Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
Tính DKXD của các căn bậc thức sau:
a)\(\sqrt{2x-4}\)
b)\(\sqrt{\dfrac{3}{-2x+1}}\)
c)\(\sqrt{\dfrac{-3x+5}{-4}}\)
d)\(\sqrt{-5\left(-2x+6\right)}\)
e)\(\sqrt{\left(x^2+2\right)\left(x-3\right)}\)
f)\(\sqrt{\dfrac{x^2+5}{-x+2}}\)
Tìm x
a)\(\sqrt{x-1}=2\left(x\ge1\right)\)
b)\(\sqrt{3-x}=4\left(x\le3\right)\)
c)\(2.\sqrt{3-2x}=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\)
d)\(4-\sqrt{x-1}=\dfrac{1}{2}\left(x\ge1\right)\)
e)\(\sqrt{x-1}-3=1\)
f)\(\dfrac{1}{2}-2.\sqrt{x+2}=\dfrac{1}{4}\)
Bài 1: Rút gọn: A left(dfrac{x}{x^2-49}-dfrac{x-7}{x^2+7x}right):dfrac{2x-7}{x^2+7x}+dfrac{x}{7-x}
Bài 2: Rút gọn: Bleft[dfrac{3}{x+1}+left(dfrac{3}{x}-dfrac{x}{x^2+2x+1}right):dfrac{2x^2+3x}{x+1}right]:dfrac{1+3x}{x^2+x}
Bài 3: Rút gọn Dleft(sqrt{a}+dfrac{b-sqrt{ab}}{sqrt{a}+b}right):left(dfrac{a}{sqrt{ab}+b}+dfrac{b}{sqrt{ab}-a}-dfrac{a+b}{sqrt{ab}}right)
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Bài 1: Rút gọn: A= \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
Bài 2: Rút gọn: B=\(\left[\dfrac{3}{x+1}+\left(\dfrac{3}{x}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}\right]:\dfrac{1+3x}{x^2+x}\)
Bài 3: Rút gọn D=\(\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+b}\right):\left(\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\)
Tìm x;yin Ntmãn : sqrt{x}+sqrt{y}sqrt{2012}
2, Rút gọn bt
Pdfrac{x}{x-sqrt{x}}+dfrac{2}{x+2sqrt{x}}+dfrac{x+2}{left(sqrt{x}-1right)left(x+2sqrt{x}right)}
b, gpt : x^2-2x-xsqrt{x}-2sqrt{x}+40
3, cho x1 ; y0 , cm
dfrac{1}{left(x+1right)^3}+left(dfrac{x-1}{y}right)^3+dfrac{1}{y^3}ge3left(dfrac{3-2x}{x-1}+dfrac{x}{y}right)
Unruly Kid
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Tìm \(x;y\in N\)tmãn : \(\sqrt{x}+\sqrt{y}=\sqrt{2012}\)
2, Rút gọn bt
\(P=\dfrac{x}{x-\sqrt{x}}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)
b, gpt : \(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\)
3, cho x>1 ; y>0 , cm
\(\dfrac{1}{\left(x+1\right)^3}+\left(\dfrac{x-1}{y}\right)^3+\dfrac{1}{y^3}\ge3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)\)
Rút gọn A= \(\left(\dfrac{2}{\sqrt{x-2}}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{3x-6\sqrt{x}}\left(x>0;x\ne4\right)\)
Cho \(x=\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\). Tính giá trị của biểu thức:
\(B=\left(x^4-2x^3-x^2+2x-1\right)^{2020}\)
Bài 1:Thu gọn và tính: a)Aleft(dfrac{asqrt{a}+bsqrt{b}}{sqrt{a}+sqrt{b}}right)left(dfrac{sqrt{a}+sqrt{b}}{a-b}right)^2 vớia^26-3sqrt{3};b^22+sqrt{3}b)Bdfrac{sqrt{2x+2sqrt{x^2-4}}}{sqrt{x^2-4}+x+2}vớix1+sqrt{5}Bài 2: Tìm GTLN GTNN của Csqrt{x-2-2sqrt{x-3}}-sqrt{x+1-4sqrt{x-3}}
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Bài 1:Thu gọn và tính:
a)A=\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\) với\(a^2=6-3\sqrt{3};b^2=2+\sqrt{3}\)
b)B=\(\dfrac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)với\(x=1+\sqrt{5}\)
Bài 2: Tìm GTLN GTNN của \(C=\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x+1-4\sqrt{x-3}}\)
rút gọn
a, \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}.\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
b,\(\left(\dfrac{\sqrt{x}-4}{x-2\sqrt{x}}-\dfrac{3}{2-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)\
c,\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
bài 1: rút gọn
A left(dfrac{sqrt{x}-1}{3sqrt{x}1}-dfrac{1}{3sqrt{x}+1}+dfrac{8sqrt{x}}{9x-1}right):left(1-dfrac{3sqrt{x}-2}{3sqrt{x}+1}right) X0; x#0
B left(dfrac{sqrt{x}-2}{x-1}-dfrac{sqrt{x}+2}{x+2sqrt{x}+1}right):left(dfrac{2}{x^2-2x+1}right) x0; xne1
bài 2:
dfrac{1}{2x-3sqrt{x}+2} x0
Tìm GTNN của A
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bài 1: rút gọn
A= \(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\) X>=0; x#0
B= \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\left(\dfrac{2}{x^2-2x+1}\right)\) x>=0; x\(\ne\)1
bài 2:
\(\dfrac{1}{2x-3\sqrt{x}+2}\) x>=0
Tìm GTNN của A