cho biểu thức \(P=\left(\frac{\sqrt{x}}{\sqrt{x^3}-1}+\frac{1}{\sqrt{x}-1}\right)\)
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17 tháng 10 2018
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
Chúc bạn học tốt ~
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
27 tháng 1 2020
;))) tớ nhớ dạng RGBT căn bậc 3 lớp 9 nhì :)))????
\(\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{\sqrt{x}}{x+\sqrt{x+1}}\right).\left(\frac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\frac{2x+1-\sqrt{x}\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left[\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right]\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x+1}\right)}.\left(1-2\sqrt{x}+x\right)\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
Ta có : \(\sqrt{x^3}-1=x\sqrt{x}-1\)
\(P=\left(\frac{\sqrt{x}}{x\sqrt{x}-1}+\frac{1}{\sqrt{x}-1}\right)\)ĐK : \(x\ne1\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}+\frac{x\sqrt{x}-1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-\sqrt{x}+x\sqrt{x}-1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}=\frac{\left(x-1\right)+\sqrt{x}\left(x-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}=\frac{x+2\sqrt{x}+1}{x\sqrt{x}-1}\)
Nếu muốn làm nhanh thì để ý \(\sqrt{x^3}-1=\sqrt{x^3}-\sqrt{1}\)là HĐT số 3
nhưng áp dụng cách đó khá dễ sai nên mình dùng cách quy đồng nhé !
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\frac{\sqrt{x}}{\sqrt{x^3}-1}+\frac{1}{\sqrt{x}-1}=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)