a. \(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{1}{1+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}\)
b/\(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}với\) x>=0
c.\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}với\) x>=0,y>=0 và x≠y
Những câu hỏi liên quan
bài 1: rút gọn các biểu thức.
a) dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-(sqrt{x}-sqrt{y})^2
b) sqrt{dfrac{x-2sqrt{x}+1}{x+2sqrt{x}+1}}(xge0)
c) dfrac{x-1}{sqrt{y}-1}sqrt{dfrac{(y-2sqrt{y}+1)^2}{(x-1)^4}}(xne1,yne1,y0)
bài 2:rút gọn và tính.
a) sqrt{dfrac{sqrt{a}-1}{sqrt{b}+1}:}sqrt{dfrac{sqrt{b}-1}{sqrt{a}+1}với}a7,25;b3,25
b) sqrt{15a^2-8asqrt{15}+16}vớiasqrt{dfrac{3}{5}}+sqrt{dfrac{5}{3}}
c) sqrt{10a^2-4asqrt{10}+4}vớiasqrt{dfrac{2}{5}}+sqrt{dfrac{5}{2}}
d) sqrt{a^2+2sqrt{a^2-1}}-sq...
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bài 1: rút gọn các biểu thức.
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-(\sqrt{x}-\sqrt{y})^2\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}(x\ge0)\)
c) \(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{(y-2\sqrt{y}+1)^2}{(x-1)^4}}(x\ne1,y\ne1,y>0)\)
bài 2:rút gọn và tính.
a) \(\sqrt{\dfrac{\sqrt{a}-1}{\sqrt{b}+1}:}\sqrt{\dfrac{\sqrt{b}-1}{\sqrt{a}+1}với}a=7,25;b=3,25\)
b) \(\sqrt{15a^2-8a\sqrt{15}+16}vớia=\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{5}{3}}\)
c) \(\sqrt{10a^2-4a\sqrt{10}+4}vớia=\sqrt{\dfrac{2}{5}}+\sqrt{\dfrac{5}{2}}\)
d) \(\sqrt{a^2+2\sqrt{a^2-1}}-\sqrt{a^2-2\sqrt{a^2-1}}(a=\sqrt{5})\)
bài 3: rút gọn các biểu thức.
a) \(\sqrt{9(x-5)^2}(x\ge5)\)
b) \(\sqrt{x^2.(x-2)^2}(x< 0)\)
c)\(\dfrac{\sqrt{108x^3}}{\sqrt{12x}}(x>0)\)
d)\(\dfrac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}(x< 0:y\ne0)\)
ai giúp mik vs ạ, cảm ơn !
Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
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\(C=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\)
a) Rút gọn
b) Tính C với x=2-\(\sqrt{3}\); y=2+\(\sqrt{3}\)
tính
\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{1}{1+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}\)
\(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\) với x>=0
\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
a: \(=3-2\sqrt{2}-\sqrt{2}+1+1+\dfrac{1}{2}\sqrt{2}\)
\(=-\dfrac{5}{2}\sqrt{2}+5\)
b: \(=\dfrac{x-4+10-x}{\sqrt{x}+2}=\dfrac{6}{\sqrt{x}+2}\)
c: \(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
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Rút gọn biểu thức 1) dfrac{sqrt{14}-sqrt{21}}{sqrt{7}} .2) dfrac{sqrt{a^2+5a+6}}{sqrt{a+3}}3) sqrt{3left(x^2-10x+25right)}.sqrt{27} với x 54)dfrac{y}{x}sqrt{dfrac{x^2}{y^4}} với x 0; y 05) dfrac{1}{x-y}.sqrt{x^6left(x-yright)^4} với x ne y
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Rút gọn biểu thức 1) \(\dfrac{\sqrt{14}-\sqrt{21}}{\sqrt{7}}\) .
