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\(đk:u\ge0\)
\(=4.5\sqrt{u}-\dfrac{15}{2}.\dfrac{4}{3}.\sqrt{u}-\dfrac{2}{u}.\dfrac{13}{2}.u\sqrt{u}\)
\(=20\sqrt{u}-10\sqrt{u}-13\sqrt{u}=-3\sqrt{u}\)
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)
b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)
a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)
\(=4-2\sqrt{3}+2\sqrt{3}\)
=4
Thay x=4 vào B, ta được:
\(B=\dfrac{2-4}{2}=-1\)
a) \(P=\dfrac{x+3\sqrt{x}+x-3\sqrt{x}}{x-9}.\dfrac{x-9}{2\sqrt{x}}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b) \(P=\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
a: \(=\dfrac{x+3\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{x-9}{2\sqrt{x}}\)
\(=\sqrt{x}\)
a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)
=2
Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
`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)`
`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)`
a: \(A=10\sqrt{x}-10\sqrt{x}-\dfrac{4}{3}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=-\dfrac{2}{3}x\sqrt{x}\)
b: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
\(=-\left(5-4\right)=-1\)
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
a: \(=10\sqrt{x}-3\cdot\dfrac{10}{3}\cdot\sqrt{x}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-2\sqrt{x}=-2\sqrt{x}\)
b: \(B=\dfrac{1}{3}\left(-a-3\right)+\dfrac{4}{3}a+5\)
\(=\dfrac{-1}{3}a-1+\dfrac{4}{3}a+5=a+4\)
c: \(=20\sqrt{u}-\dfrac{15}{2}\cdot\dfrac{4}{3}\sqrt{u}-\dfrac{2}{u}\cdot\dfrac{13}{2}u\sqrt{u}\)
\(=20\sqrt{u}-10\sqrt{u}-13\sqrt{u}=-3\sqrt{u}\)