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B=\(\dfrac{2u+\sqrt{uv}-3v}{2u-5\sqrt{uv}+3v}\) với \(u\ge\)0,\(v\ge0\) và\(u\ne\dfrac{9}{4}v\)
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16 tháng 5 2021
`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
16 tháng 5 2021
`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
HS
16 tháng 8 2018
\(A=\dfrac{u-v}{\sqrt{u}+\sqrt{v}}-\dfrac{\sqrt{u^3}+\sqrt{v^3}}{u-v}\)
\(=\sqrt{u}-\sqrt{v}-\dfrac{u\sqrt{u}+v\sqrt{v}}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)
\(=\sqrt{u}-\sqrt{v}-\dfrac{u-\sqrt{uv}+v}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)
\(=\sqrt{u}-\sqrt{v}-\dfrac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}\)
\(=\dfrac{\left(\sqrt{u}-\sqrt{v}\right)\sqrt{u}-\left(\sqrt{u}-\sqrt[]{v}\right)\sqrt{v}-\left(u-\sqrt{uv}+v\right)}{\sqrt{u}-\sqrt{v}}\)
\(=\dfrac{u-\sqrt{uv}-\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}\)
\(\Leftrightarrow\)\(-\dfrac{\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)
28 tháng 8 2023
a:
Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)
b: P>=1/2
=>P-1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)
=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)
=>\(-\sqrt{x}-15>=0\)
=>\(-\sqrt{x}>=15\)
=>căn x<=-15
=>\(x\in\varnothing\)
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>P>=-2
Dấu = xảy ra khi x=0
26 tháng 6 2021
\(P=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{x-9}:\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\left(\sqrt{x}-3\right)=\dfrac{6}{\sqrt{x}+3}\)
26 tháng 6 2021
\(P=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\sqrt{x}-3\)
\(P=\dfrac{6}{\sqrt{x}+3}\)
6 tháng 8 2020
A=\(\frac{u-v}{\sqrt{u}+\sqrt{v}}-\frac{\sqrt{u^3}+\sqrt{v^3}}{u-v}=\frac{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}{\sqrt{u}+\sqrt{v}}-\frac{\left(\sqrt{u}+\sqrt{v}\right)\left(u-\sqrt{u}\sqrt{v}+v\right)}{\left(\sqrt{u}+\sqrt{v}\right)\left(\sqrt{u}-\sqrt{v}\right)}\)
\(=\sqrt{u}-\sqrt{v}-\frac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}=\frac{u-2\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}=\frac{-\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)
QT
23 tháng 4 2021
a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)
\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)
\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)
\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)
23 tháng 4 2021
a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)
do \(x\ge0;y\ge0\)
\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)
6 tháng 8 2020
bạn vào thống kê của mình có link tham khảo
Câu hỏi của Duy Saker Hy - Toán lớp 9 - Học toán với OnlineMath
\(B=\dfrac{2u+\sqrt{uv}-3v}{2u-5\sqrt{uv}+3v}\)
\(=\dfrac{2u+3\sqrt{uv}-2\sqrt{uv}-3v}{2u-2\sqrt{uv}-3\sqrt{uv}+3v}\)
\(=\dfrac{\sqrt{u}.\left(2\sqrt{u}+3\sqrt{v}\right)-\sqrt{v}.\left(2\sqrt{u}+3\sqrt{v}\right)}{2\sqrt{u}.\left(\sqrt{u}-\sqrt{v}\right)-3\sqrt{v}.\left(\sqrt{u}-\sqrt{v}\right)}\)
\(=\dfrac{\left(2\sqrt{u}+3\sqrt{v}\right)\left(\sqrt{u}-\sqrt{v}\right)}{\left(\sqrt{u}-\sqrt{v}\right)\left(2\sqrt{u}-3\sqrt{v}\right)}\)
\(=\dfrac{2\sqrt{u}+3\sqrt{v}}{2\sqrt{u}-3\sqrt{v}}\\ =\dfrac{4u+12\sqrt{uv}+9v}{4u-9v}\)