giải phương trình \( \sqrt{ - { x }^{ 2 } +6x-9 \phantom{\tiny{!}}} + { x }^{ 3 } = 27 \)
\(\sqrt{ { \left( x-3 \right) }^{ 2 } \left( 5-3x \right) \phantom{\tiny{!}}} +2x= \sqrt{ 3x-5+4 \phantom{\tiny{!}}} \)
Những câu hỏi liên quan
\left( \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} -1 } - \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} } \right) \left( \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +1 }{ \sqrt{ x \phantom{\tiny{!}}} -2 } - \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +2 }{ \sqrt{ x \phantom{\tiny{!}}} -1 } \right)
Rút Gọn
Tìm x đễ biêu thức âm
\( \left( \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} -1 } - \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} } \right) \left( \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +1 }{ \sqrt{ x \phantom{\tiny{!}}} -2 } - \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +2 }{ \sqrt{ x \phantom{\tiny{!}}} -1 } \right) \)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(\dfrac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
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Rút gọn: A=\( \dfrac{ x \sqrt{ x \phantom{\tiny{!}}} -8 }{ x+2 \sqrt{ x \phantom{\tiny{!}}} +4 } + \dfrac{ x \sqrt{ x \phantom{\tiny{!}}} +27 }{ x-3 \sqrt{ x \phantom{\tiny{!}}} +9 } \) với x lớn hơn hoặc bằng 0
\(A=\frac{x\sqrt{x}-8}{x+2\sqrt{x}+4}+\frac{x\sqrt{x}+27}{x-3\sqrt{x}+9}\) \(=\frac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{x+2\sqrt{x}+4}+\frac{\left(\sqrt{x}+3\right)\left(x-3\sqrt{x+9}\right)}{x-3\sqrt{x}+9}\) \(=\sqrt{x}-2+\sqrt{x}+3=2\sqrt{x}+1\)
\( \displaystyle\lim_{ x \rightarrow 0 } \left( \dfrac{ \sqrt[ 3 ]{ x+1 \phantom{\tiny{!}}} - \sqrt{ 1-x \phantom{\tiny{!}}} }{ x } \right) \)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-1+1-\sqrt[]{1-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{x}{1+\sqrt[]{1-x}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[3]{\left(x+1\right)^3}+\sqrt[3]{x+1}+1}+\dfrac{1}{1+\sqrt[]{1-x}}\right)=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)
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\dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} -3 } + \dfrac{ 4 }{ \sqrt{ x \phantom{\tiny{!}}} +3 } - \dfrac{ 9- \sqrt{ x \phantom{\tiny{!}}} }{ x-9 }
\( \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} -3 } + \dfrac{ 4 }{ \sqrt{ x \phantom{\tiny{!}}} +3 } - \dfrac{ 9- \sqrt{ x \phantom{\tiny{!}}} }{ x-9 } \)(ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\))
\(=\dfrac{1}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}+3}+\dfrac{\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+3+4\left(\sqrt{x}-3\right)+\sqrt{x}-9}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}-6+4\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{6\sqrt{x}-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=6\cdot\dfrac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{6}{\sqrt{x}+3}\)
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\dfrac{ \sqrt{ y \phantom{\tiny{!}}} -2 }{ \sqrt{ y \phantom{\tiny{!}}} -3 } \times ( \dfrac{ \sqrt{ y \phantom{\tiny{!}}} }{ \sqrt{ y \phantom{\tiny{!}}} -3 } + \dfrac{ 6 \sqrt{ y \phantom{\tiny{!}}} }{ 9-y } - \dfrac{ 3 }{ \sqrt{ y \phantom{\tiny{!}}} +3 } )
Rút gọn
C= \( \dfrac{ \dfrac{ x- \sqrt{ x \phantom{\tiny{!}}} }{ \sqrt{ x \phantom{\tiny{!}}} -1 } - \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +1 }{ x+ \sqrt{ x \phantom{\tiny{!}}} } }{ \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +1 }{ x } } \)
Lời giải:
Xét tử thức:
\(\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}}=\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}=\sqrt{x}-\frac{1}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)
\(\Rightarrow C=\frac{x-1}{\sqrt{x}}: \frac{\sqrt{x}+1}{x}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{x}{\sqrt{x}+1}=\sqrt{x}(\sqrt{x}-1)\)
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\sqrt{ 5+5 \sqrt[ 4 ]{ 5 \phantom{\tiny{!}}} +3 \times \sqrt[ 4 ]{ 25 \phantom{\tiny{!}}} + \sqrt[ 4 ]{ 125 \phantom{\tiny{!}}} }
Cho biểu thức A =\(\dfrac{ \dfrac{ 1 }{ \sqrt{ 1+x \phantom{\tiny{!}}} } + \sqrt{ 1-x \phantom{\tiny{!}}} }{ \dfrac{ 1 }{ \sqrt{ 1- { x }^{ 2 } \phantom{\tiny{!}}} } +1 } \)
a)Tìm x để A có nghĩa
b) Rút gon A .
c) Tính a với \(x= \dfrac{ \sqrt{ 3 \phantom{\tiny{!}}} }{ 2+ \sqrt{ 3 \phantom{\tiny{!}}} } \)
\sqrt[ 3 ]{ x-14 \phantom{\tiny{!}}} -\sqrt{ x-1 \phantom{\tiny{!}}} = 3
\(\sqrt[ 3 ]{ x-14 \phantom{\tiny{!}}} -\sqrt{ x-1 \phantom{\tiny{!}}} = 3\)
ý bạn là cái này hả :)?
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