Câu a \(tg^2\) là gì thế ;) 

\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)

\(a,C1=744+186\\ =930\\ C2=93\times\left(8+2\right)\\ =93\times10=930\\ b,C1:=324+576\\ =900\\ C2=9\times\left(36+64\right)\\ =9\times100\\ =900\\ c,C1=456-399\\ =57\\ C2=57\times\left(8-7\right)\\ =57\)

\(a,\dfrac{x}{6}=-\dfrac{3}{4}\\ \Rightarrow x=\dfrac{\left(-3\right).6}{4}=-\dfrac{9}{2}\\ b,\dfrac{5}{x}=\dfrac{15}{-20}\)

điều kiện : \(x\ne0\)

\(\Rightarrow x=\dfrac{5.\left(-20\right)}{15}=-\dfrac{20}{3}\left(thoaman\right)\)

\(c,đk:x\ne14\\ \Rightarrow3.\left(x+11\right)=2.\left(14-x\right)\\ \Rightarrow3x+33=28-2x\\ \Rightarrow5x=-5\\ \Rightarrow x=-1\left(thoaman\right)\)

\(VT=\left(\dfrac{\sqrt{14}.\sqrt{14}}{\sqrt{14}}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right)\sqrt{5-\sqrt{21}}\\ =\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\sqrt{14}.\sqrt{5-\sqrt{2}1}+\sqrt{6}.\sqrt{5-\sqrt{21}}\\ =\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\\ =\sqrt{49-2.7.\sqrt{21}+21}+\sqrt{9-2.3.\sqrt{21}+21}\\ =\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(3-\sqrt{21}\right)^2}\\ =\left|7-\sqrt{21}\right|+\left|3-\sqrt{21}\right|\\ =7-\sqrt{21}+\sqrt{21}-3\\ =7-3=4=VP\)

C

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cau lam tutu hoi to lam voi :(