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Xét tam giác ABC vuông tại A, đường cao AH
Áp dụng hệ thức \(AB^2=HB.BC=HB\left(HC+HB\right)=HB\left(16+HB\right)\Leftrightarrow225=16HB+HB^2\)
\(\Leftrightarrow HB^2+16BH-225=0\Leftrightarrow HB=9cm\)
BC = HC + HB = 9 + 16 = 25 cm
Áp dụng hệ thức \(AH^2=HB.HC=144\Leftrightarrow AH=12cm\)
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+2}\)
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+2}\)
\(\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{6}+\sqrt{8}+2}{\sqrt{2}+\sqrt{3}+2}\)
\(\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
`=`\(\sqrt{2}+1\)
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)
\(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(\sqrt{3+2})^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=3\)
a, đk x >= 0
\(\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\)(tmđk)
b, đk x >= 0
\(\Leftrightarrow x< 2\)Kết hợp với đk vậy 0 =< x < 2
c, đk x >= 0 \(\Leftrightarrow2x< 16\Leftrightarrow x< 8\)
Kết hợp đk vậy 0 =< x < 8
a) \(2\sqrt{x}=14\)
Vì \(x\ge0\) nên bình phương hai vế ta được :
\(x=7^2\Leftrightarrow x=49\)
Vậy \(x=49\)
b) \(\sqrt{x}< \sqrt{2}\Leftrightarrow(\sqrt{x})^2< (\sqrt{2})^2\Leftrightarrow x< 4\)
c) \(\sqrt{2x}< 4\Leftrightarrow(\sqrt{2x})^2< 4^2\Leftrightarrow2x< 16\Leftrightarrow x< 8\)
\(\sqrt{\dfrac{2}{5}}\) + \(\sqrt{\dfrac{5}{2}}\) = \(\dfrac{\sqrt{2}\times\sqrt{2}}{\sqrt{10}}\)+ \(\dfrac{\sqrt{5}\times\sqrt{5}}{\sqrt{10}}\) = \(\dfrac{7}{\sqrt{10}}\)= \(\dfrac{7\sqrt{10}}{10}\)
Lời giải:
\(\sqrt{\frac{2}{5}}+\sqrt{\frac{5}{2}}=\frac{\sqrt{2}.\sqrt{2}+\sqrt{5}.\sqrt{5}}{\sqrt{5}.\sqrt{2}}=\frac{7}{\sqrt{10}}=\frac{7\sqrt{10}}{10}\)
a) \((2+\sqrt{3})\sqrt{7-4\sqrt{3}}\)
\(=(2+\sqrt{3})\sqrt{4-4\sqrt{3}+3}\)
\(=(2+\sqrt{3})\sqrt{(2-\sqrt{3})^2}\)
\(=(2+\sqrt{3})(2-\sqrt{3})=4-3=1\)
b) \(\sqrt{(1-\sqrt{2023})^2}\cdot\sqrt{2024+2\sqrt{2023}}\)
\(=|1-\sqrt{2023}|\sqrt{2023+2\sqrt{2023}+1}\)
\(=(\sqrt{2023}-1)\sqrt{(\sqrt{2023}+1)^2}\)
\(=(\sqrt{2023}-1)(\sqrt{2023}+1)\)
\(=\sqrt{2023^2}-1^2=2023-1=2022\)
`1/2(\sqrt{6}+\sqrt{5})^2-1/4\sqrt{120}-\sqrt{15}/2`
`=1/2 .(6+5+2\sqrt{30})-1/4 \sqrt{2^2 .30}-\sqrt{15}/2`
`=11/2+\sqrt{30}-1/2 \sqrt{30}-\sqrt{15}/2`
`=[11+2\sqrt{30}-\sqrt{30}-\sqrt{15}]/2`
`=[11+\sqrt{30}-\sqrt{15}]/2`
Lời giải:
$\frac{1}{2}(\sqrt{6}+\sqrt{5})^2-\frac{1}{4}\sqrt{120}-\frac{\sqrt{15}}{2}$
$=\frac{1}{2}(11+2\sqrt{30})-\frac{1}{2}\sqrt{30}-\frac{\sqrt{15}}{2}$
$=\frac{11}{2}+\frac{\sqrt{30}}{2}-\frac{\sqrt{15}}{2}$
a) Ta có : \(AB^2+AC^2=6^2+8^2=100\)
\(BC^2=10^2=100\)
\(\Rightarrow AB^2+AC^2=BC^2\)
\(\Rightarrow\bigtriangleup ABC\) vuông tại \(A\) (đpcm)
b) Từ \(AB\cdot AC=AH\cdot BC\)
\(\Rightarrow6\cdot8=AH\cdot10\)
\(\Rightarrow AH=4,8\)
c) Từ \(AB^2=BC\cdot BH\)
\(\Rightarrow6^2=10\cdot HB\)
\(\Rightarrow HB=3,6\)
Từ \(HB+HC=BC\)
\(\Rightarrow3,6+HC=10\)
\(\Rightarrow HC=6,4\)
\(S_{\bigtriangleup ABC}=\dfrac{1}{2}AB\cdot AC\) .