2) \(\dfrac{\sqrt{a^2+5a+6}}{\sqrt{a+3}}\)
3) \(\sqrt{3\left(x^2-10x+25\right)}.\sqrt{27}\) với x < 5
4)
\(\dfrac{y}{x}\sqrt{\dfrac{x^2}{y^4}}\) với x > 0; y < 0
5) \(\dfrac{1}{x-y}.\sqrt{x^6\left(x-y\right)^4}\) với x \(\ne\) y
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
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Cho hàm số \(y=\sqrt{x+\sqrt{x^2+1}}\). Tính đạo gàm của hàm số.
A. \(y'=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}}\)
B. \(y'=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^2+1}}\)
C. \(y'=\dfrac{\sqrt{x^2+1}}{2\sqrt{\sqrt{x+\sqrt{x^2+1}}}}\)
D. \(y'=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
\(y'=\dfrac{\left(x+\sqrt{x^2+1}\right)'}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{1+\dfrac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}.\sqrt{x+\sqrt{x^2+1}}}\)
\(=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
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trục căn thức
a) \(\dfrac{1}{\sqrt{x-1}};\dfrac{a+2}{\sqrt{a^2-4}};\dfrac{x-y}{\sqrt{x^2-y^2}};\dfrac{a}{\sqrt{x^2}}\) (n lẻ)
b) \(\dfrac{\sqrt{x^2-1}+1}{\sqrt{x^2-1}-1}\)
c) \(\dfrac{2}{\sqrt{7-2\sqrt{6}}}\)
a) \(\dfrac{1}{\sqrt{x-1}}=\dfrac{\sqrt{x-1}}{x-1}\)
\(\dfrac{a+2}{\sqrt{a^2-4}}=\dfrac{\sqrt{a+2}}{\sqrt{a-2}}=\dfrac{\sqrt{a^2-4}}{a-2}\)
\(\dfrac{x-y}{\sqrt{x^2-y^2}}=\dfrac{x-y}{\sqrt{\left(x-y\right)\left(x+y\right)}}=\dfrac{\sqrt{x-y}}{\sqrt{x+y}}=\dfrac{\sqrt{x^2-y^2}}{x+y}\)
\(\dfrac{a}{\sqrt{x^2}}=\dfrac{a}{\left|x\right|}\)
b) \(\dfrac{\sqrt{x^2-1}+1}{\sqrt{x^2-1}-1}=\dfrac{\left(\sqrt{x^2-1}+1\right)^2}{x^2-2}\)
c) \(\dfrac{2}{\sqrt{7-2\sqrt{6}}}=\dfrac{2}{\sqrt{6}-1}=\dfrac{2\left(\sqrt{6}+1\right)}{5}\)
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Bài 1: Cho Aleft(dfrac{x-y}{sqrt{x}-sqrt{y}}+dfrac{sqrt{x^3}-sqrt{y^3}}{y-x}right):dfrac{left(sqrt{x}-sqrt{y}right)^2+sqrt{xy}}{sqrt{x}+sqrt{y}}với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A dfrac{xsqrt{x}-1}{x-sqrt{x}}-dfrac{xsqrt{x}+1}{x+sqrt{x}}+left(sqrt{x}-dfrac{1}{sqrt{x}}right).left(dfrac{sqrt{x}+1}{sqrt{x}-1}+dfrac{sqrt{x}-1}{sqrt{x}+1}right)
với x0; x≠1
a) Rút gọn A
b)Tìm x để A6
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Bài 1: Cho A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A= \(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
với x>0; x≠1
a) Rút gọn A
b)Tìm x để A=6
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
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CM:
\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{6}}{6}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}+\sqrt{y}}=x-y\) với x.0, y>0, x≠y
\(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{x}}{y-\sqrt{xy}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)với x>0, y>0, x≠y
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
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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)
\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\dfrac{x+1}{\sqrt{x}}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}/\dfrac{\sqrt{x}-\sqrt{y}}{x-y}\)
\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(\left(\dfrac{\sqrt{x}+2}{3\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}-3\right)/\dfrac{2-4\sqrt{x}}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}\)
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
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