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\(\frac{x^2}{\sqrt{5}}-2\sqrt{5}=0\\ \frac{x^2}{\sqrt{5}}=2\sqrt{5}\\ \frac{x^2\sqrt{5}}{\sqrt{5}}=2\sqrt{5}.\sqrt{5}\\ x^2=10\\ x=+-\sqrt{10}\)
2)\(\sqrt{25\left(2x+1\right)^2}=0\\ 5\left(2x+1\right)=0\\ x=\frac{-1}{2}\)
Thiên Thư mk cx hk lp 7 nek
a\ \(\sqrt{x^2-4x+4}=6\)
\(x^2-4x+4=6^2=36\)
\(x\left(x-4\right)=32\)
ta có \(32=8.4=\left(-8\right)\left(-4\right)\)
\(\Rightarrow x\in\left\{8;-4\right\}\)
b)\(\sqrt{2x+5}=2x-1\)
\(2x+4=4x^2-4x\)
\(2\left(x+2\right)=4x\left(4x-1\right)\)
\(........................\)
e bí mất r a ạ
1.
\(x+4\sqrt{x}+3=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow x+\sqrt{x}+3\sqrt{x}+3=0\\ \Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=0\\ \Rightarrow x\in\varnothing\)
2.
\(x^2+3x\sqrt{x}+2x=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow x^2+x\sqrt{x}+2x\sqrt{x}+2x=0\\ \Leftrightarrow x\sqrt{x}\left(\sqrt{x}+1\right)+2x\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\)
3.
\(x+2\sqrt{x}-8=0\\ \Leftrightarrow x-2\sqrt{x}+4\sqrt{x}-8=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)+4\left(\sqrt{x}-2\right)=0\\ \Leftrightarrow\left(\sqrt{x}+4\right)\left(\sqrt{x}-2\right)=0\\ \Leftrightarrow\sqrt{x}-2=0\\ \Leftrightarrow x=4\)
4.
\(x+\sqrt{9x}-\sqrt{100}=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow x+3\sqrt{x}-10=0\\ \Leftrightarrow x+5\sqrt{x}-2\sqrt{x}-10=0\\ \Leftrightarrow\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)=0\\ \Leftrightarrow\sqrt{x}-2=0\\ \Leftrightarrow x=4\)
5.
\(x+\sqrt{3x}-\sqrt{2x}-\sqrt{6}=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+\sqrt{3}\right)-\sqrt{2}\left(\sqrt{x}+\sqrt{3}\right)=0\\ \Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-\sqrt{2}\right)=0\\ \Leftrightarrow\sqrt{x}-\sqrt{2}=0\Leftrightarrow x=2\)
6.
\(\sqrt{5x}-x-\sqrt{15}+\sqrt{3x}=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow\sqrt{x}\left(\sqrt{5}-\sqrt{x}\right)-\sqrt{3}\left(\sqrt{5}-\sqrt{x}\right)=0\\ \Leftrightarrow\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\sqrt{3}=0\\\sqrt{5}-\sqrt{x}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
Ta có: \(x^4+16x^2+32=0\Leftrightarrow\left(x^2-8\right)^2-32=0\left(1\right)\)
Với \(x=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)\(\Leftrightarrow x=\sqrt{3}\sqrt{2-\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)
\(\Rightarrow x^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3}\sqrt{2-\sqrt{3}}\)
Thay x vào vế phải của (1) ta được:
\(\left(x^2-8\right)^2-32=\left(8-2\sqrt{2+\sqrt{3}}-2\sqrt{3}\sqrt{2-\sqrt{3}}-8\right)^2-32\)
\(=4\left(2+\sqrt{3}\right)+4\sqrt{3}+12\left(2-\sqrt{3}\right)-32\)
\(=8+4\sqrt{3}+8\sqrt{3}+24-12\sqrt{3}-32=0\)= vế phải
Vậy \(x-\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)là 1 nghiệm của phương trình đã cho(đpcm)
a) \(\sqrt{x^2-6x+9}+x=11\)
\(\Rightarrow\sqrt{\left(x-3\right)^2}+x=11\)
\(\Rightarrow x-3+x=11\)
\(\Rightarrow2x=14\Rightarrow x=7\)
Vậy........
b) \(\sqrt{3x^2-4x+3}=1-2x\)
\(3x^2-4x+3=1-4x+4x^2\)
\(3x^2-4x^2-4x+4x=-2\)
\(-x^2=-2\)
\(2=x^2\Rightarrow\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
Vậy.........
d) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
\(\Rightarrow2x-1=x-3\)
\(\Rightarrow x=1-3\)
\(\Rightarrow x=-2\)
Vậy x=-2
a)A<=>\(\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2\right)^2}\)=0(đk -2<=x)
<=>\(\sqrt{x+2}\left(1+\sqrt{x+2}\right)\)=0
vì 1+\(\sqrt{x+2}\) >=1 nên để A=0 thì \(\sqrt{x+2}\)=0
=>x+2=0
=>x=-2
a/ \(\sqrt{4x^2}=6\Rightarrow\left|2x\right|=6\Rightarrow\orbr{\begin{cases}2x=6\\2x=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}}\)
b/ \(x^2-2\sqrt{11}x+11=0\Rightarrow\left(x-\sqrt{11}\right)^2=0\Rightarrow x=\sqrt{11}\)
c/ \(\sqrt{16x}=8\Rightarrow4\sqrt{x}=8\Rightarrow\sqrt{x}=2\Rightarrow x=4\) (ĐKXĐ : x>=0)