\(\frac{x+3}{x^2-4}.\frac{8-12x+6x^2-x^3}{9x+27}\)
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TH
0
LP
0
4 tháng 7 2015
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
29 tháng 7 2019
\(M=\frac{2\sqrt{x}-3}{\sqrt{x}-4}-\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{2-3\sqrt{x}}{x-3\sqrt{x}-4}\)
\(=\frac{2\sqrt{x}-3}{\sqrt{x}-4}-\frac{\sqrt{x}+2}{\sqrt{x}+1}\)\(+\frac{3\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+4\right)+3\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{2x-\sqrt{x}-3-x+2\sqrt{x}+8+3\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}-4}\)
V
0
18 tháng 7 2017
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
\(\frac{x+3}{x^2-4}.\frac{8-12x+6^2-x^3}{9x+27}\)
\(=\frac{x+3}{x^2-4}.\frac{-x^3+6x^2-4}{9x+27}\)
\(=\frac{\left(x+3\right)\left(-x^3+6x^2-4\right)}{\left(x^2-4\right)\left(9x+27\right)}\)
\(=\frac{\left(x+3\right)\left(-x^3+6x-4\right)}{9\left(x+3\right)\left(x^2-4\right)}\)
\(=\frac{-x^3+6x^2-4}{9\left(x^2-4\right)}\)
Mk ko chắc
(x+3 )/ (x-2)(x+2) . [(2-x)^3 / 9(x+3)]
= -(x-2)^2 / [(x+2).9]