Cho a,b>0. Rút gọn bt K = \(9\sqrt{ab}-6b\sqrt{\dfrac{a}{b}}+\dfrac{1}{b}\sqrt{9ab^3}\)
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NT
1
2 tháng 1 2023
\(=9\sqrt{ab}-6b\cdot\dfrac{\sqrt{a}}{\sqrt{b}}-\dfrac{1}{b}\cdot3b\sqrt{ab}\)
\(=9\sqrt{ab}-6\sqrt{ab}-3\sqrt{ab}=0\)
30 tháng 5 2023
2:
\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)
=căn ab(6+7/b-5/a)
5 tháng 9 2022
a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)
b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)
\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)
\(=19ab\sqrt{ab}\)
c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(=\sqrt{ab}\left(\dfrac{1}{b}+1\right)-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\sqrt{ab}\)
d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)
3 tháng 5 2021
a) \(ĐK:x\ne9;x\ne0\)
\(B=\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}+\dfrac{\sqrt{a}}{\sqrt{a}+3}\right).\dfrac{a-9}{\sqrt{a}}\\ =\dfrac{\sqrt{a}.\left(\sqrt{a}+3\right)+\sqrt{a}.\left(\sqrt{a}-3\right)}{a-9}.\dfrac{a-9}{\sqrt{a}}\\ =\dfrac{2a.\left(a-9\right)}{\left(a-9\right).\sqrt{a}}\\ =\dfrac{2a}{\sqrt{a}}=2\sqrt{a}\)
Theo giả thiết:
B=a-8
<=> \(2\sqrt{a}=\left(a-8\right)\\ \Leftrightarrow4a=\left(a-8\right)^2\\ \Leftrightarrow a^2-20a+64=0\\ \Leftrightarrow\left[{}\begin{matrix}a=4\left(n\right)\\a=16\left(n\right)\end{matrix}\right.\)
13 tháng 5 2022
\(a,\) ta có :
\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)
\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)
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\(b,\) với \(x>0\) và \(x\ne1\) . ta có :
\(B=\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}+\dfrac{3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\left(\sqrt{x}-1\right)+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\sqrt{x}+1+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\) \(B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)
vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)
để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
vậy để \(B=2\) thì \(x=4\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 4 2021
a. \(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)
b. \(=\dfrac{1-\left(2\sqrt{a}\right)^3}{1-2\sqrt{a}}=\dfrac{\left(1-2\sqrt{a}\right)\left(1+2\sqrt{a}+4a\right)}{1-2\sqrt{a}}=1+2\sqrt{a}+4a\)
c. \(=\dfrac{1-\left(\sqrt{a}\right)^2}{1+\sqrt{a}}=\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}{1+\sqrt{a}}=1-\sqrt{a}\)
d. \(=\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}=\sqrt{a}\)
4 tháng 9 2022
1: \(Q=\dfrac{ab\left(a-b\right)}{ab}\cdot\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\left(\sqrt{a}+\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
2: \(=\dfrac{-1+\sqrt{5}-\sqrt{5}+\sqrt{9}-...-\sqrt{2001}+\sqrt{2005}}{4}\)
\(=\dfrac{\sqrt{2005}-1}{4}\)
2
27 tháng 8 2023
Với `a > 0,b >= 0` có:
`Bth=[a\sqrt{b}+b]/[a-b] . \sqrt{[b(a+b-2\sqrt{ab})]/[a^2+2a\sqrt{b}+b]} . (\sqrt{a}+\sqrt{b})`
`=[\sqrt{b}(a+\sqrt{b})]/[a-b].\sqrt{[b(\sqrt{a}-\sqrt{b})^2]/[(a+\sqrt{b})^2]}.(\sqrt{a}+\sqrt{b})`
`=[\sqrt{b}(a+\sqrt{b})|\sqrt{a}-\sqrt{b}|.\sqrt{b}.(\sqrt{a}+\sqrt{b})]/[(a-b)(a+\sqrt{b})]`
`=[b|\sqrt{a}-\sqrt{b}|]/[\sqrt{a}-\sqrt{b}]`
`={(b\text{ nếu }\sqrt{a} >= \sqrt{b}),(-b\text{ nếu }\sqrt{a} < \sqrt{b}):}`
H9
HT.Phong (9A5)
CTVHS
26 tháng 6 2023
a) \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\)
\(=\dfrac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}-\dfrac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}\)
\(=\dfrac{3-\sqrt{a}}{1}-\dfrac{\sqrt{a}-3}{1}\)
\(=3-\sqrt{a}-\sqrt{a}+3\)
\(=-2\sqrt{a}+6\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}-\sqrt{b}}{1}-\dfrac{\sqrt{a}-\sqrt{b}}{1}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
Y
26 tháng 6 2023
\(a,\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\left(dkxd:a\ne9,a\ge0\right)\)
\(=\dfrac{-\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}{\sqrt{a}+3}-\dfrac{\left(3-\sqrt{a}\right)^2}{3-\sqrt{a}}\)
\(=-\left(\sqrt{a}-3\right)+\left(3-\sqrt{a}\right)\)
\(=-\sqrt{a}+3+3-\sqrt{a}\)
\(=6-2\sqrt{a}\)
\(b,\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\left(dkxd:a\ne b,a\ge0,b\ge0\right)\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
\(=9\sqrt{ab}-6\sqrt{ab}+\dfrac{1}{b}\cdot3b\sqrt{ab}\)
\(=3\sqrt{ab}+3\sqrt{ab}=6\sqrt{ab}\)