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14dm5cm=14,5dm;3dm7cm=3,7dm
chu vi hình chữ nhật đó là:
(14,5+3,7)x2=36,4(dm)
ĐS:36,4dm
14 dm 5 cm = 14,5 dm
3 dm 7 cm = 3,7 dm
Chiều rộng HCN là :
14,5 - 3,7 = 10,8 ( dm )
chu vi HCN là :
( 14,5 + 10,8 ) x 2 = 50,6 ( dm )
ĐS:..
![](https://rs.olm.vn/images/avt/0.png?1311)
tách 2,3 câu ra làm 1 câu hỏi đi. bạn đăng cả đóng thế này k ai tl cho đâu. khi nào tách thì gửi link mình tl cho
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\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{2\sqrt{2}}{2+\sqrt{2}}-\dfrac{4\sqrt{2}+4}{2+\sqrt{2}}\)
\(=\sqrt{3}+\sqrt{3}+\dfrac{-2\sqrt{2}-4}{2+\sqrt{2}}\)
\(=2\sqrt{3}+\dfrac{-2\left(2+\sqrt{2}\right)}{2+\sqrt{2}}\)
\(=2\sqrt{3}-2\)
\(------\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
\(=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}+\dfrac{5\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\dfrac{5\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{4\sqrt{5}-4}{5-1}+\dfrac{5\sqrt{5}-10}{5-4}+\dfrac{5\sqrt{5}-15}{5-9}\)
\(=5\sqrt{5}-10+\left(\dfrac{4\sqrt{5}-4}{4}+\dfrac{5\sqrt{5}-15}{-4}\right)\)
\(=\dfrac{4\cdot\left(5\sqrt{5}-10\right)}{4}+\left(\dfrac{4\sqrt{5}-4}{4}-\dfrac{5\sqrt{5}-15}{4}\right)\)
\(=\dfrac{20\sqrt{5}-40}{4}+\dfrac{-\sqrt{5}+11}{4}\)
\(=\dfrac{19\sqrt{5}-29}{4}\)
#Ayumu
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)2sqrt{48}-4sqrt{27}+sqrt{75}+sqrt{12}`
`=8sqrt3-12sqrt3+5sqrt3+2sqrt3`
`=3sqrt3`
`b)sqrt{(3-sqrt5)^2}-sqrt{20}`
`=3-sqrt5-2sqrt5`
`=3-3sqrt5`
2 câu cuối không rõ đề :v
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Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
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\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{\sqrt{35}.\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{35}\)
\(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
\(=\frac{\sqrt{4}}{\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{\sqrt{4}}\)
\(=\frac{2\sqrt{3}}{\sqrt{3}.\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{2}\)
\(=\frac{2\sqrt{3}}{3}+2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
\(=2\sqrt{3}\left(\frac{1}{3}+1-\frac{1}{3}\right)\)
\(=2\sqrt{3}\)
a)\(\sqrt{x+1}-\sqrt{x-2}=1\)
Đk:\(x\ge2\)
\(pt\Leftrightarrow\left(\sqrt{x+1}-2\right)-\left(\sqrt{x-2}-1\right)=0\)
\(\Leftrightarrow\dfrac{x+1-4}{\sqrt{x+1}+2}-\dfrac{x-2-1}{\sqrt{x-2}+1}=0\)
\(\Leftrightarrow\dfrac{x-3}{\sqrt{x+1}+2}-\dfrac{x-3}{\sqrt{x-2}+1}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{x+1}+2}-\dfrac{1}{\sqrt{x-2}+1}\right)=0\)
Dễ thấy:\(\dfrac{1}{\sqrt{x+1}+2}-\dfrac{1}{\sqrt{x-2}+1}< 0\)
Nên \(x-3=0\Rightarrow x=3\)
b)\(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)
Đk:\(x\ge1\)
\(pt\Leftrightarrow\sqrt{x-1}=\sqrt{5x-1}+\sqrt{3x-2}\)
\(\Leftrightarrow x-1=5x-1+3x-2+2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
\(\Leftrightarrow2-7x=2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
\(\Leftrightarrow49x^2-28x+4=4\left(5x-1\right)\left(3x-2\right)\)
\(\Leftrightarrow49x^2-28x+4=60x^2-52x+8\)
\(\Leftrightarrow-11x^2+24x-4=0\Leftrightarrow\left(2-x\right)\left(11x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{11}\end{matrix}\right.\) (loại hết)
c)\(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0\)
Đk:\(x\ge0\)
\(pt\Leftrightarrow\sqrt{x}-\left(\sqrt{x+1}+1\right)-\left(\sqrt{x+4}+2\right)+\left(\sqrt{x+9}-3\right)=0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{x+1-1}{\sqrt{x+1}+1}-\dfrac{x+4-4}{\sqrt{x+4}+2}+\dfrac{x+9-9}{\sqrt{x+9}-3}=0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{x}{\sqrt{x+1}+1}-\dfrac{x}{\sqrt{x+4}+2}+\dfrac{x}{\sqrt{x+9}-3}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}+1}-\dfrac{1}{\sqrt{x+4}+2}+\dfrac{1}{\sqrt{x+9}-3}\right)=0\)
Dễ thấy:\(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}+1}-\dfrac{1}{\sqrt{x+4}+2}+\dfrac{1}{\sqrt{x+9}-3}>0\)
Nên \(x=0\)