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1) \(x-2\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
2) \(x=\sqrt{x}\Rightarrow x-\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
3) \(2x+5\sqrt{x}=0\Rightarrow\sqrt{x}\left(2\sqrt{x}+5\right)=0\Rightarrow\sqrt{x}=0\)(Vì \(\sqrt{x}\ge0\Rightarrow2\sqrt{x}+5>0\))\(\Rightarrow x=0\)
a) \(2\sqrt{x}+3=0\)
\(2\sqrt{x}=-3\)
\(\sqrt{x}=\frac{-3}{2}\)
\(x=\frac{9}{4}\)
vậy \(x=\frac{9}{4}\)
b) \(\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
\(\frac{5}{12}\sqrt{x}=\frac{1}{3}+\frac{1}{6}\)
\(\frac{5}{12}\sqrt{x}=\frac{1}{2}\)
\(\sqrt{x}=\frac{1}{2}:\frac{5}{12}\)
\(\sqrt{x}=\frac{6}{5}\)
\(x=\frac{36}{25}\)
vậy \(x=\frac{36}{25}\)
c) \(\sqrt{x+3}+3=0\)
\(\sqrt{x+3}=-3\)
\(\Rightarrow x\in\varnothing\) vì ko thỏa mãn ĐKXĐ của căn thức \(x\ge0\)
hay nói khác đi căn thức \(\sqrt{x+3}\) ko có nghĩa
vậy \(x\in\varnothing\)
\(a,ĐK:x\ge-2\)
\(\sqrt{x+2}=3\)
\(\Leftrightarrow x+2=9\Rightarrow x=7\left(Tm\right)\)
\(b,\sqrt{x^2+3}=\sqrt{7}\)
\(\Leftrightarrow x^2+3=7\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
\(c,\sqrt{x}=0\Rightarrow x=0\)
\(d,\sqrt{x}=-3\)
Vì \(\sqrt{x}\ge0;-3< 0\)=> pt vô nghiệm
\(e,3\sqrt{x}=1\)
\(\Rightarrow\sqrt{x}=\frac{1}{3}\Rightarrow x=\frac{1}{9}\)
\(g,4-5\sqrt{x}=-1\)
\(\Rightarrow5\sqrt{x}=5\)
\(\Rightarrow\sqrt{x}=1\Rightarrow x=1\)
a,\(\sqrt{x+2}=3\Leftrightarrow x+2=3^2\Leftrightarrow x=9-2=7\)
b,\(\sqrt{x^2+3}=\sqrt{7}\Leftrightarrow x^2+3=7\Leftrightarrow x^2=4\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
c,\(\sqrt{x}=0\Leftrightarrow x=0\)
d,\(\sqrt{x}=-3\Leftrightarrow x=\left(-3\right)^2\Leftrightarrow x=9\)
e,g tương tự các câu trên bạn tự làm ik mk mỏi tay lắm r
\(a,2\sqrt{x}+3=0\)
\(\Leftrightarrow2\sqrt{x}=-3\)
\(\Leftrightarrow\sqrt{x}=-\frac{3}{2}\)( loại )
\(b,\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow\frac{5}{12}\sqrt{x}=\frac{1}{2}\Leftrightarrow\sqrt{x}=\frac{6}{5}\Leftrightarrow x=\frac{36}{25}\)
\(c,\sqrt{x+3}+3=0\Leftrightarrow\sqrt{x+3}=-3\)( loại )
\(a.\)
\(x-3\sqrt{x}=0\)
\(\Rightarrow\left(\sqrt{x}\right)^2-3\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-3\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\\sqrt{x}-3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\\sqrt{x}=3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)
Vậy : \(x\in\left\{0;9\right\}\)
1. a) x^2=16=>x=+_4
b)x^2=36=>x=+_6
c)x^2=49=>x=+_7
d) x-1=+_5
+) x-1=5
=>x=6
+)x-1=-5
=>x=-4
e) (x+3)^2=-1( vô lý)
ko cs gtri của x
f) (2x+7)^2=36=>2x+7=+_6
+) 2x+7=6
x=-1/2
+) 2x+7=-6
=>x=-13/2
a) \(\sqrt{16}x+\frac{3}{4}=2\sqrt{\frac{4}{25}}+0,01.\sqrt{100}\)
=> \(4x+\frac{3}{4}=2\cdot\frac{2}{5}+0,01\cdot10\)
=> \(4x+\frac{3}{4}=\frac{4}{5}+0,1\)
=> \(4x+\frac{3}{4}=0,9\)
=> \(4x=0,9-\frac{3}{4}\)
=> \(4x=0,15\)
=> \(x=0,15:4=0,0375\)
b) \(\left(x-\frac{2}{5}\right)\left(x+\frac{3}{7}\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{2}{5}=0\\x+\frac{3}{7}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{3}{7}\end{cases}}\)
a) \(x-2\sqrt{x}=0\)
\(\Rightarrow x=2\sqrt{x}\)\(\Rightarrow x^2=4x\)\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow x=0\)hoặc \(x=4\)
Vậy \(x=0\)hoặc \(x=4\)
b) \(x=\sqrt{x}\)\(\Rightarrow x^2=x\)\(\Rightarrow x\left(x-1\right)=0\)
\(\Rightarrow\)\(x=0\)hoặc \(x=1\)
Vậy \(x=0\)hoặc \(x=1\)
\(b,\text{ }x=\sqrt{x}\)
\(x^2=x\)
\(x^2-x=0\)
\(x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=0+1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }1\right\}\)
đk x>=0
<=>\(x=\sqrt[5]{x}\)
<=>\(x^5=x\)
<=>x(x^4-1)=0
<=>\(\orbr{\begin{cases}x=0\\x^4=1\end{cases}}\)