Giải các phương trình:
a) 3 + x − 2 = 0 ; b) − x + 2 + 1 = 0 ;
c) 1 − 2 x = 3 x + 1 ; d) x + 1 = 3 2 − x
Giải các phương trình:
a) \(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)
b) \(\left(x^2-4x+4\right)\left(x^2-3x\right)=0\)
c) \(1,2x^3-x^2-0,2x=0\)
a.\(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)
Ta có: \(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\ge4>0;\forall x\)
\(\Rightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
b.\(\left(x^2-4x+4\right)\left(x^2-3x\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)
c.\(1,2x^3-x^2-0,2x=0\)
\(\Leftrightarrow x\left(1,2x^2-x-0,2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{1}{6}\end{matrix}\right.\)
giải các phương trình:
a)(x2+x+1)2-2x2-2x=5
b)\(\dfrac{1}{\left(x-1\right)\left(x-3\right)}\)+x2-4x+5=0
giải phương trình:
a) \(\sqrt{x+6}-\sqrt{x-2}=2\)
b) \(2\sqrt{x-3}-2x+3=0\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x+6>=0\\x-2>=0\end{matrix}\right.\Leftrightarrow x>=2\)
\(\sqrt{x+6}-\sqrt{x-2}=2\)
=>\(\left(\sqrt{x+6}-\sqrt{x-2}\right)^2=4\)
=>\(x+6+x-2-2\sqrt{\left(x+6\right)\left(x-2\right)}=4\)
=>\(2\sqrt{\left(x+6\right)\left(x-2\right)}=2x+4-4=2x\)
=>\(\sqrt{\left(x+6\right)\left(x-2\right)}=x\)
=>\(\left\{{}\begin{matrix}x>=0\\\left(x+6\right)\left(x-2\right)=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2+4x-12=x^2\end{matrix}\right.\)
=>x=3
b: ĐKXĐ: \(x-3>=0\)
=>x>=3
\(2\sqrt{x-3}-2x+3=0\)
=>\(\sqrt{4x-12}=2x-3\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\4x-12=4x^2-12x+9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\4x^2-12x+9-4x+12=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\4x^2-16x+21=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Giải phương trình:
a.2x+x-3=5
b.3(2-x)-(x+2)=0
a: =>3x=8
hay x=8/3
b: =>6-3x-x-2=0
=>-4x+4=0
hay x=1
\(a,2x+x-3=5\\ \Rightarrow3x=5+3\\ \Rightarrow3x=8\\ \Rightarrow x=\dfrac{8}{3}\\ b,3\left(2-x\right)-\left(x+2\right)=0\\ \Rightarrow6-3x-x-2=0\\ \Rightarrow4-4x=0\\ \Rightarrow4x=4\\ \Rightarrow x=1\)
a,2x+x-3=5
3x-3=5
=>x=\(\dfrac{8}{3}\)
b.3(2-x)-(x+2)=0
6-3x-x-2=0
4-4x=0
=>x=1
Giải các phương trình:
a) \(\left|\sin x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
b) \(\tan^2\left(x+\dfrac{\pi}{6}\right)=3\)
c) \(2\sin\left(4x-\dfrac{\pi}{3}\right)-1=0\)
a, \(\left|sinx+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow sin^2x+sinx+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow sin^2x+sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
b, \(tan^2\left(x+\dfrac{\pi}{6}\right)=3\)
\(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=\pm\sqrt{3}\)
\(\Leftrightarrow x+\dfrac{\pi}{6}=\pm\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
c, \(2sin\left(4x-\dfrac{\pi}{3}\right)-1=0\)
\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\4x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Giải phương trình:
a) \(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\)
b \(2x^4-5x^3+6x^2-5x+2=0\)
\(a,\left(đk:x\ge0\right)\)
\(x=0\Rightarrow\sqrt{0+3}+0=0\left(vô-nghiệm\right)\)
\(x>0\)
\(\)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}+\dfrac{4\sqrt{x}}{\sqrt{x+3}}=4\)
\(VT\ge2\sqrt{\dfrac{\sqrt{x+3}}{\sqrt{x}}.\dfrac{4\sqrt{x}}{\sqrt{x+3}}}=4\)
