sin2x=sin2x-cos2x\)
\(=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)
Bạn ghi ko đúng đề
\(x=1\)