với x>9, x>0 , x\(\ne\)4, x\(\ne\) 9 tìm MinP = \(\dfrac{4x}{\sqrt[]{x}-3}\)
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10 tháng 10 2020
Ta có: \(\frac{x+\sqrt{x}-6}{x-9}+\frac{x-7\sqrt{x}+19}{x+\sqrt{x}-12}-\frac{x-5\sqrt{x}}{x+4\sqrt{x}}\)
\(=\frac{x+3\sqrt{x}-2\sqrt{x}-6}{x-9}+\frac{x-7\sqrt{x}+19}{x+4\sqrt{x}-3\sqrt{x}-12}-\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\sqrt{x}\left(\sqrt{x}+4\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)}{x-9}+\frac{x-7\sqrt{x}+19}{\sqrt{x}\left(\sqrt{x}+4\right)-3\left(\sqrt{x}+4\right)}-\frac{\sqrt{x}-5}{\sqrt{x}+4}\)
\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{x-7\sqrt{x}+19}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}+\frac{x-7\sqrt{x}+19}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}-\frac{x-8\sqrt{x}+15}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{x+2\sqrt{x}-8+x-7\sqrt{x}+19-x+8\sqrt{x}-15}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{x+3\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{x+4\sqrt{x}-\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+4\right)-\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-3}\)
24 tháng 7 2018
Ta có :
a , \(M=2\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\left[\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(M=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}-\dfrac{2\left(x+9\right)}{x-9}\right]:\left[\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(M=\left(\dfrac{2x-6\sqrt{x}-2x-18}{x-9}\right).\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\right]\)
\(M=\dfrac{-6\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(2\sqrt{x}+4\right)}\)
\(M=\dfrac{-6\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(M=-\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b , mik ko chắc chắn nên mik chưa làm nhé !
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)`
24 tháng 9 2022
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
T
19 tháng 4 2021
(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)
(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)
(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)
(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)
10 tháng 7 2022
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
12 tháng 6 2022
a: \(P=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
b: Để P<-1 thì P+1<0
\(\Leftrightarrow-\sqrt{x}+4< 0\)
=>0<x<16 và x<>9
TP
21 tháng 6 2019
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
TP
21 tháng 6 2019
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)