\(dấu"="xảy-ra\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}=\dfrac{4\sqrt{x}}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Leftrightarrow x=1\left(tm\right)\)
\(b.2x^4-5x^3+6x^2-5x+2=0\Leftrightarrow\left(x-1\right)^2\left(2x^2-2x+2\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+2=0\left(vô-nghiệm\right)\end{matrix}\right.\)
a) ĐKXĐ : \(x\ge0\)
PT <=> \(x+3-4\sqrt{x}\sqrt{x+3}+4x=0\)
<=> \(\left(\sqrt{x+3}-2\sqrt{x}\right)^2=0\)
<=> \(\sqrt{x+3}=2\sqrt{x}\)
<=> \(x+3=4x\)
<=> x = 1
Vậy x = 1 là nghiệm phương trình
Giải phương trình:
a)\(2x^3+4x^2+10x=0\)
b)\(\dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)
a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)
=>x=0
b: \(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{x+1}{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow x^2-4x+3=2x\left(x-3\right)-2\left(x^2-1\right)\)
\(\Leftrightarrow x^2-4x+3=2x^2-6x-2x^2+2=-6x+2\)
\(\Leftrightarrow x^2+2x+1=0\)
=>x=-1(nhận)
\(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)
\(\Leftrightarrow x=0\) ( vì \(x^2+2x+5>0;\forall x\)
b.\(\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)
\(ĐK:x\ne1;3;4\)
\(\Leftrightarrow\dfrac{x}{\left(x-1\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)
\(\Leftrightarrow\dfrac{x\left(x-3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-1\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)\)
\(\Leftrightarrow x^2-3x-\left(x^2-3x-x+3\right)=x^2-1\)
\(\Leftrightarrow x^2-3x-x^2+4x-3=x^2-1\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-2\right\}\)
\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)
\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2x^2-2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-6x-x^2+4x-3-2x^2+2}{2\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow-x^2-2x-1=0\)
\(\Leftrightarrow x^2+2x+1=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\left(tm\right)\)
Câu 5: Giải phương trình:
a. \(x\)\(\sqrt{3}\) - \(\sqrt{3}\) = \(1-x\)
b. \(7-\sqrt{x^2-6x+9}=0\)
c. \(\sqrt{9\left(x-2\right)^2}\) - 45 = 0
a) \(\Leftrightarrow\sqrt{3}\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\sqrt{3}-1\right)=0\Leftrightarrow x=1\)
b) \(\Leftrightarrow\sqrt{\left(x-3\right)^2}=7\)
\(\Leftrightarrow\left|x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
c) \(\Leftrightarrow3\left|x-2\right|=45\)
\(\Leftrightarrow\left|x-2\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=15\\x-2=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-13\end{matrix}\right.\)
\(a,PT\Leftrightarrow\sqrt{3}\left(x-1\right)=1-x\\ \Leftrightarrow\sqrt{3}\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(\sqrt{3}+1\right)=0\\ \Leftrightarrow x=1\left(\sqrt{3}+1\ne0\right)\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}x-3=7\\3-x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\\ c,ĐK:x\in R\\ PT\Leftrightarrow3\left|x-2\right|=45\Leftrightarrow\left|x-2\right|=15\\ \Leftrightarrow\left[{}\begin{matrix}x-2=15\\2-x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-13\end{matrix}\right.\)
Giải các bất phương trình:
a,\(\frac{x+1}{X-1}\)>0
b,\(\frac{x^2+x-2}{x-9}\)<0
a.
\(\dfrac{x+1}{x-1}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
b.
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)}{x-9}< 0\Rightarrow\left[{}\begin{matrix}x< -2\\1< x< 9\end{matrix}\right.\)