Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
giải pt
1. \(\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt[]{x}}=1\)
2.\(\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}=\dfrac{7}{\sqrt{x-3}}\)
3.\(\sqrt{14-x}-\sqrt{x-4}\sqrt{x-1}\)
4. \(3+\sqrt{x+2\sqrt{x-1}}=2\sqrt{x-2\sqrt{x-1}}\)
tính giá trị biểu thức
1) A frac{15sqrt{x}-11}{x-2sqrt{x}-3}+frac{3sqrt{x}-2}{1-sqrt{x}}-frac{2sqrt{x}+3}{sqrt{x}+3} tại x3-2sqrt{2}
2) Bfrac{x+2}{xsqrt{x}-1}+frac{sqrt{x}+1}{x+sqrt{x}+1}-frac{sqrt{x}+1}{x-1} tại x7-2sqrt{6}
3) Cleft(frac{sqrt{x}}{3+sqrt{x}}+frac{x+9}{9-x}right):left(frac{3sqrt{x}+1}{x-3sqrt{x}}-frac{1}{sqrt{x}}right) tại x7-4sqrt{3}
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tính giá trị biểu thức
1) A = \(\frac{15\sqrt{x}-11}{x-2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\) tại \(x=3-2\sqrt{2}\)
2) \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\) tại \(x=7-2\sqrt{6}\)
3) \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) tại \(x=7-4\sqrt{3}\)
rút gọn
dfrac{9-x}{sqrt{x}+3}-dfrac{9-6sqrt{x}+x}{sqrt{x}-3}-6 (với x_9)
left(dfrac{2sqrt{x}}{xsqrt{x}+x+sqrt{x}+1}-dfrac{1}{sqrt{x}+1}right)/left(dfrac{2sqrt{x}}{sqrt{x}+1}-1right) (với x0, x#1)
sqrt{x+12+6sqrt{x+3}}-sqrt{x+12-6sqrt{x+3}} ( với x_6)
sqrt{m^2+6m+9}+sqrt{m^2-6m+9} (m bát kì)
dfrac{xsqrt{x}-1}{x-sqrt{x}}-dfrac{xsqrt{x}+1}{x+sqrt{x}}dfrac{x+1}{sqrt{x}}
dfrac{xsqrt{y}+ysqrt{x}}{sqrt{xy}}/dfrac{sqrt{x}-sqrt{y}}{x-y}
left(dfrac{1}{sqrt{x}-1}+dfrac{1}{sqrt{x}+1}right)left...
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rút gọn
\(\dfrac{9-x}{\sqrt{x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6\) (với x>_9)
\(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)/\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\) (với x>=0, x#1)
\(\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}\) ( với x>_6)
\(\sqrt{m^2+6m+9}+\sqrt{m^2-6m+9}\) (m bát kì)
\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\dfrac{x+1}{\sqrt{x}}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}/\dfrac{\sqrt{x}-\sqrt{y}}{x-y}\)
\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(\left(\dfrac{\sqrt{x}+2}{3\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}-3\right)/\dfrac{2-4\sqrt{x}}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}\)
18 tháng 10 2021 lúc 15:16
\(Q=\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\right)+\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\)
Rút gọn Q với x>0, x≠0, x≠9
\(Q=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+x}+\frac{8\sqrt{x}}{9x-1}\right)\div\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
a)\(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)
b)\(\sqrt[3]{x+1}+\sqrt[3]{x^2}+\sqrt[3]{x}+\sqrt[3]{x^2+x}\)
c)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\)
giải phương trình :a,\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3-4\sqrt{x-1}}=1\)
b,\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
c,\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
d, \(3+\sqrt{x+2\sqrt{x-1}}=2\sqrt{x-2\sqrt{x-1}}\)
\(A=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
Giải pt:
a) xsqrt{1-dfrac{1}{x}}+sqrt{x-dfrac{1}{x}}
b) sqrt{x^2+x}+sqrt{x-x^2}x+1
c) sqrt{x^2-x}+sqrt{x^2+2x}2sqrt{x^2}
d)sqrt{dfrac{x^3+1}{x+3}}+sqrt{x+1}sqrt{x^2-x+1}+sqrt{x+3}
e) sqrt{sqrt{3}-x}xsqrt{sqrt{3}+x}
f) 4xsqrt{x+7}+3xsqrt{7x-3}6x^2+2sqrt{7x^2+46x-21}
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Giải pt:
a) x=\(\sqrt{1-\dfrac{1}{x}}+\sqrt{x-\dfrac{1}{x}}\)
b) \(\sqrt{x^2+x}+\sqrt{x-x^2}=x+1\)
c) \(\sqrt{x^2-x}+\sqrt{x^2+2x}=2\sqrt{x^2}\)
d)\(\sqrt{\dfrac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
e) \(\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}\)
f) \(4x\sqrt{x+7}+3x\sqrt{7x-3}=6x^2+2\sqrt{7x^2+46x-21}\)
giải pt: sqrt{x+3-4sqrt{x-1}}+sqrt{x+8-6sqrt{x-1}}1
làm thế này mà chả hiểu sao lại bị gạch, ai biết chỉ với, cảm ơn nak:
+ ĐK:left{{}begin{matrix}xge1x+3-4sqrt{x-1}ge0x+8-6sqrt{x-1}ge0end{matrix}right. Leftrightarrow xge1
+ pt đã cho Leftrightarrowsqrt{left(sqrt{x-1}-2right)^2}+sqrt{left(sqrt{x-1}-3right)^2}1
Leftrightarrowleft|sqrt{x-1}-2right|+left|sqrt{x-1}-3right|1 (*)
Th1: left{{}begin{matrix}sqrt{x-1}-2 0sqrt{x-1}-3 0end{matrix}right.
(*) Leftrightarrow2-sqrt{x-1}+3-sqrt{x-1}1Left...
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giải pt: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
làm thế này mà chả hiểu sao lại bị gạch, ai biết chỉ với, cảm ơn nak:
+ ĐK:\(\left\{{}\begin{matrix}x\ge1\\x+3-4\sqrt{x-1}\ge0\\x+8-6\sqrt{x-1}\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)
+ pt đã cho \(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\) (*)
Th1: \(\left\{{}\begin{matrix}\sqrt{x-1}-2< 0\\\sqrt{x-1}-3< 0\end{matrix}\right.\)
(*) \(\Leftrightarrow2-\sqrt{x-1}+3-\sqrt{x-1}=1\Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\Leftrightarrow x=5\left(N\right)\)
Th2: \(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\ge0\end{matrix}\right.\)
(*) \(\Leftrightarrow\sqrt{x-1}-2+\sqrt{x-1}-3=1\Leftrightarrow2\sqrt{x-1}=6\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x=10\left(N\right)\)
Th3: \(\sqrt{x-1}-3< 0\le\sqrt{x-1}-2\)
(*) \(\Leftrightarrow\sqrt{x-1}-2+3-\sqrt{x-1}=1\Leftrightarrow1=1\left(đúng\right)\)
Kl: \(x\ge1\)