a) \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{a}{b}}\) với \(a>0;b>0\)
b) \(\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}\) vớ \(m>0;x\ne1\)
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30 tháng 8 2022
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
13 tháng 7 2018
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
17 tháng 11 2021
\(=\dfrac{\sqrt{ab}}{b}+\sqrt{\dfrac{a^2b}{b^2a}}=\dfrac{\sqrt{ab}}{b}+\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{ab}}{b}+\dfrac{\sqrt{ab}}{b}=\dfrac{2\sqrt{ab}}{b}\left(B\right)\)
17 tháng 6 2021
`(asqrtb-bsqrta)/sqrt{ab}-(a-b)/(sqrta-sqrtb)`
`=(sqrt{ab}(\sqrta-sqrtb))/sqrt{ab}-((sqrta-sqrtb)(sqrta+sqrtb))/(sqrta-sqrtb)`
`=sqrta-sqrtb-(sqrta-sqrtb)`
`=-2sqrtb`
TA
17 tháng 6 2021
`(a\sqrtb-b\sqrta)/(\sqrt(ab)) -(a-b)/(\sqrta-\sqrtb)`
`=(\sqrt(ab) (\sqrta-\sqrtb))/(\sqrt(ab)) - ((\sqrta-\sqrtb)(\sqrta+\sqrtb))/(\sqrta-\sqrtb)`
`=(\sqrta-\sqrtb) - (\sqrta+\sqrtb)`
`=-2\sqrtb`
13 tháng 11 2021
Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
13 tháng 11 2021
a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
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}):}`
Y
28 tháng 6 2023
\(VT=\left(\sqrt{\dfrac{a}{b}}-\sqrt{\dfrac{b}{a}}\right):\left(a-b\right)\\ =\left(\dfrac{\sqrt{a}}{\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}}\right).\dfrac{1}{a-b}\\ =\dfrac{\sqrt{a}.\sqrt{a}-\sqrt{b}.\sqrt{b}}{\sqrt{ab}}.\dfrac{1}{a-b}\\ =\dfrac{\sqrt{a^2}-\sqrt{b^2}}{\sqrt{ab}}.\dfrac{1}{a-b}\\ =\dfrac{a-b}{\sqrt{ab}}.\dfrac{1}{a-b}\\ =\dfrac{1}{\sqrt{ab}}=VP\left(dpcm\right)\)
28 tháng 6 2023
\(VT=\dfrac{a-b}{\sqrt{ab}}\cdot\dfrac{1}{a-b}=\dfrac{1}{\sqrt{ab}}=VP\)
H9
HT.Phong (9A5)
CTVHS
7 tháng 11 2023
\(\dfrac{\sqrt{a}+\sqrt{ab}}{a-b}-\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\left(a,b\ge0;a\ne b\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{ab}-\sqrt{ab}-b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}-b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}-b}{a-b}\)
a) \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{a}{b}}\) với a>0 và b>0
b) \(\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m\left(2-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{4m^2\left(1-2x+x^2\right)}{81\left(1-2x+x^2\right)}}=\sqrt{\dfrac{4m^2}{81}}=\sqrt{\dfrac{2m}{9}}\)