Tìm GTLN của biểu thức: \(A=\frac{2}{\sqrt{x}+3}\)
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a) Có \(BC^2=15^2=225\)
\(AB^2+AC^2=9^2+12^2=81+144=225\)
do đó \(BC^2=AB^2+AC^2\)
Theo định lí Pythaogre đảo suy ra tam giác \(ABC\)vuông tại \(A\).
b) \(AH=\frac{AB.AC}{BC}=\frac{9.12}{15}=7,2\left(cm\right)\)
\(HB=\frac{AB^2}{BC}=\frac{9^2}{15}=5,4\left(cm\right)\)
\(HC=BC-HB=15-5,4=9,6\left(cm\right)\)
Ta có: E = \(\frac{x}{\sqrt{x}-1}=\frac{x-1+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}-1}\)
E = \(\sqrt{x}-1+\frac{1}{\sqrt{x}-1}+2\ge2\sqrt{\left(\sqrt{x}-1\right)\cdot\frac{1}{\sqrt{x}-1}}+2=2+2=4\)(bđt cosi)
Dấu "=" xảy ra <=> \(\sqrt{x}-1=\frac{1}{\sqrt{x}-1}\) <=> \(\left(\sqrt{x}-1\right)^2=1\)
<=> \(\orbr{\begin{cases}\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{cases}}\) <=> \(\orbr{\begin{cases}x=4\left(tm\right)\\x=0\left(ktm\right)\end{cases}}\)
Vậy MinE = 4 <=>. x = 4
\(\sqrt{x-3}-2\ne0\)
\(\sqrt{x-3}\ne2\)
\(x\ne7\left(1\right)\)
\(\sqrt{x-3}\ge0\)
\(x\ge3\left(2\right)\)
\(\left(1\right);\left(2\right)< =>x\ge3;x\ne7\)
\(15.a,\sqrt{3}\left(\sqrt{4}+2\sqrt{9}\right)\frac{\sqrt{3}}{2}-\sqrt{150}\)
\(\left(2+2.3\right)\frac{3}{2}-\sqrt{150}\)
\(12-\sqrt{150}\)
\(6\left(2-\sqrt{25}\right)=6\left(-3\right)=-18\)
\(b,\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)
\(\sqrt{196}-\sqrt{84}-7+2\sqrt{21}\)
\(14-\sqrt{21}\left(\sqrt{4}+2\right)-7\)
\(7-\sqrt{21}.4\)
\(7-\sqrt{84}\)
\(c,\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)
\(1+\sqrt{2}-\sqrt{3}+\sqrt{2}+2-\sqrt{6}+\sqrt{3}+\sqrt{6}-3\)
\(=2\sqrt{2}\)
\(d,\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)^2-\sqrt{3}+\sqrt{2}\)
\(\sqrt{3}\left(2-\sqrt{6}+3\right)-\sqrt{3}+\sqrt{2}\)
\(\sqrt{3}\left(5-\sqrt{6}-1\right)+\sqrt{2}\)
\(\sqrt{3}\left(4-\sqrt{6}\right)+\sqrt{2}\)
\(4\sqrt{3}-\sqrt{18}+\sqrt{2}\)
\(\sqrt{2}\left(2\sqrt{3}-3+1\right)\)
\(\sqrt{2}\left(2\sqrt{3}-2\right)\)
\(2\sqrt{6}-\sqrt{6}=\sqrt{6}\)
\(\frac{\sqrt{10}+5\sqrt{2}}{\sqrt{5}+1}-6\sqrt{\frac{5}{2}}+\frac{12}{4-\sqrt{10}}\)
\(=\frac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\frac{6\sqrt{10}}{2}+\frac{12\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}\)
\(=\sqrt{10}-3\sqrt{10}+2\left(4+\sqrt{10}\right)=-2\sqrt{10}+8+2\sqrt{10}=8\)
2. \(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{2}-1}+4\sqrt{\frac{3}{2}}+\frac{9}{3+\sqrt{3}}\)
\(=\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\frac{4\sqrt{6}}{2}+\frac{9\left(3-\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\)
\(=\sqrt{6}+2\sqrt{6}+\frac{9\left(3-\sqrt{6}\right)}{9-6}=3\sqrt{6}+3\left(3-\sqrt{6}\right)=9\)
vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\)\(\frac{2}{\sqrt{x}+3}\le\frac{2}{3}\)
vậy A đạt GTLN ki và chỉ khi \(\sqrt{x}=0\)suy ra x=0