49.(2x+1)^2:25-49=0
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a: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)
=>\(5-\sqrt{2x-1}=0\)
=>\(\sqrt{2x-1}=5\)
=>2x-1=25
=>2x=26
=>x=13(nhận)
c: \(\sqrt{x^2-10x+25}=2\)
=>\(\sqrt{\left(x-5\right)^2}=2\)
=>\(\left|x-5\right|=2\)
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
d: \(\sqrt{x^2-14x+49}-5=0\)
=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)
=>\(\sqrt{\left(x-7\right)^2}=5\)
=>|x-7|=5
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
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https://www.youtube.com/channel/UCT23clmdY5azigRNMRDxGfw
a) \(\left(x^2+5\right).\left(x^2-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+5=0\\x^2-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-5\left(vl\right)\\x^2=25\end{cases}\Rightarrow}\orbr{\begin{cases}\\x=\pm5\end{cases}}}\)
b) \(\left(x^2-5\right)\left(x^2-25\right)< 0\)
\(\Rightarrow\left(x^2-5\right)\)và \(\left(x^2-25\right)\)trái dấu
Vì \(\left(x^2-5\right)>\left(x^2-25\right)\)
\(\Rightarrow\hept{\begin{cases}x^2-5>0\\x^2-25< 25\end{cases}\Rightarrow\hept{\begin{cases}x^2>5\\x^2< 50\end{cases}}}\)
\(\Rightarrow5< x^2< 50\)
\(\Rightarrow x^2\in\left\{0;1;4;9;16;25;36;49\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5;\pm6;\pm7\right\}\)
c) \(\left(x-2\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
các câu còn lại lm tương tự nhé!! hok tốt!!
1) \(...\Rightarrow x-7=0\Rightarrow x=7\)
2) \(...\Rightarrow x-4=0\Rightarrow x=4\)
3) \(...\Rightarrow6x-12=0\Rightarrow6x=12\Rightarrow x=12:6=2\)
4) \(...\Rightarrow9x-27=0\Rightarrow9x=27\Rightarrow x=27:9=3\)
5) \(...\Rightarrow15-x=30-25\Rightarrow15-x=5\Rightarrow x=15-5=10\)
6) \(...\Rightarrow43-24+x=20\Rightarrow19+x=20\Rightarrow x=20-19=1\)
7) \(...\Rightarrow2x+14-17=25\Rightarrow2x-3=25\Rightarrow2x=28\Rightarrow x=28:2=14\)
8) \(...\Rightarrow3x+21-15=27\Rightarrow3x-6=27\Rightarrow3x=33\Rightarrow x=33:3=11\)
9) \(...\Rightarrow15+4x-8=95\Rightarrow4x+7=95\Rightarrow4x=88\Rightarrow x=88:4=22\)
10) \(...\Rightarrow20-x-14=5\Rightarrow6-x=5\Rightarrow x=6-5=1\)
11) \(...\Rightarrow24+15-3x=27\Rightarrow39-3x=27\Rightarrow3x=39-27\Rightarrow3x=12\Rightarrow x=12:3=4\)
\(\left(2x-1\right)^2-49-4x^2=0.\Leftrightarrow4x^2-4x+1-49-4x^2=0.\Leftrightarrow-4x=48.\Leftrightarrow x=-12.\)
\(\Leftrightarrow4x^2-4x+1-49-4x^2=0\)
=>4x+48=0
hay x=-12