\(P=\frac{7}{\sqrt{x}-3}\)
Với \(x\ge0;x\ne9\)tìm GTLN của P
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19 tháng 8 2021
mình làm mẫu thôi, bên dưới tương tự bạn nhé
a, \(\frac{\sqrt{x}+6}{\sqrt{x}-3}=\frac{\sqrt{x}-3+9}{\sqrt{x}-3}=1+\frac{9}{\sqrt{x}-3}\)ĐK : \(x\ge0;x\ne9\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
| \(\sqrt{x}-3\) | 1 | -1 | 3 | -3 | 9 | -9 |
| x | 16 | 4 | 36 | 0 | 144 | loại |
B
1 tháng 8 2018
\(\left(\dfrac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\right)\) : \(\dfrac{1}{\sqrt{x}+2}\)
=\(\dfrac{3x-3\sqrt{x}-3+\sqrt{x}+2-\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}\) .\(\sqrt{x}+2\)
=\(\dfrac{(3x-3\sqrt{x})(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\)
=\(\dfrac{3\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\) =\(3\sqrt{x}\)
19 tháng 10 2021
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
19 tháng 10 2021
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
5 tháng 9 2020
a) \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}=27-4\sqrt{3x}\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28=3\sqrt{2x}+2\sqrt{8x}+28=3\sqrt{2x}+4\sqrt{2x}+28=7\sqrt{2x}+28\)
c) \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)
d) \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4x+4a^2\right)}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.\sqrt{5}\left|a\left(2a-1\right)\right|=2a\sqrt{5}\)
Thiếu ĐKXĐ : ..............
5 tháng 9 2020
a) Ta có: \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}\)
\(=27-4\sqrt{3x}\)
b) Ta có: \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28\)
\(=3\sqrt{2x}-5.2\sqrt{2x}+7.2\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+14\sqrt{2x}+28\)
\(=7\sqrt{2x}+28\)
c) Ta có: \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{4}{\left(x-y\right)^2.\left(x+y\right)^2}.\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{2.3}{\left(x-y\right)^2}}\)
\(=\frac{1}{x-y}.\sqrt{6}\)
d) Ta có: \(\frac{2}{2a-1}.\sqrt{5a^2.\left(1-4a+4a^2\right)}\)
\(=\sqrt{\frac{4}{\left(2a-1\right)^2}.5a^2.\left(2a-1\right)^2}\)
\(=2a.\sqrt{5}\)
20 tháng 10 2016
\(P=\frac{2}{\sqrt{x}-1}+\frac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\frac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\)
\(=\frac{2\left(x+\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+2\sqrt{x}+2+2x-2+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
20 tháng 10 2016
Với \(x\ge0;x\ne1\), ta có:
\(P=\frac{2}{\sqrt{x}-1}+\frac{2.\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\frac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\)
\(P=\frac{2.\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}+\frac{2.\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}+\frac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{2x+2\sqrt{x}+2+2.\left(x-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{3x-8\sqrt{x}+5+2x-2}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{5x-\sqrt{8x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{5x-5\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right).\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
Vậy với \(x\ge0;x\ne1\) ta có: \(P=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
P lớn nhất <=> \(\sqrt{x}\)- 3 bé nhất
<=> \(\sqrt{x}\)bé nhất
<=> x=0
Vậy MaxP=\(\frac{-7}{3}\) <=> x=0
Ta có \(P=\frac{7}{\sqrt{x}-3}\)
Do \(\sqrt{x}-3\ge-3;\sqrt{x}\ge0\forall x\)
\(\Rightarrow\frac{7}{\sqrt{x}-3}\le\frac{7}{-3}\)
Dấu ''='' xảy ra khi \(\sqrt{x}=0\Leftrightarrow x=0\)( tm đk \(x\ge0;x\ne9\))
Vậy GTLN P là 7/-3 khi x = 0