\(\sqrt{5x^2-20x+20}=2\sqrt{5}\\ \Leftrightarrow5x^2-20x+20=20\\ \Leftrightarrow5x^2-20x=0\\5x\left(x-4\right)=0\\ \Leftrightarrow \left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Gải phương trình \(\sqrt{4x+2}+\sqrt{x^2+5x+6}=\sqrt{5x^2+20x+15}\)
a)\(\sqrt{4-5x}=12\) tìm x
b)\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
c)\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(\sqrt{x^2+x+1}=x+1\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{x^2-2x+1}=4\)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)
cho f(x) = \(\sqrt{5x^2+20}+\sqrt{5x^2-32x+64}+\sqrt{5x^2-40x+100}+\sqrt{5x^2-8x+16}\) Tìm giá trị nhỏ nhất của f(x)
rút gọn Q= ($\frac{\sqrt{x+2} }{x-2\sqrt{x}+4 }$ - $\frac{x-\sqrt{x} }{x\sqrt{x} +8 }$ ). $\frac{5x-10\sqrt{x}+20 }{5\sqrt{x}+4}$
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
b)\(\sqrt{4x-20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
c) \(\sqrt{\dfrac{3x-2}{x+1}}=3\)
d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\)
\(\sqrt{5x^2-14x+9}-\sqrt{x^2-x-20}=5\sqrt{x-1}\)
a)
\(\sqrt{\left(5x+2\right)^2}\)=3
b)
\(\sqrt{4x+20}\)+3\(\sqrt{5+x}\)-4=6
