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Bài 15:
a) Ta có: \(A=\cos^252^0\cdot\sin45^0+\sin^252^0\cdot\cos45^0\)
\(=\dfrac{\sqrt{2}}{2}\left(\sin^252^0+\cos^252^0\right)\)
\(=\dfrac{\sqrt{2}}{2}\)
b) Ta có: \(B=\tan60^0\cdot\cos^247^0+\sin^247^0\cdot\cot30^0\)
\(=\sqrt{3}\cdot\left(\sin^247^0+\cos^247^0\right)\)
\(=\sqrt{3}\)
Bài 17:
c) Ta có: \(C=\tan1^0\cdot\tan2^0\cdot\tan3^0\cdot\tan4^0\cdot...\cdot\tan89^0\)
\(=\left(\tan1^0\cdot\tan89^0\right)\cdot\left(\tan2^0\cdot\tan88^0\right)\cdot...\cdot\tan45^0\)
\(=1\cdot1\cdot...\cdot1=1\)
\(\sqrt{8-2\sqrt{15}}+\sqrt{48+6\sqrt{15}}\\ =\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{45+2\cdot3\sqrt{5}\cdot\sqrt{3}+3}\\ =\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{5}+\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}+3\sqrt{5}+\sqrt{3}=4\sqrt{5}\)
\(\sqrt{8-\sqrt{60}}-\sqrt{23-\sqrt{240}}\\ =\sqrt{8-\sqrt{4\cdot15}}-\sqrt{23-\sqrt{4\cdot60}}\\ =\sqrt{8-2\sqrt{15}}-\sqrt{23-2\sqrt{60}}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{20-2\cdot\sqrt{20}\cdot\sqrt{3}+3}\\ =\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}-\sqrt{20}+\sqrt{3}\\ =\sqrt{5}-2\sqrt{5}=-\sqrt{5}\)
=\(\sqrt{\left(5+2\sqrt{6}\right)+\left(2\sqrt{10}+2\sqrt{15}\right)+5}\)
=\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2+2\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{5}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)^2}\)
=\(\sqrt{3}+\sqrt{2}+\sqrt{5}\)
\(\left(4+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{5}\right)\left(8-2\sqrt{15}\right)\)
\(a.\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}\)
\(=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=\sqrt{\left(16-15\right)}=\sqrt{1}=1\)
\(b.\sqrt{7-\sqrt{47}}.\sqrt{14+2\sqrt{47}}\)
\(=\sqrt{7-\sqrt{47}}.\sqrt{2\left(7-\sqrt{47}\right)}\)
\(=\sqrt{2\left(7-\sqrt{47}\right)\left(7+\sqrt{47}\right)}=\sqrt{2\left(49-47\right)}=\sqrt{2^2}=\sqrt{4}=2\)
\(c.\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=\sqrt{16-\left(\sqrt{10+2\sqrt{5}}\right)^2}\)
\(=\sqrt{16-10-2\sqrt{5}}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
Lời giải:
Lần sau bạn nhớ ghi đầy đủ đề. $ABC$ là tam giác vuông tại $A$.
$\frac{AB}{AC}=\frac{3}{4}$
$\Rightarrow AC=\frac{4AB}{3}=\frac{4.15}{3}=20$ (cm)
Áp dụng định lý Pitago:
$y=BC=\sqrt{AB^2+AC^2}=\sqrt{15^2+20^2}=25$ (cm)
$S_{ABC}=AB.AC:2=AH.BC:2$
$\Rightarrow AB.AC=AH.BC$
$\Rightarrow x=AH=\frac{AB.AC}{BC}=\frac{15.20}{25}=12$ (cm)
a: \(\sqrt{12}-\sqrt{27}+\sqrt{3}\)
\(=2\sqrt{3}-3\sqrt{3}+\sqrt{3}\)
=0
b: \(\left(\sqrt{12}-3\sqrt{15}-4\sqrt{135}\right)\cdot\sqrt{3}\)
\(=\left(2\sqrt{3}-3\sqrt{15}-12\sqrt{15}\right)\cdot\sqrt{3}\)
\(=6-45\sqrt{5}\)
\(\sqrt{15-10\sqrt{2}}\\ =\sqrt{15-2\cdot\sqrt{5}\cdot\sqrt{5}\cdot\sqrt{2}}\\ =\sqrt{\left(\sqrt{10}\right)^2-2\cdot\sqrt{10}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}\\ =\sqrt{10}-\sqrt{5}\\ =\sqrt{5}\left(\sqrt{2}-1\right)\)