Đề dọa trẻ con à
\(\sqrt{x^2\sqrt{x^4\sqrt{x^8\sqrt{x^{16}}}}}=\sqrt{3^{14}}\)
\(\Leftrightarrow\sqrt{x^2\sqrt{x^4\sqrt{x^8\cdot x^8}}}=\sqrt{3^{14}}\)
\(\Leftrightarrow\sqrt{x^2\sqrt{x^4\cdot x^8}}=\sqrt{3^{14}}\)
\(\Leftrightarrow\sqrt{x^2\cdot x^6}=\sqrt{3^{14}}\)
\(\Leftrightarrow x^4=2187\)\(\Rightarrow x=\pm3\sqrt[4]{27}\)
Tìm x biết:
a)\(\sqrt{x}=4\)
b)\(\sqrt{x-2}=3\)
c)\(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\)
d)\(x^2=7v\text{ới}x< 0\)
e)\(x^2-4=0v\text{ới}x>0\)
f)\(\left(2x+7\sqrt{7}\right)^2=7\)
Tìm x biết :a) \(\sqrt{x}\) = \(x\)
b) \(x-2\sqrt{x}=0\)
c) \(\sqrt{x+1}=1-x\)
d) \(\sqrt{3x^2+4}+\sqrt{2004x^2+1}=3-4x^2\)
e) \(\sqrt{3x^2+4}+\sqrt{2007x^2+25}=7-69x^2\)
Tìm x
\(\sqrt{1+\sqrt{1+\sqrt{x-16}}}\) = 2
Bài 1 : Kí hiệu [x] là số nguyên lớn nhất không vượt qua x. Tìm [x] biết :
a) x = \(\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2}}}}\) ( n dấu căn )
b) x = \(\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+...+\left[\sqrt{100}\right]\)
Bài 2 : Tìm x để A có giá trị nguyên:
a) A = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) A = \(\dfrac{\sqrt{x}+3}{2\sqrt{x}+1}\)
c) A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x\) thuộc Z
Tìm x bt
a) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
b)\(2\sqrt{3x}+11x-18=5x 3+6\sqrt{3x}-21\)
Tìm x biết :
\(3x\sqrt{x+1}=40\)
\(\sqrt{x+1}+2=0\)
\(\sqrt{\left(x+1\right)^2}=3\)
\(\sqrt{x-3}=4\)
x-4\(\sqrt[]{\text{x}}=0\)
(x mũ 2 - 6)(\(\sqrt[]{\text{x}}\)+3)=0
Tìm x, biết:
\(\left(x^2-8\right)\left(\sqrt{x}-5\right)=0\)
