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3 tháng 8 2020
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có :
\(A=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{x-1}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+1}\)
\(=1\)
Vậy...
b/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có :
\(B=\left(\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}+6\right)\left(\frac{x\sqrt{x}-1}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}+6\right)\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\sqrt{x}-2+6\right)\left(\sqrt{x}-1-3\right)\)
\(=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)\)
\(=x-16\)
Vậy..
c/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có :
\(C=\frac{2\sqrt{x}}{x-1}+\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}-x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2}{\sqrt{x}}\)
Vậy..
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
5 tháng 9 2018
Chậc :))) T còn cách khác đây =)))
\(\sqrt{x-1+2\sqrt{x-2}}-\sqrt{x-1-2\sqrt{x-2}}=1\)
\(\Leftrightarrow\left(\sqrt{x-1+2\sqrt{x-1}}\right)^2=\left(1+\sqrt{x-1-2\sqrt{x-2}}\right)^2\)
\(\Leftrightarrow x-1+2\sqrt{x-2}-x=2\sqrt{x-1-2\sqrt{x-2}}+x-2\sqrt{x-2}-x\)
\(\Leftrightarrow2\sqrt{x-2}-1=2\sqrt{x-1-2\sqrt{x-2}}-2\sqrt{x-2}\)
\(\Leftrightarrow4x-4\sqrt{x-2}-7=-8\sqrt{x-2}-8\sqrt{x-2}.\sqrt{x-2\sqrt{x-2}-1}+8x-12\)
\(\Leftrightarrow5-4\sqrt{x-2}-4x=-8\sqrt{x-2}-8\sqrt{x-2}.\sqrt{x-2\sqrt{x-2}-1}\)
\(\Leftrightarrow x=\frac{9}{4}\) (tmyk)
2 tháng 8 2017
ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có \(A=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{x-\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\frac{\sqrt{x}+2+x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\frac{x-1-x+\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+3}=\frac{x+3}{\sqrt{x}+3}\)
30 tháng 7 2020
mình sẽ xóa câu này mong bạn gửi lại câu hỏi khác để rõ ràng cho các bạn khác tham khảo nha
\(\sqrt{1-\sqrt{x^4-x^2}}=x-1\)
\(\Leftrightarrow1-\sqrt{x^4-x^2}=\left(x-1\right)^2\)
\(\Leftrightarrow-\sqrt{x^4-x^2}=x^2-2x+1-1\)
\(\Leftrightarrow x^4-x^2=\left(x^2-2x\right)^2\)
\(\Leftrightarrow x^4-x^2=x^4-4x^3+4x^2\)
\(\Leftrightarrow4x^3-5x^2=0\)
\(\Leftrightarrow x^2\left(4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\4x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{4}\end{matrix}\right.\)